Answer:
56,522 seconds = 942 minutes
Explanation:
n(Mg) = m/M(Mg) = 25 g / 24.31 g/mol = 1.03 mol of Mg
Q = n × 2 F = 1.03 mol × 2 × 96,485 C/mol = 197,820 C
t = Q / I = 197,820 C / 3.5 A = 56,522 s
The deltaG for the reaction below implies that the decomposition of H2O2 is thermodynamically favorable. However, H2O2 is typically stable for up to a year when stored in a dark bottle at 298K. The best explanation for this observation is the decomposition reaction
2H2O2 (aq) ⇒2H2O (l) +O2 (g) ΔG°= -234 KJ/molrxn
A. is only thermodynamically favored in the presence of a catalyst
B. occurs with an increase in entropy because O2 is a product
C. is reversible and H2O2 is produced almost as fast as it decomposes
D. has a slow rate at 298 K because the activation energy is relatively high
the stability of hydrogen peroxide under normal storage conditions despite its negative ΔG° for decomposition can be attributed to the relatively high activation energy required for the reaction to occur.
How to solve the question?
The fact that the ΔG° for the decomposition of hydrogen peroxide (H2O2) is negative (-234 KJ/molrxn) indicates that the reaction is thermodynamically favorable, meaning that the products (water and oxygen gas) are more stable than the reactant (hydrogen peroxide). However, as mentioned in the question, hydrogen peroxide is typically stable for up to a year when stored in a dark bottle at 298K, which seems to contradict the thermodynamic prediction.
The best explanation for this observation is option D: the decomposition of hydrogen peroxide has a slow rate at 298 K because the activation energy is relatively high. Even though the reaction is thermodynamically favorable, it may not occur spontaneously at room temperature because the activation energy required to break the O-O bond in hydrogen peroxide is relatively high. This means that a significant amount of energy is needed to initiate the reaction and form the products, and at room temperature, the rate of the reaction is too slow for any significant decomposition to occur over a period of one year.
Additionally, the reaction is not reversible (option C) because the entropy of the system decreases due to the formation of liquid water from hydrogen peroxide and the release of oxygen gas. Option A is also incorrect because the decomposition of hydrogen peroxide does not require a catalyst to occur, although a catalyst can increase the rate of the reaction by lowering the activation energy.
In summary, the stability of hydrogen peroxide under normal storage conditions despite its negative ΔG° for decomposition can be attributed to the relatively high activation energy required for the reaction to occur.
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A researcher is studying gold complexes. She makes
a 125 mL solution of gold(III) chloride by dissolving
0.144 moles of the solute into water. Which of the
following would cause the concentration of the
solution to increase?
Additional gold(III) chloride is added to the
beaker.
Part of the solution is poured into the chemical
waste.
O Water is added to the beaker.
We do not have enough information to answer
this question.
Additional gold(III) chloride is added to the beaker. This would cause the concentration of the solution to increase. Therefore, the correct option is option A.
Concentration in chemistry refers to the quantity of a material in a certain area. The ratio of the solute within a solution to the solvent or whole solution is another way to define concentration. In order to express concentration, mass / unit volume is typically used.
The solute concentration can, however, alternatively be stated in moles or volumetric units. Concentration may be expressed as per unit mass rather than volume. Although concentration is typically used to describe chemical solutions, it may be computed for any mixture. Additional gold(III) chloride is added to the beaker. This would cause the concentration of the solution to increase.
Therefore, the correct option is option A.
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How would the percent yield of the reaction be affected (higher, lower or no change) if some sodium bicarbonate is left unreacted? Explain.
Answer:
Explanation:
Ah, the unpredictable nature of chemical reactions - always keeping us on our toes! If some sodium bicarbonate is left unreacted in a chemical reaction, it can certainly have an impact on the percent yield of the reaction.
Now, let's get a little bit technical. Percent yield is a measure of the efficiency of a chemical reaction, and it's calculated by comparing the actual yield of the reaction (i.e., the amount of product actually obtained) to the theoretical yield of the reaction (i.e., the amount of product that should be obtained based on stoichiometry). The formula for percent yield is:
Percent yield = (actual yield / theoretical yield) x 100%
If some sodium bicarbonate is left unreacted, it means that there was not enough reactant available to fully consume all of the other reactants and produce the maximum possible amount of product. As a result, the actual yield of the reaction will be lower than the theoretical yield, and the percent yield will also be lower.
