Answer:
A.The kernel dissolved in the oil.
a polymer with which type of tacticity is expected to exhibit the lowest degree of crystallinity? atactic isotactic syndiotactic
A polymer which is expected to exhibit the lowest degree of crystallinity is an atactic tacticity (Option A).
What is tacticity?Tacticity is the configuration or spatial arrangement of monomer units in a polymer chain with respect to each other. It defines the regularity of the orientation of the pendant groups that are attached to the main chain of the polymer. Tacticity is a critical element that determines the physical and chemical properties of the polymer.
The three types of tacticities are atactic, isotactic, and syndiotactic. Atactic polymers have no specific tacticity because their pendant groups are randomly arranged with respect to each other. Since atactic polymers lack long-range order, they have the lowest degree of crystallinity.
Isotactic polymer have all of their pendant groups arranged in the same orientation, with respect to the main chain. As a result, these polymers have a high degree of crystallinity. Syndiotactic polymers have pendant groups that alternate in orientation along the main chain. The degree of crystallinity in syndiotactic polymers is moderate.
The degree of crystallinity for each type of polymer depends on the extent to which their pendant groups align, which affects their chain packing and orientation in the solid-state.
Thus, the correct option is A.
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Identify the variables that must remain constant to maintain an equilibrium constant and equilibrium position. : Color Variables that Must Remain Constant Variables that Do Not Need to Remain Constant : Concentration :
Equilibrium in a chemical reaction is reached when the rates of the forward and reverse reactions become equal. For a reversible reaction, the equilibrium constant is the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium.
The following variables need to remain constant for an equilibrium constant and position to remain constant:
Temperature: The temperature of the reaction vessel needs to remain constant. An increase in temperature would cause the forward reaction to speed up and the reverse reaction to slow down. As a result, the equilibrium constant would change. A decrease in temperature would cause the forward reaction to slow down and the reverse reaction to speed up, resulting in a change in the equilibrium constant. Pressure: For gaseous reactions, the pressure of the system must remain constant to maintain the equilibrium constant and position. A change in pressure would result in a shift in the equilibrium position to balance out the pressures. Concentration: The concentrations of the reactants and products must remain constant to maintain the equilibrium constant and position. A change in concentration would cause the equilibrium position to shift. For example, increasing the concentration of reactants would cause the equilibrium position to shift towards the products to consume the excess reactants. Colour: Color doesn't have any influence on the equilibrium constant and position, therefore, colour doesn't have to remain constant.know more about equilibrium constant
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consider the energy diagram for an exothermic reaction. what would you expect for the reverse reaction? a) to be exothermic and require the same activation energy b) to be endothermic and require the same activation energy c) to be endothermic and require a higher activation energy d) to be endothermic and require a lower activation energy e) to be exothermic and require a lower activation energy
Considering the energy diagram for an exothermic reaction, the expectations for the reverse reaction is: (c) The reverse reaction would be endothermic and require a higher activation energy.
In an exothermic reaction, the products have lower energy than the reactants, resulting in the release of energy. The energy diagram for an exothermic reaction typically shows a downward slope from the reactants to the products, with the energy of the products being lower than the energy of the reactants.
When considering the reverse reaction of an exothermic reaction, it would involve the conversion of products back into reactants. Since the energy diagram for the forward reaction shows a decrease in energy from reactants to products, the reverse reaction would be expected to have an increase in energy from products to reactants.
In the reverse reaction, energy would need to be absorbed from the surroundings, making it an endothermic process. Additionally, the activation energy for the reverse reaction is expected to be higher because the reverse reaction has to overcome a higher energy barrier to convert the products back into the reactants.
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Lymph nodes are...
