Answer:
[tex]m_{O_2}=6gO_2[/tex]
Explanation:
Hello,
In this case, given the reaction:
[tex]2S+3O_2\rightarrow 2SO_3[/tex]
Thus, for 4.00 g of sulfur (atomic mass = 32 g/mol), we apply the shown 2:3 mole ratio with oxygen (molar mass = 32 g/mol) in order to compute its requirement:
[tex]m_{O_2}=4.00gS*\frac{1molS}{32gS}*\frac{3molO_2}{2molS}*\frac{32gO_2}{1molO_2}\\ \\m_{O_2}=6gO_2[/tex]
Best regards.
The mass of oxygen would be required is [tex]6gO_2[/tex]
The calculation is as follows;[tex]= 4.00 \times \frac{1molS}{32gS} \times \frac{3molO_2}{2molS} \times \frac{32gO_2}{1molO_2} \\\\= 6gO_2[/tex]
Here for 4.00 g of sulfur (atomic mass = 32 g/mol), we used the shown 2:3 mole ratio with oxygen (molar mass = 32 g/mol).
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Calculate the kinetic energy, in joules of a 1250-kg automobile moving at 20.0 m/s.
Answer:
250000 J.
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 1250 kg
Velocity (v) = 20.0 m/s.
Kinetic energy (K.E) =?
Kinetic energy is simply defined as energy possed by a body in motion. Mathematically, it is expressed as:
K.E = ½mv²
Where:
K.E is the kinetic energy
m is the mass of the object
v is the velocity of the object.
Thus, we can obtain the kinetic energy of the automobile by using the above formula as illustrated below:
Mass (m) = 1250 kg
Velocity (v) = 20.0 m/s.
Kinetic energy (K.E) =?
K.E = ½mv²
K.E = ½ × 1250 × 20²
K.E = 625 × 400
K.E = 250000 J
Therefore, the kinetic energy of the automobile is 250000 J
How many significant figures does the number 0.00650
have?
Answer:
3
Explanation:
For decimals, you always start at the first non-zero number and count all the way to the end of the number. To count the number of significant figures this number has, you start at 6 and count every number after that for a total of 3.
Which of the following does not involve chemical processes? A. cooking of food B.Changing of fruit color when it ripens C.Burning of Wood D.None of the above
Answer:
c because a chemical process is when something changes into another thing which the wood would be doing if you burned it
What kind of questions CANNOT be answered by chemistry?
Answer:
Why does matter exist?
Explanation:
Under what set of conditions does HCl(g) deviate the most from ideal behavior?a) high temperature and low pressure.b) low temperature and high pressure.c) high temperature and high pressure.d) low temperature and low pressure. e) standard temperatore and pressure.
Answer:
b) low temperature and high pressure
Explanation:
Deviation of gases from their ideal behavior could be as a result of two things which include a very small volume for the gases and the collisions not being elastic enough.
Small volume will decrease the frequency of elastic collisions a gas will experience as a result of the tiny space.
Low temperature will decrease the amount of elastic collisions and energy of the gas and high pressure will decrease the volume which is why there will be a deviation in the ideal behavior of Hcl(g)
Write 5 facts you know about weather and climate.
1) You can tell the temperature by counting a cricket's chirps! 2) Sandstorms can swallow up entire cities. 3) Dirt mixed with wind can make dust storms called black blizzards. ... 5) The coldest temperature ever officially recorded was -89.2°C.
Convert 355 km/s into mm/s.
Answer:
the answer would be, 355000 mm/s.
A student measures the mass of an 11.2cm3 block of sugar to be 17.8 g. What is the density of the sugar?
Answer:
21 cm 3 that is the answer
Use the virtual lab to prepare 150.0 ml of an iodine solution with a concentration of 0.06 g/ ml from the bottle of 0.12g/ml iodine solution
Answer:
Explanation:
In 150 ml of .06 g / ml solution , gram of iodine = 150 x .06 g = 9 g
Let volume of given concentration of .12 g / ml required be V
In volume V , gram of iodine = V x .12 g
According to question
V x .12 = 9 g
V = 9 / .12 = 75 ml
So, 75 ml of .12 g/ml will be taken and it is diluted to the volume of 150 ml to get the solution of required concentration .