In other words, if there is unreacted sodium bicarbonate in a reaction mixture, the percent yield of the reaction will be lower than it would be if all of the reactants were consumed completely. This is because the presence of unreacted sodium bicarbonate means that some of the other reactants did not have the opportunity to react fully and produce the maximum amount of product.
So, in short, if some sodium bicarbonate is left unreacted in a chemical reaction, the percent yield of the reaction will be lower. But hey, chemistry is all about experimentation and learning from our mistakes, so there's always room for improvement in the next reaction!
Reaction A: consider a solution of acetophenone (AKA methyl phenyl ketone) and sodium trifluoroperacetate (deprotonated trifluoroperacetic acid). Draw these two reactants and then show a full arrow-pushing mechanism providing the flow of electrons, showing how the two react with one another and the resulting product
The concept Beckmann rearrangement is used here in order to show the reaction between acetophenone (AKA methyl phenyl ketone) and sodium trifluoroperacetate. The product formed is acetaminophen.
In simple Beckmann rearrangement is a reaction in which oxime is changed to an amide under some acidic conditions. The oxime is obtained by treating an aldehyde or a ketone with hydroxylamine.
This Beckmann rearrangement reaction is named after Ernst Otto Beckmann who is a German scientist. It is used for producing various drugs and steroids.
The mechanism of the Beckmann reaction is given below:
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Calculation: 3. A new alloy of steel is 525 g at 100-C. It is dropped into 375 grams of water at 25 C. The final temperature changes to 55C, what is the specific heat of steel?
The specific heat capacity of the steel is 2 g/°C.
What is the specific heat capacity of the steel?To determine the specific heat capacity of a particular sample of steel, the sample can be heated to a known temperature and then placed in contact with a known amount of water at a lower temperature.
We know that the heat lost by the steel is equal to the heat gained by the water.
Thus we have that;
-(525 * c * (55- 100)) = (375 * 4.2 * (55- 25)
23625c = 47250
c = 47250/23625
c = 2 g/°C
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The following reaction take place in a container where CONDITIONS ARE NOT STP! Calculate the volume nitogen dioxide that will be produced when 4,86 dm3 N2O5 decompose. 2N2O5(g) → 4NO2(g) + O2(g)
9.77 litres of NO2 are generated on average.
Calculation-The balanced equation for the breakdown of N2O5 is as follows:
[tex]2N_2O_5(g) -- > 4NO_2(g) + O_2(g)[/tex]
determine how many moles of N2O5 decompose:
[tex]V(N_2O_5) / Vm = n(N_2O_5)(N_2O_5)[/tex]
where V(N2O5) = 4.86 dm3 is N2O5's volume and Vm(N2O5) is N2O5's molar volume under the circumstances stated in the ideal gas law:
[tex](R*T)/P = Vm = V/n[/tex]
when the gas constant R is used.
the kelvin scale of temperature, T
The pressure is P.
The ideal gas law:
[tex]n(N_2O_5) = V(N2O5) / Vm(N_2O_5) = 4.86 dm3 / (24.46 L/mol) = 0.1982 mol[/tex]
the number of moles of NO2 is:
[tex]n(NO_2 = 4/2 * n(N_2O_5) = 0.3964 mol[/tex]
then,
[tex]n(NO_2 = 4/2 * n(N_2O_5) = 0.3964 mol[/tex]
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Select the correct answer. A certain reaction has this form:aA bB. At a particular temperature and [A]0 = 2.00 x 10-2 Molar, concentration versus time data were collected for this reaction and a plot of ln[A]t versus time resulted in a straight line with a slope value of -2.97 x 10-2 min-1. What is the reaction order and rate law for this reaction? A. first, rate = k[A] B. first, rate = k[A]2 C. second, rate = k[A] D. second, rate = k[A]2 E. third, rate = k[A]
The reaction order and rate law for this reaction is first, rate = k[A]. Option A is correct.
The reaction order and rate law for a reaction can be determined from the slope of a plot of ln[A] versus time.