A. bean shaped.
B. circular.
C. carrot shaped.
D. asparagus shaped.
Answer:
a
Explanation:
H2SO4 + 2NaOH → Na2SO4 + 2H2O what volume of water vapor will be produced if you start with 71.5 g of sulfuric acid and excess of sodium hydroxide
Answer:
Explanation:
Find the molar mass of H2SO4
2H = 2 * 1 = 2
S = 1 * 32 =32
O4 = 4*16 = 64
total 98
Find the number of mols in 71.5 grams
mols = given mass / molar mass.
given mass = 71.5
molar mass = 98
mols = 71.5/98
mols = 0.7296 mols of H2SO4
Find the moles of H2O
From the Balanced equation, every mol of H2SO4 produces 2 moles of H2O
mols water = 2 * 0.7296
mols water = 1.4592
That's as far as I can take you. I have to know a great deal more to get the volume of H2O
Use the spuces prooided for the ansvrs and additional paper f eceosary 1. Compare the densities of your two rubbing alcohol samples a) Were they identical? lf not, why do you think they were d (b) Should these densities have been identical? Briefly explain why or why not. 2. Suppose you had been asked to share your rubber stopper with your laboratory partner. To do this, you cut the stopper into two pieces, and determined the density of your piece. Should you report your experimentally determined density as the density of the stopper, or should you add the density you determined to the one your partner determined, and report this total density? Briefly explain. 3. (a) Suppose that when you added your rubber stopper to the graduated cylinder containing water, some of the water splashed out. Due to this procedural error, would your experimentally determined density of the stopper be erroneously high or erroneously low? Briefly explain. (b) Suppose that after a student added his unknown object to the cylinder containing water, the top of the object remained above the surface of the water. After reading the new water level, he calculated the volume of his object. Would his calculated object volume be correct, too high, or too low? Briefly explain.
(1)The densities of the two rubbing alcohol samples may not be identical. (2)The density of the rubber stopper should not be reported based on the density determined from only one piece. (a) Due to the procedural error of water splashing out when adding the rubber stopper, the experimentally determined density of the stopper would be erroneously low. (b) If the top of an unknown object remains above the water surface, the calculated volume would be too low, as it does not account for the portion of the object above the water level.
The densities of rubbing alcohol samples may differ due to variations in the concentration or impurities present in each sample. Differences in the manufacturing process or storage conditions can result in slight variations in concentration, which can impact the density. Additionally, impurities present in the samples can affect their densities, as different impurities have different molecular masses and structures.
When sharing the rubber stopper, the experimentally determined density of the individual pieces should not be reported as the density of the entire stopper. Instead, the densities determined by both partners should be added together to obtain the total density of the original stopper. This is because cutting the stopper into two pieces alters its overall volume, and reporting the density based on only one piece would not provide an accurate representation of the original stopper's density.
(a) If water splashes out when adding the rubber stopper to the graduated cylinder, the displaced water volume would be erroneously low. This would result in a lower overall density calculation for the stopper since density is calculated by dividing mass by volume. Therefore, the experimentally determined density of the stopper would be erroneously low.
(b) If the top of an unknown object remains above the water surface, the calculated volume would be too low. The volume of an object should include its entire submerged portion to obtain an accurate measurement. As the portion above the water surface is not accounted for, the calculated volume would underestimate the actual volume, leading to a calculated object volume that is too low.
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If 0.9% saline solution is isotonic to a cell, 0.5% saline solution Answers is hypertonic will cause crenation would result in water osmosing into the cell would result in water osmosing outside of the cell
If a 0.9% saline solution is isotonic to a cell, a 0.5% saline solution would be hypotonic. A hypotonic solution has a lower solute concentration compared to the cell. Hence, the correct answer is:
A 0.5% saline solution would result in water osmosing into the cell.
In a hypotonic solution, water tends to move from an area of lower solute concentration (outside the cell) to an area of higher solute concentration (inside the cell) in order to equalize the concentration on both sides. This can lead to an influx of water into the cell, potentially causing it to swell or burst (lysis) depending on the cell type.
It's important to note that a hypertonic solution has a higher solute concentration compared to the cell, which would result in water osmosing outside of the cell. In the given scenario, a 0.5% saline solution is hypotonic, not hypertonic. Therefore, the correct answer is:
A 0.5% saline solution would result in water osmosing into the cell.
The given question is incomplete and the completed question is given as,
If 0.9% saline solution is isotonic to a cell, 0.5% saline solution
Answers
is hypertonic will cause crenation would result in water osmosing into the cell would result in water osmosing outside of the cellLearn more about Osmosis from the link given below.
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why is the molar specific heat of a gas at constant pressure larger than the molar specific heat at constant volume?