Rank these hydrocarbons in order from predicted highest boiling point to lowest boiling point: a. CH3CH3b. CH3CH2CH3c. CH3CH2CH2CH3d. CH3CH(CH3)2
Answer:
c. CH3CH2CH2CH3 > d. CH3CH(CH3)2 > b. CH3CH2CH3 > a. CH3CH3
Explanation:
Hello,
In this case, considering the physical properties of hydrocarbons we find that the longer the carbon chain, the higher the boiling point as more energy is required for the phase transition from liquid to gas, thus, we see that c. CH3CH2CH2CH3 and d. CH3CH(CH3)2 have the highest boiling points since they have four carbons both. Then, we have b. CH3CH2CH3 which has three carbon atoms and the lowest boiling point is had by a. CH3CH3.
Nevertheless for both c. and d. we can see that c. has a straight chain whereas d. a branched chain, it means that c, has the highest boiling point due to the higher surface area for energy transfer.
Therefore, the order from highest to lowest is:
c. CH3CH2CH2CH3 > d. CH3CH(CH3)2 > b. CH3CH2CH3 > a. CH3CH3
Regards.
What is the empirical formula of an oxide of chromium that is 48% oxygen
Answer:
[tex]CrO_3[/tex]
Explanation:
Hello,
In this case, considering that the compound is 48% oxygen, we infer it is 52% chromium, therefore, by using their atomic masses we compute the moles by assuming those percentages as masses:
[tex]n_{Cr}=52g*\frac{1mol}{52g}=1mol\\ \\n_O=48g*\frac{1mol}{16g}=3mol[/tex]
Next, by dividing by the smallest moles, we find the subscripts in the empirical formula:
[tex]Cr=\frac{1}{1} =1\\\\O=\frac{3}{1} =3[/tex]
Which are also expressed in the smallest whole numbers, therefore the empirical formula is:
[tex]CrO_3[/tex]
Which corresponds to the chromic oxide or chromium (IV) oxide.
Regards.
What are non-examples of a nucleus
Answer:n a red blood cell, the control center is the nucleus. A mitochondria is not a nucleus. It is another organelle that produces energy for the cell.
Explanation:
0.660 moles of NaCl are dissolved in 95.0 mL of water. Calculate the molarity of the NaCl solution. A) 62.7 M B) 0.069 M C) 6.95 M D) 0.0069 M
Answer:
C) 6.95 M.
Explanation:
Hello,
In this case, considering that the molarity is computed via the ratio of the moles of solute to the volume of the solution in liters:
[tex]M=\frac{n_{solute}}{V_{solution}}[/tex]
Since the volume here is 95.0 mL or also 0.0950 L, the molarity turns out:
[tex]M=\frac{0.660mol}{0.0950L} \\\\M=6.95 M[/tex]
Therefore, answer is C) 6.95M.
Best regards.
What obligations do scientist have to society
name of element flaurine what is latin name
Answer:
Fluere
i hope this helps!
How many grams of NH4OH are present in 37.0 grams of 6.00 M NH4OH which has a density of 0.960 g/mL?
Answer:
The mass of NH₄OH present in the solution is 8.085 g
Explanation:
Given;
molarity of NH₄OH = 6 M
density of NH₄OH = 0.960 g/mL
mass of NH₄OH = 37.0 grams
volume of NH₄OH = mass /density
= 37/0.96
= 38.54 mL
molarity = moles of solute / Liters of solution
moles of solute = molarity x Liters of solution
moles of solute = 6 x 38.54 x 10⁻³
moles of solute = 0.231 moles
Reacting mass = number of moles x molecular mass
Molecular mass of NH₄OH = 35 g/mol
Reacting mass = 0.231 mol x 35 g/mol
Reacting mass = 8.085 g
Therefore, the mass of NH₄OH present in the solution is 8.085 g
Which explains why mixtures can be separated
Answer: A: the components have different properties
Explanation:
The guy above is wrong I’m right got it correct on my unit review on edge
Answer:
A.