Given that the plot of ln[A] versus time resulted in a straight line with a slope value of -2.97 x 10⁻² min⁻¹, we can determine the reaction order and rate law as follows;
If the slope is equal to 1, the reaction order is 1st order.
If the slope is equal to 2, the reaction order is 2nd order.
If the slope is equal to 3, the reaction order is 3rd order.
From the given slope of -2.97 x 10⁻² min⁻¹, we can conclude that the reaction order is 1st order, because the slope value is equal to -1 times the reaction order. So, the reaction order for this reaction is 1st order.
The rate law for a 1st order reaction will be given by;
rate = k[A]
where [A] is the concentration of the reactant, and k is the rate constant.
Therefore, first, rate = k[A].
Hence, A. is the correct option.
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What is the molarity of an unknown H2C2O4 solution if 10.00 mL of the H2C2O4 solution required 15.85 mL of 0.0198 M KMnO4 solution to obtain a stoichiometric endpoint?
2 KMnO4 + 5 H2C2O4 + 3 H2SO4 → 2 MnSO4 + 10 CO2 + K2SO4 + 8 H2O
Answer:
Explanation:
to balance the given chemical equation using the oxidation state change method, we need to follow these steps:Step 1: Write the unbalanced equationKMnO4 + KCl + H2SO4 → K2SO4 + MnSO4 + Cl2Step 2: Identify the elements that undergo oxidation and reductionIn this equation, the oxidation state of Mn changes from +7 to +2, which means it undergoes reduction, while the oxidation state of Cl changes from -1 to 0, which means it undergoes oxidation.Step 3: Write the half-reactionsReduction half-reaction: MnO4^- → Mn^2+Oxidation half-reaction: Cl^- → Cl2Step 4: Balance the atoms and charges in each half-reactionReduction half-reaction: 8H+ + MnO4^- → Mn^2+ + 4H2OOxidation half-reaction: 2Cl^- → Cl2 + 2e^-Step 5: Balance the electrons in each half-reactionReduction half-reaction: 5e^- + 8H+ + MnO4^- → Mn^2+ + 4H2OOxidation half-reaction: 2Cl^- → Cl2 + 2e^-Step 6: Multiply each half-reaction by a factor to equalize the number of electrons transferredReduction half-reaction: 10e^- + 16H+ + 2MnO4^- → 2Mn^2+ + 8H2OOxidation half-reaction: 14Cl^- → 7Cl2 + 14e^-Step 7: Add the balanced half-reactions together10e^- + 16H+ + 2MnO4^- + 14Cl^- → 2Mn^2+ + 8H2O + 7Cl2Step 8: Cancel out the common terms on both sides of the equation2KMnO4 + 16KCl + 8H2SO4 → 2K2SO4 + 2MnSO4 + 7Cl2 + 8H2OTherefore, the balanced equation using the oxidation state change method is:2KMnO4 + 16KCl + 8H2SO4 → 2K2SO4 + 2MnSO4 + 7Cl2 + 8H2O.
The cell diagram for a silver‑zinc button battery is Zn(s),ZnO(s)∣∣KOH(aq)∣∣Ag2O(s),Ag(s) Write the balanced half‑reaction that occurs at the anode during discharge. anode half-reaction: Write the balanced half‑reaction that occurs at the cathode during discharge. cathode half-reaction: Write the balanced overall cell reaction that occurs during discharge. overall cell reaction:
Voltaic or galvanic (spontaneous) cells are denoted by cell notations in shorthand is cathode.
Thus, This special shorthand describes the anode, cathode, and electrode components as well as the reaction circumstances.
Recall that reduction occurs at the cathode whereas oxidation occurs at the anode. Electrons go from the anode to the cathode when the anode and cathode are connected by a wire.
First, the cathode half-cell is detailed, then the anode half-cell. The reactants are mentioned first and the products are specified last within a specific half-cell.
Thus, Voltaic or galvanic (spontaneous) cells are denoted by cell notations in shorthand is cathode.
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A compound contains 0.58 m Ag, 0.58 mol N, and 1.79 mol O. What is its empirical formula?
Explanation:
To find the empirical formula of the compound, we need to determine the simplest whole-number ratio of the atoms in the compound.