Match the words in the left column to the appropriate blanks in the sentences on the right. options: on done by isochoric isobaric
volume pressure According to the first law of thermodynamics, the heat transferred to the gas equals the sum of the change in thermal energy and the work________ the gas. During an _______ process the work is zero. In process with a constant _________ with increasing temperature some energy leaves the system as work.
The heat transferred to the gas equals the total of the change in thermal energy and the work done on the gas, according to the first law of thermodynamics. During an isochoric process the work is zero. In process with a constant pressure with increasing temperature some energy leaves the system as work.
The heat provided to a gas (Q) equals the change in thermal energy (U) plus the work performed by the gas (W), as stated by the first law of thermodynamics. During an isochoric (constant volume) process, the work done by the gas is zero (W = 0) because there is no change in volume. In this case, all the heat energy transferred to the gas increases its thermal energy, resulting in a larger temperature increase and a larger molar specific heat at constant volume.
In a process with a constant pressure (isobaric), the gas can expand and do work against the external pressure. As the gas expands, it does work by pushing against the external pressure, and some of the heat energy is used to perform this work. Therefore, less heat energy is available to solely increase the thermal energy of the gas. This results in a smaller temperature increase compared to the constant volume process, leading to a smaller molar specific heat at constant pressure.
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Which of the following nuclides has the largest nuclear binding energy per nucleon? O 44 Ca 20 206 Pb 82 {Li 58 28 5 Ni 138 56 Ba
The correct answer is 5 Ni 138 56.
Explanation :
The nucleotide which has the largest nuclear binding energy per nucleon is nickel-56 (56Ni). It is important to note that nuclear binding energy per nucleon refers to the energy released when a nucleus is formed by assembling its nucleons.
The nuclear binding energy is the amount of energy required to separate an atomic nucleus into its individual nucleons. It is the sum of the energy of the individual nucleons in the nucleus. It's important to remember that when atomic nuclei are formed, they release a certain amount of energy in the process.
The nuclear binding energy per nucleon depends on the total number of nucleons present. In general, the larger the number of nucleons in an atomic nucleus, the greater the amount of nuclear binding energy per nucleon.
Therefore, among the given nucleotides, nickel-56 (56Ni) has the largest nuclear binding energy per nucleon. The amount of energy released when nickel-56 (56Ni) is formed is greater than that released by any other nuclide. The high energy released by nickel-56 (56Ni) makes it important for the process of nuclear fusion in stars.
Thus, the answer to this question is nickel-56 (56Ni).
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what is the ph at the equivalence point when 35.0 ml of a 0.200 m solution of acetic acid (ch3cooh) is titrated with 0.100 m naoh to its end point?
The pH at the equivalence point when 35.0 mL of a 0.200 M solution of acetic acid (CH3COOH) is titrated with 0.100 M NaOH to its end point is 4.76.
The pH at the equivalence point when 35.0 mL of a 0.200 M solution of acetic acid (CH3COOH) is titrated with 0.100 M NaOH to its end point can be calculated using the formula derived from the Henderson-Hasselbalch equation.
The balanced chemical equation for this reaction is as follows: CH3COOH + NaOH → CH3COONa + H2OSince the reaction between the acetic acid and sodium hydroxide is a 1:1 ratio, the number of moles of NaOH required to reach the equivalence point will be equal to the number of moles of CH3COOH present in the 35.0 mL solution of 0.200 M acetic acid.
This can be calculated using the following formula: moles of CH3COOH = M x V moles of CH3COOH = 0.200 M x 0.0350 L = 0.00700 mol of CH3COOHTo calculate the volume of NaOH required to reach the equivalence point, we can use the formula: moles of NaOH = moles of CH3COOHmoles of NaOH = 0.00700 mol of CH3COOHNow that we know the number of moles of NaOH, we can calculate the volume of NaOH required to reach the equivalence point using the following formula: V = n / CV = 0.00700 mol / 0.100 M = 0.0700 L = 70.0 mL.