Explanation:
why is copper cobalt sulphide concentrates treated via roast rather than smelting
Explanation:
Cobalt extraction refers to the techniques used to extract cobalt from its ores and other compound ores. Several methods exist for the separation of cobalt from copper and nickel. They depend on the concentration of cobalt and the exact composition of the used ore.
Cobalt ore
Contents:
The ores are treated by a sulfatizing roast in a fluidized bed furnace to convert copper and cobalt sulfides into soluble sulfates and iron into insoluble hematite. The calcine is subsequently leached with sulfuric acid from the spent copper recovery electrolyte. Oxide concentrates are introduced at this leaching step to maintain the acid balance in the circuit. Iron and aluminum are removed from the leach solution by the addition of lime, and copper is electrowon on copper cathodes. A part of the spent electrolyte enters the cobalt recovery circuit and is purified by the removal of iron, copper, nickel, and zinc prior to the precipitation of cobalt as its hydroxide. This is accomplished by the addition of more lime to raise the pH until the remaining copper precipitates. This copper is sent back to the copper circuit. As more lime is then added, a copper-cobaltite precipitates and is fed back to the leaching process. Sodium hydrosulfide (NaHS) is added (along with some metallic cobalt as a catalyst) to precipitate nickel sulfide (NiS). Hydrogen sulfide (H2S) and sodium carbonate (Na2CO3) are then added to precipitate zinc sulfide (ZnS). Lime is then added to saturation to precipitate cobalt(II) hydroxide (Ca(OH)2). In the final stages, this cobalt hydroxide is redissolved and the metal is refined by electrolysis. The resulting cobalt cathodes are crushed and vacuum degassed to obtain a pure cobalt metal.
A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL of 0.150 M NiSO4. (a) Write the balanced chemical equation for the reaction that occurs. (b) What precipitate forms? (c) What is the limiting reactant? (d) How many grams of this precipitate form? (e) What is the concentration of each ion that remains in solution?
Answer:
a) 2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂
b) Ni(OH)₂
c) KOH
d) 0.927 g
e) K⁺=0.067 M, SO₄²⁻=0.1 M, Ni²⁺=0.067 M
Explanation:
a) The equation is:
2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂ (1)
b) The precipitate formed is Ni(OH)₂
c) The limiting reactant is:
[tex] n_{KOH} = V*M = 100.0 \cdot 10^{-3} L*0.200 mol/L = 0.020 moles [/tex]
[tex] n_{NiSO_{4}} = V*M = 200.0 \cdot 10^{-3} L*0.150 mol/L = 0.030 moles [/tex]
From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:
[tex] n = \frac{2}{1}*0.030 moles = 0.060 moles [/tex]
Hence, the limiting reactant is KOH.
d) The mass of the precipitate formed is:
[tex] n_{Ni(OH)_{2}} = \frac{1}{2}*n_{KOH} = \frac{1}{2}*0.020 moles = 0.010 moles [/tex]
[tex] m = n*M = 0.010 moles*92.72 g/mol = 0.927 g [/tex]
e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:
[tex]C_{K^{+}} = \frac{2*\frac{1}{2}*n_{KOH}}{V} = \frac{0.020 moles}{0.300 L} = 0.067 M[/tex]
[tex]C_{SO_{4}^{2-}} = \frac{\frac{1}{2}*n_{KOH + (0.03 - 0.01)}}{V} = \frac{0.030 moles}{0.300 L} = 0.1 M[/tex]
[tex]C_{Ni^{2+}} = \frac{0.020 moles}{0.300 L} = 0.067 M[/tex]
I hope it helps you!
The mass of precipitate formed is 0.93 g .
The equation of the reaction is;
NiSO4(aq) + 2KOH(aq) ---->K2SO4(aq) + Ni(OH)2(s)
The precipitate that forms is Ni(OH)2(s).