First, we need to find the number of moles of Ag, N, and O relative to each other. The smallest number of moles is used as a reference point, and the other moles are divided by this value to get their ratios:
0.58 mol Ag : 0.58 mol N : 1.79 mol O
Dividing each mole value by 0.58 mol (the smallest number of moles), we get:
1 mol Ag : 1 mol N : 3.09 mol O
We can see that the ratio of Ag:N:O is 1:1:3.09. However, we need to express the ratio in whole numbers, so we divide each number by the smallest number in the set of ratios:
1 : 1 : 3.09/1.00 ≈ 3.
Rounding to the nearest whole number, we get the empirical formula of the compound:
AgN3O3
Therefore, the empirical formula of the compound is AgN3O3.
What is the name of the covalent compound Cl₂ O?
Answer:
The name of the covalent compound Cl₂O is dichlorine monoxide.
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A compound has a similar molecular mass of 180grams/mol. It contains 40.8% carbon,5.8% hydrogen and 53.4% oxygen. Calculate its empirical formula and molecular formula, and give its common name.
Answer:
To find the empirical formula of the compound, we need to determine the simplest whole-number ratio of atoms in the molecule.
First, we can assume that we have 100 grams of the compound, and calculate the number of moles of each element present in it:
Carbon: 40.8 g / 12.011 g/mol = 3.398 mol
Hydrogen: 5.8 g / 1.008 g/mol = 5.753 mol
Oxygen: 53.4 g / 15.999 g/mol = 3.337 mol
Next, we divide each of these values by the smallest one, which is 3.337, to get the mole ratio:
Carbon: 3.398 mol / 3.337 mol = 1.019
Hydrogen: 5.753 mol / 3.337 mol = 1.723
Oxygen: 3.337 mol / 3.337 mol = 1.000
Rounding these values to the nearest whole number, we get the empirical formula: C1.0H1.7O1.0, which can be simplified to CH1.7O.
To find the molecular formula, we need to know the actual molecular mass of the compound. We can estimate it by adding the atomic masses of the elements in the empirical formula:
Carbon: 1 x 12.011 = 12.011
Hydrogen: 1.7 x 1.008 = 1.714
Oxygen: 1 x 15.999 = 15.999
Total = 29.724 g/mol (approx.)
The molecular mass is close to 180 g/mol, which suggests that the actual molecular formula is a multiple of the empirical formula. Dividing 180 by 29.724, we get a value of about 6.05. Multiplying the subscripts in the empirical formula by this value, we get the molecular formula: C6H10.2O6, which can be further simplified to C3H5.1O3.
The compound with the empirical formula CH1.7O and the molecular formula C3H5.1O3 is commonly known as glyoxylic acid.
1. A new 125 g alloy of brass at 100°C is dropped into 76 g of water at 25 "C The final temperature of the water and brass is 35 "C. what is the specific heat of the sample of brass? The specific heat of water - 4.184 J/g.
We can solve this problem using the principle of conservation of energy, which states that the total energy of an isolated system remains constant.
The heat lost by the brass is equal to the heat gained by the water:
heat lost by brass = heat gained by water
The heat lost by the brass can be calculated as:
Qbrass = mbrass × cbrass × ΔT
where mbrass is the mass of the brass, cbrass is its specific heat, and ΔT is the change in temperature of the brass.
The heat gained by the water can be calculated as:
Qwater = mwater × cwater × ΔT
where mwater is the mass of the water, cwater is its specific heat, and ΔT is the change in temperature of the water.
Since the brass and water reach a final temperature of 35°C, we know that:
ΔT = Tfinal - Tinitial = 35°C - 25°C = 10°C
Plugging in the given values, we get:
Qbrass = mbrass × cbrass × ΔT = (0.125 kg) × cbrass × (10°C) = 1.25 cbrass J
Qwater = mwater × cwater × ΔT = (0.076 kg) × (4.184 J/g°C) × (10°C) = 31.97 J
Since Qbrass = Qwater, we can set the two equations equal to each other and solve for cbrass:
1.25 cbrass J = 31.97 J
cbrass = 31.97 J / 1.25 J/kg°C = 25.58 J/kg°C
Therefore, the specific heat of the brass sample is 25.58 J/kg°C.