Since we have added 70.0 mL of 0.100 M NaOH to the 35.0 mL of 0.200 M acetic acid solution, the total volume of the solution at the equivalence point is: Vtotal = 35.0 mL + 70.0 mL = 105.0 mL. Now that we have determined the volume and concentration of the NaOH solution required to reach the equivalence point, we can use the Henderson-Hasselbalch equation to calculate the pH at the equivalence point: pH = pKa + log([A-]/[HA])At the equivalence point, the concentration of CH3COOH and CH3COO- are equal.
Therefore, the concentration of [HA] and [A-] are equal, and we can simplify the equation to:pH = pKa + log(1)Since the pKa of acetic acid is 4.76, we can substitute this value into the equation and solve for pH:pH = 4.76 + log(1)pH = 4.76.
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Would glucose be considered a natural resource of synthetic product?
a sample of an unknown gas is placed in a vessel with a volume of 4,372 ml at a temperature of 32.3 °c. if the pressure is 5.7 atm, how many moles of gas are present?
There are 0.893 moles of gas present in the sample.
The ideal gas law is PV = nRT, which expresses the relationship between the pressure, volume, temperature, and amount of gas present. We can use this equation to find the number of moles of gas present in a sample. Here's how: PV = nRTn = PV / RT, where P is the pressure in atm, V is the volume in litres, T is the temperature in Kelvin, R is the universal gas constant (0.0821 L atm/mol K), and n is the number of moles of gas. Let's convert the volume to litres and the temperature to Kelvin. 4,372 ml = 4.372 L, and 32.3 °C + 273.15 = 305.45 K. Now we can plug in the values: n = (5.7 atm) (4.372 L) / (0.0821 L atm/mol K) (305.45 K)n = 0.893 mol.
Therefore, there are 0.893 moles of gas present in the sample.
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A solution of 6.2 M H2SO4 is diluted from 2.0 L to 3.0 L. What is the molarity of the resulting solution?
Answer:
Here's link to the answer:
bittuly/3a8Nt8
Explanation:
According to molar concentration, molarity of the resulting solution is 4.32 molar.
What is molar concentration?Molar concentration is defined as a measure by which concentration of chemical substances present in a solution are determined. It is defined in particular reference to solute concentration in a solution . Most commonly used unit for molar concentration is moles/liter.
The molar concentration depends on change in volume of the solution which is mainly due to thermal expansion. Molar concentration is calculated by the formula, molar concentration=mass/ molar mass ×1/volume of solution in liters.
In terms of moles, it's formula is given as molar concentration= number of moles /volume of solution in liters.Substitution in formula M₁V₁=M₂V₂ gives M₂=6.2×2/3=4.32 M.
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The annual production of sulfur dioxide from burning coal, fossil fuels,
auto exhaust, and other sources is about 130 million tons. When solid sulfur
reacts with oxygen gas it produces gaseous sulfur dioxide. What is the mass
in grams of sulfur that is needed to produce 2.30 Kg of sulfur dioxide?
*Don’t forget from your metric you were given that 1 Kg = 1000 g.
Answer:
1150g
Explanation:
The equation for producing sulphur dioxide is [tex]S + O_{2} ==> SO_{2}[/tex]
Looking at the ratios, you need 1 mole of sulphur for every 1 mole of sulphur dioxide
Sulphur dioxide has a molar mass of 32 + (2 x 16) = 64g
2.30 x 1000 = 2300
2300 / 64 = 35.9375
35.9375 x 32 = 1150g of sulphur
3Al + 3 NH4ClO4 ---> Al2O3 + AlCl3 + 3 NO + 6H20
How many liters of nitrogen monoxide (NO) are formed by completely reacting 50.0 grams of ammonium perchlorate (NH4C104)?
Answer:
9.63 L of NO
Explanation:
We'll begin by calculating the number of mole in 50.0 g of NH₄ClO₄. This can be obtained as follow:
Mass of NH₄ClO₄ = 50 g
Molar mass of NH₄ClO₄ = 14 + (4×1) + 35.5 + (16×4)
= 14 + 4 + 35.5 + 64
= 117.5 g/mol
Mole of NH₄ClO₄ =?