To obtain the limiting reactant;
Number of moles of NiSO4 = 200/1000 × 0.150 M = 0.03 moles
Number of moles of KOH = 100/1000 × 0.200 M = 0.02 moles
Since 1 mole of NiSO4 reacts with 2 moles of KOH
x moles of NiSO4 reacts with 0.02 moles of KOH
x = 1 mole × 0.02 moles/2 moles = 0.01 moles of NiSO4
Hence, KOH is the limiting reactant.
Now;
Since 2 mole of KOH yields 1 mole of Ni(OH)2
0.02 moles of KOH yields 0.02 moles × 1 mole/ 2 mole = 0.01 moles of Ni(OH)2
Mass of Ni(OH)2 produced = 0.01 moles of Ni(OH)2 × 93 g/mol = 0.93 g
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Compare and contrast the results between the commercial and homemade pH test strips. Which test strips were more accurate? Explain your answer.
Answer:
See explanation
Explanation:
Homemade pH test strips make it a bit difficult to identity the colours that indicate various values of pH on a pH colour scale. These homemade pH strips make the results of the experiment more prone to human error. Also the creator of the test strip is also prone to human error during the creation of the strip.
The commercial pH test strips are more accurate when used.
Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor: 4 NH3 + 5 O2 → 4 NO + 6 H2O When 40.0 g NH3 and 50.0 g O2 are allowed to react, what is the mass of the remaining excess reagent?
Answer:
18.75 g of NH3.
Explanation:
The balanced equation for the reaction is given below:
4NH3 + 5O2 → 4NO + 6H2O
Next, we shall determine the masses of NH3 and O2 that reacted from the balanced equation.
This can be obtained as follow:
Molar mass of NH3 = 14 + (3x1) = 17 g/mol
Mass of NH3 from the balanced equation = 4 x 17 = 68 g
Molar mass of O2 = 16x2 = 32 g/mol
Mass of O2 from the balanced equation = 5 x 32 = 160 g
From the balanced equation above,
68 g if NH3 reacted with 160 g of O2.
Next, we shall determine the excess reactant. This can be obtained as follow:
From the balanced equation above,
68 g if NH3 reacted with 160 g of O2.
Therefore, 40 g of NH3 will react with = (40 × 160)/68 = 94.12 g of O2.
From the calculations made above, we can see that it will take a higher amount of O2 i.e 94.12g than what was given i.e 50g to react completely with 40 g of NH3.
Therefore, O2 is the limiting reactant and NH3 is the excess reactant.
Next we shall determine the mass of excess reactant that reacted. This can be obtained as follow:
From the balanced equation above,
68 g if NH3 reacted with 160 g of O2.
Therefore, Xg of NH3 will react with 50 g of O2 i.e
Xg of NH3 = (68 × 50)/160
Xg of NH3 = 21.25 g
Therefore, 21.25 g of NH3 (excess reactant) were consumed in the reaction.
Finally, we shall determine mass of the remaining excess reactant as follow:
Mass of excess reactant = 40 g
Mass of excess reactant that reacted = 21.25 g
Mass of excess reactant remainig =?
Mass of excess reactant remainig = (Mass of excess reactant) – (Mass of excess reactant that reacted)
Mass of excess reactant remainig
= 40 – 21.25
= 18.75 g
Therefore, the mass of excess reactant remaining is 18.75 g of NH3.
Compound A, C7H13Br, is a tertiary alkyl bromide. On treatment with CH3CH2ONa, A is converted into B, C7H12. Ozonolysis of B with O3 gives C as the only product. What are the structures of compounds A and B
Answer:
Compound A: 1-bromo-1-methylcyclohexane
Compound B: 1-methylcyclohex-1-ene
Explanation:
In this question, we can start with the I.D.H (hydrogen deficiency index):
[tex]I.D.H~=~\frac{(2C)+2+(N)-(H)-(X)}{2}[/tex]
In the formula we have 7 carbons, 13 hydrogens, and 1 Br, so:
[tex]I.D.H~=~\frac{(2*7)+2+(0)-(13)-(1)}{2}=1[/tex]
We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.