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Exercise 4. Some diamonds appear yellow because they contain nitrogenous compounds
that absorb purple light of frequency 7.23×1014 s–1
. Calculate the wavelength (in nm) of
the absorbed light. 2. The FM station broadcasts traditional music at 102 MHz on your
radio. Units for FM frequencies are given in megahertz (MHz). Find the wavelength of
these radio waves in meters (m), nanometers (nm), and angstrom (Å).
Purple light that is absorbed has a frequency of 7.23 1014 s-1.
What is the wavelength (in nm) of the absorbed light?1. The wavelength of the absorbed light can be determined using the equation c = v, where c is the speed of light, is the wavelength, and v is the frequency. We arrive at = c/v by rearranging the equation to account for. By substituting the values, we arrive at = 4.14 10-7 m, or 414 nm.
2. The FM station's frequency is 102 MHz, or 102 106 Hz. The wavelength of the radio waves can be determined using the equation v = f, where v is the speed of light, f is the frequency, and is the wavelength. We arrive at = v/f by rearranging the equation to account for. When the values are substituted, we obtain = 2.94 m, 2.94 109 nm, or 29400.
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A researcher is studying gold complexes. She makes a 125mL solution of gold (III) chloride by dissolving 0.144 moles of the solute into water. Which of the following would cause the concentration of the solution to increase?
A additional gold (III) chloride is added to the baker
B part of the solution is poured into the chemical waste.
C water is added to the beaker
D we do not have enough information to answer this question.
Adding more gold (III) chloride to the beaker would increase the number of moles of solute in the solution while keeping the volume of the solution constant, option A is correct.
Adding more solute (gold (III) chloride) to a fixed volume of solvent (water) increases the concentration of the solution would result in an increase in the concentration of the solution.
The concentration of a solution is defined as the amount of solute dissolved in a given amount of solvent. When more solute is added to the solution, the amount of solute per unit volume of the solution increases, leading to an increase in concentration, option A is correct.
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A 346.9 mL sample of carbon dioxide was heated to 373 K. If the volume of the carbon dioxide sample at 373 K is 596.2 mL,
what was its temperature at 346.9 mL?
T =
Answer:
216.9 K
Explanation:
We can use the combined gas law to solve this problem:
(P1V1/T1) = (P2V2/T2)
where P is pressure, V is volume, and T is temperature. We can assume that the pressure is constant, since the problem doesn't mention any changes in pressure.
Let's label the initial temperature T1 and the final temperature T2, and plug in the given values:
(P1V1/T1) = (P2V2/T2)
V1 = 346.9 mL
V2 = 596.2 mL
T2 = 373 K
(P1 * 346.9 mL / T1) = (P1 * 596.2 mL / 373 K)
Simplifying and solving for T1:
T1 = (P1 * 346.9 mL * 373 K) / (P1 * 596.2 mL)
T1 = (346.9 * 373) / 596.2
T1 = 216.9 K
Therefore, the temperature of the carbon dioxide sample at 346.9 mL was 216.9 K.
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pls. help for chemistry hw. ASAP
6 moles will have 12 equivalents of PO₄²⁻
The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter.
A mole is defined as the amount of substance containing the same number of atoms, molecules, ions, etc. as the number of atoms in a sample of pure 12C weighing exactly 12 g.
1 mole of Mg₃(PO₄)₂ has 2 equivalents of PO₄²⁻.
Thus, 6 moles will have 2 × 6 = 12 equivalents of PO₄²⁻
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In order to understand how digestion of nutrients occurs, or how nutrients are used to provide cellular energy, it is necessary to understand
Multiple Choice
anatomy
radioactivity
chemistry
cytology
In order to understand how digestion of nutrients occurs, or how nutrients are used to provide cellular energy, it is necessary to understand C. chemistry.
What is chemistry ?Chemistry is the branch of science that deals with the composition, properties, and interactions of matter. In the context of digestion and cellular energy, chemistry is essential because it provides the foundation for understanding the chemical reactions that occur within the body.
For example, the digestion of nutrients involves the breakdown of complex molecules such as carbohydrates, proteins, and fats into simpler molecules that can be absorbed and utilized by the body.
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Consider a sample containing 0.300 mol of a substance. How many atoms are in the sample if the substance is iron?