Mole = mass /molar mass
Mole of NH₄ClO₄ = 50/117.5
Mole of NH₄ClO₄ = 0.43 mole
Next, we shall determine the number of mole of NO produced by the reaction of 50 g (i.e 0.43 mole) of NH₄ClO₄. This can be obtained as follow:
3Al + 3NH₄ClO₄ –> Al₂O₃ + AlCl₃ + 3NO + 6H₂O
From the balanced equation above,
3 moles of NH₄ClO₄ reacted to produce 3 moles of NO.
Therefore, 0.43 mole of NH₄ClO₄ will also react to produce 0.43 mole of NO.
Finally, we shall determine the volume occupied by 0.43 mole of NO. This can be obtained as follow:
1 mole of NO = 22.4 L
Therefore,
0.43 mole of NO = 0.43 × 22.4
0.43 mole of NO = 9.63 L
Thus, 9.63 L of NO were obtained from the reaction.
Were any paint chip cards missing from your set? How did you determine they were or were not missing? How would you describe the missing cards to someone?
Answer:
A continuous color sequence is a pattern that can be used to determine if paint chips are missing from the set. If the color change from chip to chip is not continuous, a chip is probably missing. This pattern is useful to display paint chips for customers.
a diagram of the electromagnetic spectrum is shown. which wavelength is the longest safest wavelengths for humans？ Helppp pleaseee
Answer:
radio waves
Explanation:
the answer is simple, radio waves. it is radio waves because when you look at the diagram the waves below Radio are very long. if you look at the waves under gamma ray, they are very short. radio is the longest wave while gamma is the shortest. if you are still confused it also says longer wavelength and low frequency( meaning it is safe) right under the wavelength.
What volume of o2 is produced when 28.5 g of hydrogen peroxide decomposes to form water and oxygen at 150 degrees c and 2.0 atm?
Taking into account the reaction stoichiometry, the volume of O₂ is 7.28406 L when 28.5 g of hydrogen peroxide decomposes to form water and oxygen at 150 degrees c and 2.0 atm.
Reaction stoichiometryThe balanced reaction is:
2 H₂O₂ → 2 H₂O + O₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
H₂O₂: 2 molesH₂O: 2 molesO₂: 1 moleThe molar mass of the compounds is:
H₂O₂: 34 g/moleH₂O: 18 g/moleO₂: 32 g/moleBy reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
H₂O₂: 2 mole× 34 g/mole= 68 gramsH₂O: 2 moles× 18 g/mole= 36 gramsO₂: 1 mole× 32 g/mole= 32 gramsMass of O₂ formedThe following rule of three can be applied: if by reaction stoichiometry 68 grams of H₂O₂ form 1 mole of O₂, 28.5 grams of H₂O₂ form how many moles of O₂?
moles of O₂= (28.5 grams of H₂O₂×1 mole of O₂)÷68 grams of H₂O₂
moles of O₂= 0.42 grams
Then, 0.42 moles of O₂ are formed.
Definition of ideal gas lawIdeal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P×V = n×R×T
Where:
P is the gas pressure.V is the volume that occupies.T is its temperature.R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances. n is the number of moles of the gas. Volume of O₂In this case, you know:
P= 2 atmV = ?n= 0.42 molesR= 0.082 (atmL)÷(molK)T= 150 C= 423 KReplacing in the definition of the ideal gas law:
2 atm×V = 0.42 moles×0.082 (atmL)÷(molK)× 423 K
Solving:
V = (0.42 moles×0.082 (atmL)÷(molK)× 423 K)÷ 2 atm
V= 7.28406 L
Finally, the volume of O₂ is 7.28406 L.
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Both HC2O4− and HS−HS are amphoteric.
A. Enter an equation to show how HC2O4−can act as a base with HS−acting as an acid.
B. Enter an equation to show how HC2O4− can act as an acid with HS− acting as a base.
A. [tex]HC_{2} O^{-} _{4[/tex] (oxalate ion) acts as a base by accepting a proton from [tex]HS^{-}[/tex] (hydrosulfide ion) to form [tex]H_{2}C_{2} O_{4}[/tex] (oxalic acid).
B. [tex]HC_{2} O^{-} _{4[/tex] (oxalate ion) acts as an acid by donating a proton to [tex]HS^{-}[/tex](hydrosulfide ion) to form [tex]C_{2} O^{-} _{4[/tex] (oxalate ion).