The number of atoms in a sample containing 0.300 mol of iron is 1.8066 x 10²³ atoms.
The number of atoms in a sample of a substance can be determined using Avogadro's number, which is equal to 6.022 x 10²³ particles per mole. The atomic mass of iron is 55.85 g/mol.
Therefore, 0.300 mol of iron will contain:
0.300 mol x 6.022 x 10³ atoms/mol = 1.8066 x 10²³ atoms
Avogadro's number is a fundamental constant in chemistry that represents the number of particles (atoms or molecules) in one mole of a substance.
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Even though plutonium−239 (t1/2 = 2.41 × 104 yr) is one of the main fission fuels, it is still a radiation hazard present in spent uranium fuel from nuclear power plants. How many years does it take for 95% of the plutonium−239 in spent fuel to decay?
It takes about 1.41 × 10⁵ years for 95% of the Plutonium-239 (Pu-239) in spent fuel to decay.
To calculate the time it takes for 95% of Plutonium-239 (Pu-239) to decay, we can use the formula for radioactive decay:
[tex]N_{t} =N_{o}[/tex] × e^(-λt)
where [tex]N_{t}[/tex] is the amount of radioactive material remaining after time t, [tex]N_{o}[/tex] is the initial amount of radioactive material, λ is the decay constant, and e is the mathematical constant equal to about 2.718.
We can solve for t when [tex]N_{t}[/tex] is 0.05N0 (95% decay) and substitute the values for λ and the half-life of Pu-239:
0.05N0 = N × [tex]e^{(-0.693/t1/2 * t)}[/tex]
0.05 = [tex]e^{(-0.693/2.41 * 10^4 * t)}[/tex]
ln(0.05) = -0.693/2.41 × 10⁴ × t
t = 1.41 × 10⁵ years
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when 56.6 g of calcium and 30.5g of nitrogen gas under go a reaction that has 90% yield, what mass of calcium nitride is formed?
Answer:
152.57 g
Explanation:
The balanced chemical equation for the reaction between calcium and nitrogen gas to form calcium nitride is:
3Ca + N2 → Ca3N2
From the balanced equation, we can see that the mole ratio between calcium and calcium nitride is 3:1. This means that 3 moles of calcium react to form 1 mole of calcium nitride.
First, let's calculate the number of moles of calcium and nitrogen gas given:
Mass of calcium = 56.6 g
Molar mass of calcium (Ca) = 40.08 g/mol
Moles of calcium = Mass of calcium / Molar mass of calcium = 56.6 g / 40.08 g/mol = 1.41 mol
Mass of nitrogen gas = 30.5 g
Molar mass of nitrogen gas (N2) = 28.02 g/mol
Moles of nitrogen gas = Mass of nitrogen gas / Molar mass of nitrogen gas = 30.5 g / 28.02 g/mol = 1.09 mol
Since the reaction has a 90% yield, only 90% of the limiting reactant (which is calcium in this case) will be converted to product. Therefore, we need to multiply the moles of calcium by 0.90 to account for the yield:
Moles of calcium nitride formed = Moles of calcium x Yield = 1.41 mol x 0.90 = 1.27 mol
Now, using the mole ratio from the balanced equation, we can determine the mass of calcium nitride formed:
Molar mass of calcium nitride (Ca3N2) = 40.08 g/mol (molar mass of calcium) x 3 + 14.01 g/mol (molar mass of nitrogen) x 2 = 120.25 g/mol
Mass of calcium nitride formed = Moles of calcium nitride formed x Molar mass of calcium nitride = 1.27 mol x 120.25 g/mol = 152.57 g
So, the mass of calcium nitride formed in the reaction is 152.57 g.
A is a solution of 0.1mol/dm³ and B contains 5.0gram anhydrous X₂CO₃/dm³ of solution. if 25.50cm³ of acid is used to neutralize 25cm³ of solution B. Calculate; 1. the molar concentration of solution B 2. the relative molar mass of X₂CO₃ 3. the atomic mass of the element X
the molar concentration of solution B is 0.102 mol/dm³, the relative molar mass of X₂CO₃ is approximately 106 g/mol, and the atomic mass of the element X is approximately 24 g/mol.