A. At the point when [tex]HC_{2} O^{-} _{4[/tex] goes about as a base, it acknowledges a proton ([tex]H^{+}[/tex]) from [tex]HS^{-}[/tex], which goes about as a corrosive. The condition can be addressed as:
[tex]HC_{2} O^{-} _{4[/tex]+ [tex]HS^{-}[/tex] → [tex]H_{2}C_{2} O_{4}[/tex] +[tex]S^{-} _{2}[/tex]
In this response,[tex]HC_{2} O^{-} _{4[/tex] goes about as a base by tolerating a proton from the corrosive [tex]HS^{-}[/tex], shaping the impartial particle [tex]H_{2}C_{2} O_{4}[/tex] (oxalic corrosive) and the form base [tex]S^{-} _{2}[/tex].
B. At the point when [tex]HC_{2} O^{-} _{4[/tex] goes about as a corrosive, it gives a proton ([tex]H^{+}[/tex]) to [tex]HS^{-}[/tex], which goes about as a base. The condition can be addressed as:
[tex]HC_{2} O^{-} _{4[/tex] +[tex]HS^{-}[/tex] → [tex]HC_{2} O^{-} _{4[/tex]+ [tex]H_{2} S[/tex]
In this response, [tex]HC_{2} O^{-} _{4[/tex]goes about as a corrosive by giving a proton to the base [tex]HS^{-}[/tex], bringing about the development of the form base [tex]HC_{2} O^{-} _{4[/tex](oxalate particle) and the unbiased atom [tex]H_{2} S[/tex] (hydrogen sulfide).
In the two cases, [tex]HC_{2} O^{-} _{4[/tex] displays its amphoteric nature by having the option to go about as both a corrosive and a base, contingent upon the idea of the responding species.
[tex]HS^{-}[/tex], thus, exhibits its capacity to go about as a corrosive while responding with [tex]HC_{2} O^{-} _{4[/tex] as a base and as a base while responding with [tex]HC_{2} O^{-} _{4[/tex] as a corrosive.
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Which element, when it gains two electrons, will have the electron configuration 1s22s22p63s23p64s 23d104p65 Select the correct answer below: a) S. b) Se. c) Kr. d) Te.
The required correct answer is option D) Te.
An atom of Tellurium (Te) has 52 electrons and its electron configuration is 1s22s22p63s23p64s23d104p65s24d105p4. It has six electrons in its outermost shell and it requires two more electrons to complete its octet. Upon gaining two electrons, the electron configuration of Te will become 1s22s22p63s23p64s23d104p65s24d105p6.Te (Tellurium) is a chemical element with an atomic number of 52. It is a rare, silvery-white metalloid that is widely used in various industries. The majority of tellurium is used in alloys and as an additive to copper, steel, and lead to increase their strength and durability. It's also used in the production of solar panels and other electronic devices.
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Balancing chemical equation
_CoBr3 + _CaSO4 -> _CaBr2 + _Co2 (SO4)2
- solution please
Answer:
2 CoBr3 + 3 CaSO4 = 3 CaBr2 + Co2(SO4)2
Explanation:
Complete and balance the following half-reaction in acidic solution
Sn²+(aq) → Sn** (aq)
The complete and balanced half-reaction for [tex]Sn^{2+}[/tex] → Sn** in the acidic solution can be given as follows: [tex]Sn^{2+}[/tex](aq) → [tex]Sn^{4+}[/tex](aq) + 2[tex]e^-[/tex] (oxidation half-reaction)
Here, [tex]Sn^{2+}[/tex] undergoes oxidation to [tex]Sn^{4+}[/tex]. In this reaction, the oxidation state of Sn changes from +2 to +4. It means Sn loses 2 electrons in this reaction. Hence, it is an oxidation half-reaction.
2[tex]H^+[/tex](aq) + 2e- → [tex]H_2[/tex](g) (reduction half-reaction)
This half-reaction represents the reduction of [tex]H^+[/tex] ions. In this reaction, [tex]H^+[/tex] ions gain 2 electrons to form [tex]H_2[/tex]. This reaction occurs in an acidic solution.
Therefore, to obtain a completely balanced reaction, we have to combine both half-reactions as follows:
[tex]Sn^{2+}[/tex](aq) + 2[tex]H^+[/tex](aq) → [tex]Sn^{4+}[/tex](aq) + [tex]H_2[/tex](g)
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HELP MEE PLEASEEEEE
Answer to question #1.
Answer to question #2.
Answer:
1. A
2. B
Explanation:
assuming the compound is dissolved in water, what is the formula for phosphorous acid? name symbol phosphate phosphite phosphorus p h3p h3po4 hp h3po3
The formula for phosphorous acid is H₃PO₃. It can also be written as HPO(OH)₂, which is the hydrated form of the compound.
Phosphorous acid is a diprotic acid that contains one P(III) center and two -OH groups. It is a colorless and odorless solid that is highly soluble in water. When it is dissolved in water, it can act as a reducing agent because it can easily donate electrons. The formula for phosphate is PO₄³⁻. It is a polyatomic ion that contains one central phosphorus atom and four oxygen atoms arranged in a tetrahedral structure.
Phosphate has a negative three charge and is commonly found in minerals such as apatite. The formula for phosphite is PO₃³⁻. It is a polyatomic ion that contains one central phosphorus atom and three oxygen atoms arranged in a trigonal pyramidal structure. Phosphite has a negative two charge and is commonly used as a reducing agent and a chelating agent in chemical reactions involving metal ions.
The formula for phosphorus is P. It is a chemical element that has the atomic number 15 and the symbol P. Phosphorus is a nonmetal that is essential for life and is found in DNA, RNA, and ATP. It is also used in the production of fertilizers, detergents, and other industrial chemicals.
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Energy pyramid worksheet
Answer:
From top to bottom coyote, crow, squirrel, then acorn
Explanation:
The coyote has the least amount of energy and its the biggest predator so it belongs at the top. The crows eat squirrels and the squirrels eat acorns.
When excess water is added to 0.50 g of Mg3N2(s), 3.4 kJ are released as the reaction below occurs at standard state conditions. Find H˚ for the reaction
½ Mg3N2(s) + 3H2O(l) ½ Mg(OH)2 (s) + NH3(g)
The ∆H° for the reaction [tex]1/2 Mg_3N_2(s) + 3H_2O(l)[/tex]→[tex]1/2 Mg(OH)_2 (s) + NH_3(g)[/tex] is approximately -1372.54 kJ/mol
To calculate the ∆H° for the reaction[tex]1/2 Mg_3N_2(s) + 3H_2O(l)[/tex] → [tex]1/2 Mg(OH)_2 (s) + NH_3(g)[/tex], we need to use the given information that 3.4 kJ of energy are released when excess water is added to 0.50 g of Mg3N2(s).
First, let's calculate the moles of [tex]Mg_3N_2[/tex](s):
Molar mass of [tex]Mg_3N_2[/tex] = 3 * molar mass of Mg + 2 * molar mass of N
Molar mass of [tex]Mg_3N_2[/tex] = 3 * 24.31 g/mol + 2 * 14.01 g/mol = 100.95 g/mol
Moles of [tex]Mg_3N_2[/tex](s) = Mass / Molar mass
Moles of [tex]Mg_3N_2[/tex](s) = 0.50 g / 100.95 g/mol = 0.004955 mol
Next, we can determine the ∆H° per mole of [tex]Mg_3N_2[/tex](s):
∆H° per mole = Energy released / Moles of [tex]Mg_3N_2[/tex](s)
∆H° per mole = -3.4 kJ / 0.004955 mol = -686.27 kJ/mol
Since the balanced equation shows that only half a mole of [tex]Mg_3N_2[/tex](s) is involved, we multiply the ∆H° per mole by 2:
∆H° = 2 * ∆H° per mole
∆H° = 2 * (-686.27 kJ/mol) = -1372.54 kJ/mol
Therefore, the ∆H° for the reaction [tex]1/2 Mg_3N_2(s) + 3H_2O(l)[/tex]→[tex]1/2 Mg(OH)_2 (s) + NH_3(g)[/tex] is approximately -1372.54 kJ/mol.
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Question 2
1 pts
Which of the following best describes what families/groups of elements have in common?