How to solve the question?
To solve this problem, we will use the concept of neutralization reaction. In a neutralization reaction, an acid reacts with a base to form a salt and water. The balanced equation for the reaction between an acid and a carbonate is:
H₂SO₄ + X₂CO₃ → X₂(SO₄) + CO₂ + H₂O
where H₂SO₄ is the acid and X₂CO₃ is the carbonate.
We are given that 25.50 cm³ of the acid solution reacts with 25 cm³ of the carbonate solution B. From this, we can calculate the moles of acid used:
Moles of acid = concentration × volume = 0.1 mol/dm³ × 0.02550 dm³ = 0.00255 mol
Since the acid reacts with the carbonate in a 1:1 ratio, we know that 0.00255 mol of X₂CO₃ were present in the 25 cm³ of solution B. We can use this information to find the molar concentration of solution B:
Molar concentration of solution B = moles of X₂CO₃ / volume of solution B
= 0.00255 mol / 0.025 dm³ = 0.102 mol/dm³
To find the relative molar mass of X₂CO₃, we need to know its molecular formula. X₂CO₃ is a binary compound containing two atoms of X, one atom of carbon, and three atoms of oxygen. The relative molecular mass of X₂CO₃ can be calculated as follows:
Relative molecular mass of X₂CO₃ = 2 × relative atomic mass of X + relative atomic mass of C + 3 × relative atomic mass of O
= 2X + 12 + 3(16)
= 2X + 60
To find the atomic mass of X, we need to solve for X in the equation above. We can do this by rearranging the equation:
2X + 60 = relative molecular mass of X₂CO₃
2X = relative molecular mass of X₂CO₃ - 60
X = (relative molecular mass of X₂CO₃ - 60) / 2
Substituting the given value for the relative molecular mass of X₂CO₃, we get:
X = (2 × 105.99 - 60) / 2 = 23.995
Therefore, the atomic mass of the element X is approximately 24 g/mol.
In summary, the molar concentration of solution B is 0.102 mol/dm³, the relative molar mass of X₂CO₃ is approximately 106 g/mol, and the atomic mass of the element X is approximately 24 g/mol.
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3) The following reaction is at equilibrium, when pressure is increased:
H₂(g) + CO2(g) + heat H₂O(g) + CO(g)
10
A. In order to restore equilibrium, the reaction shifts left, toward reactants
B. In order to restore equilibrium, the reaction shifts right, toward products
C. No change occurs
The effect of increasing pressure on the given equilibrium reaction:
H₂(g) + CO₂(g) + heat ⇌ H₂O(g) + CO(g)
is determined by the stoichiometry of the reaction and the number of moles of gas on each side of the equation.
On the left-hand side, there are two moles of gas (H₂ and CO₂), whereas on the right-hand side, there are only one mole of gas (CO). Therefore, increasing the pressure will shift the equilibrium towards the side with fewer moles of gas, in order to decrease the overall pressure.
In this case, the reaction will shift towards the right, towards the products, to decrease the pressure. Therefore, the correct answer is:
B. In order to restore equilibrium, the reaction shifts right, toward products.
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Write the balanced molecular chemical equation for the reaction in aqueous solution for sodium hydroxide and tin (IV) acetate. If no reaction occurs, simply write only NR.
Be sure to include the proper phases for all species within the reaction.
The balanced molecular chemical equation for the reaction in aqueous solution for sodium hydroxide and tin (IV) acetate is 4 NaOH (aq) + Sn (CH[tex]_3[/tex]COO) (aq) → Sn (OH)(s) + 4 CH[tex]_3[/tex]COONa (aq) .
An equation per a chemical reaction is said to be balanced if both the reactants plus the products have the same number of atoms and total charge for each component of the reaction. In other words, each component of the reaction have an equal balance of mass and charge. The balanced molecular chemical equation for the reaction in aqueous solution for sodium hydroxide and tin (IV) acetate is 4 NaOH (aq) + Sn (CH[tex]_3[/tex]COO) (aq) → Sn (OH)(s) + 4 CH[tex]_3[/tex]COONa (aq) .