O Same number of electrons
O Same number of valence electrons
O Same number of electron shells
O Same number of protons
U
Question 3
1 pts
For the reaction CH3 COOH → CH3 COO^- + H+, which statement is true?
O CH3 COO^- is a Brønsted-Lowry acid.
O CH3 COO^- is a conjugate base.
O CH3 COOH is a Brønsted-Lowry base.
O CH3 COO^- is an Arrhenius base.
In the given reaction CH3COOH → CH3COO^- + H+, the statement "CH3COO^- is a conjugate base" is true.The statement "CH3COO^- is a conjugate base" is true for the given reaction.
In the Brønsted-Lowry theory of acids and bases, an acid is a species that donates a proton (H+), while a base is a species that accepts a proton. In this reaction, CH3COOH donates a proton to form CH3COO^- and H+. The species that remains after the acid donates a proton is called the conjugate base.
In the reaction, CH3COOH acts as the acid by donating a proton, and CH3COO^- is the species that forms after the donation. Since CH3COO^- accepts the proton, it is considered the conjugate base of CH3COOH.
To determine whether a species is an acid or a base, we need to consider its behavior in a given reaction. In this case, CH3COOH donates a proton, making it the Brønsted-Lowry acid, while CH3COO^- accepts the proton, making it the conjugate base.
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Balanced Equations and Reaction Types. Write balanced, net ionic equations for each of the five reactions and, using the scheme outlined in the introduction, indicate the reaction type for each: redox, acid/base, or precipitation. Part 1: Cu(s) + 4 HNO_3(aq) rightarrow Cu(NO_3)_2(aq) + 2 NO_2(g) + 2 H_2O(l) Part 2: Cu(NO_3)_2(aq) + 2NaOH(aq) rightarrow Cu(OH)_2(s) + 2NaNO_3(aq) Part 3: Cu(OH)_2(s) rightarrow CuO(s) + H_2O(l) Part 4: CuO(s) + H_2SO_4(aq) rightarrow CuSO_4(aq) + H_2O(l) Part 5: Zn(.s) + CuSO_4(aq) rightarrow Cu(s) + ZnSO_4(aq)
Part 1: Cu(s) + 4 HNO₃(aq) ⇒ Cu(NO₃)2(aq) + 2 NO₂(g) + 2 H₂O(l) is a redοx reactiοn
Hοw dο yοu cοnstruct a reactiοn-specific balanced equatiοn?The reactants and prοducts are placed οn the left and right sides οf the arrοw, respectively, tο create a balanced equatiοn. Cοefficients, which appear as a number befοre a chemical fοrmula, represent mοles οf a cοmpοund. The number οf atοms in a single mοlecule is indicated by the subscripts (numbers belοw an atοm).
When catiοns and aniοns in an aqueοus sοlutiοn react tο generate a precipitate, an insοluble iοnic sοlid, precipitatiοn prοcesses take place. The sοlubility principles fοr typical iοnic sοlids can be used tο determine whether οr nοt such a reactiοn takes place.
Part 2: Cu(NO₃)₂(aq) + 2NaOH(aq) ⇒ Cu(OH)2(s) + 2NaNO₃(aq) is a dοuble displacement reactiοn οr a precipitatiοn reactiοn.
Part 3: Cu(OH)₂ (s) ⇒ CuO(s) + H₂O(l) is a Decοmpοsitiοn reactiοn
Part 4: CuO(s) + H₂SO₄(aq) ⇒ CuSO₄(aq) + H₂O(l) is an acid-base neutralizatiοn reactiοn.
Part 5: Zn(.s) + CuSO₄(aq) ⇒ Cu(s) + ZnSO₄(aq) is a displacement reactiοn
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Raul needs to ensure that when users enter an order into the tblOrders, the shipping date is at least two days after the order date on the data entry. Which option should he use?
Answer:
Input Mask
Explanation:
In this scenario, the best tool for Raul to use would be an Input Mask. When dealing with Microsoft Access or any other database tool, Input Masks are a great way of validating data inputs so that they always match the desired format. In this case, Raul would need to use input mask and apply the desired format while comparing it to the order date, therefore, making sure that it is always a minimum of 2 days after before it is accepted as a valid entry in the database.