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6) Apply the rules for canceling complex units to simplify and find the units in an answer:
a)
=
atm / mol • (atm •L/
mol • K)
b) (mol • (atm•L / mol • K) • K ) / L
Cancelling units, also referred to to be "unit analysis" as well as "dimensional analysis," atm / mol × (atm ×L/mol × K) = atm / (atm ×L/ K)
A conversion factor (described in the form of a fractions, ) is considered as possessing a value of "1" for the purposes of transforming between units. Cancelling units, also referred to to be "unit analysis" as well as "dimensional analysis," depends on the principles that multiplying a thing by "1" does not alter the value, that any amount divided through a single value equals "1," and that any value allocated by itself.
A) atm / mol × (atm ×L/mol × K)
cancel the mole by mole
atm / (atm ×L/ K)
B) (mol × (atm×L / mol × K) × K ) / L
cancel the mole by mole
(atm×L / K) × K ) / L
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Compare and contrast the four different types of solar activity above the photosphere. (Select all that apply)
A. Prominences are huge loops of the Sun’s ionized but cool material that are pushed by magnetic forces from the chromosphere into the corona.
B. There are regions of higher density and temperature than the surrounding material in the chromosphere called plages.
C. Flares are long lived phenomena above the photosphere beginning at the solar minimum and ending at solar maximum.
D. Coronal mass ejections happen when a flare is so violent that the material exceeds the escape speed of the sun. It is connected to sunspots.
There are up to four different kinds of features that can be seen when viewing the photosphere. Sunspots, faculae, granulation, and super-granulation are listed in decreasing order of observational ease. All the given statement are true.
Solar activityIn the chromosphere, plateaus are areas that are more dense and hotter than the surrounding material. Large, chilly, ionized loops of material known as prominences are slowly pushed into the corona by magnetic energy from the chromosphere. Flares are erratic, fiery, and intensely energetic bursts that last only a moment or two. When a flare is extremely powerful, its material is thrown into the solar system beyond the Sun's escape velocity and causes coronal mass ejections. These are all essentially connected to sunspots.For more information on solar activity above photosphere kindly visit to
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What is the temperature of 0.59 mol of gas at a pressure of 1.0 atm and a volume of 11.0 L? (Express the temperature in kelvins to two significant figures).
Answer:
T = 230 K
Explanation:
Simply use the equation: PV = nRT, where P is pressure, V is volume, n is moles, R is 0.0821, and T is temperature in kelvin
Place the values in
(1.0 atm)(11.0 L)=(0.59 moles)(0.0821 Lxatm/Kxmol)(T)
Solve for T = 230 K
Which of the diagrams show a beaker in which electrolysis takes place?
The beakers that we can see that electrolysis is going on are in beakers B and C
What is electrolysis?Electrolysis is a process in which an electric current is passed through a substance to produce a chemical reaction. The substance being electrolyzed is usually an electrolyte, which is a liquid or solution that contains ions.
The electric current causes the ions in the electrolyte to move towards the electrodes (conducting surfaces) where they either gain or lose electrons, depending on the nature of the electrode and the ions. This causes chemical reactions to occur at the electrodes, leading to the formation of new substances.
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If a high altitude balloon is filled with 14,100 L of hydrogen at a temperature of 21 degrees celsius and a pressure of 0.98 atm. What's the volume of the balloon at a height of 20 km, where the temperature is -48 degrees celsius and the pressure is 0.08 atm?
The volume of the balloon at a height of 20 km is approximately 296,837 L.
To find the new volume of the balloon, we can use the combined gas law equation:
(P1V1/T1) = (P2V2/T2)
Where P1 is the initial pressure, V1 is the initial volume, T1 is the initial temperature, P2 is the final pressure, V2 is the final volume, and T2 is the final temperature.
Substituting the given values:
(P1) (V1) / (T1) = (P2) (V2) / (T2)
(0.98 atm) (14,100 L) / (294 K) = (0.08 atm) (V2) / (225 K)
Simplifying:
V2 = [(0.98 atm) (14,100 L) (225 K)] / [(0.08 atm) (294 K)]
V2 ≈ 296,837 L
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___________ is described as a metal, while __________ is described as a metalloid?
A) nitrogen, iron
B) fluorine, lithium
C) argon, sodium
D) strontium, silicon
Answer:
D) strontium, silicon
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