We have to keep in mind that the Br atom must be bonded to a tertiary carbon. We can not have a double bond because in the ozonolysis reaction we have only 1 product, therefore, we can not have a double bond in the initial molecule (if we have a double bond in the initial molecule we will have more than 1 product in the ozonolysis reaction).
With this in mind, we will have a cyclic structure. If we have 7 carbons and we need a tertiary alkyl halide. We can have a cyclic structure of 6 members and a methyl group bonded to a carbon that also is bonded to a Br atom (1-bromo-1-methylcyclohexane).
In the reaction with [tex]CH_3CH_2ONa[/tex] we will have an elimination reaction. In other words, we have the production of a double bond inside of the cyclic structure (1-methylcyclohex-1-ene).
See figure 1 for further explanations.
I hope it helps!
100. cal of heat are added to 18.0 g of ethanol (0.581 cal/g °C) originally at 23 °C. The final temperature is ____________.
Answer:
Final temperature is 32.56 °C
Explanation:
The specific heat of a substance is the amount of heat required to raise the temperature of 1g of the substance by 1°C.
The following equation/formula is used;
Q = m × c × ΔT
Where; Q= amount of heat supplied
(cal)
M= mass of ethanol (g)
C= specific heat of ethanol
(cal/g °C)
ΔT= change in temperature (°C)
i.e. (final temperature - initial
temperature)
According to the question, Q= 100 calories (cal), M= 18g, C= 0.581 cal/g °C, initial temperature = 23°C, final temperature = ?
Hence, we insert our values into the equation;
Q = m × c × ΔT
ΔT = Q/mc
(Final T - Initial T) = Q/mc
(Final T - 23) = 100/ 18 × 0.581
(Final T - 23) = 100/10.458
Final T - 23 = 9.562
Final T = 23 + 9.562
Final T = 32.562
Hence, the final temperature of ethanol is 32.56°C
Atomic Structure and Bonding_quiz Q12 Points The principal, angular momentum, magnetic and spin quantum numbers for an atomic system are designated as nn, ll, m_lm l , and m_sm s , respectively. Which of the following statements is true?
a. nn represents the shape of the electron cloud
b. m_lm l represents the orientation of the electron spin an electron at n=1
c. n=1 classifies as a "free electron" ll represents the shape of an orbital
Answer:
ll represents the shape of an orbital
Explanation:
There are four quantum numbers which give information about various orbitals in an atom. They describe the probability of finding the electron in an atom.
The orbital or azimuthal quantum number is represented as l. This quantum number gives information about the shape of an orbital.
WORTH A LOT OF POINTS! just copy what on the picture for notes so i can copy and paste i do not feel like writing all of that down
Answer:
A base pair is a pair of bases that form hydrogen bonds in the double stranded DNA molecule.
- Adenine-thymine: A-T
- Guanine-cytosine: G-C
Replication Process:
- Double strand unwinds.
- New nucleorides line up via base pairing.
- Colvalent bonds link nucleotides together in the new strands.
Explanation:
please answer these true false
Answer:
False
False
they both will stay on the water
Hope this helped
The process of making proteins using information from DNA is called ___ a. mitosis b. transcription c. replication d. translation e. mitosis
Answer:
Transcription
Explanation:
Transcription is when you make a copy of the RNA from DNA. This is how the mRNA can leave the nucleus which exists to the cytoplasm and into ribosomes where protein synthesis occurs.
What is the meaning of a positive or negative percent error?
Answer:
positive--real negative--not real
If this container of gas is heated,
then the pressure inside the container will
increase because pressure increases as
temperature increases.
Hypothesis
Theory
Law
Answer:
theory
Explanation:
kinetic theory of gases
Answer:
Hypothesis
Explanation:
Just did it
The diagram shows the layers formed when 10 mL each of honey, maple syrup, and corn syrup were slowly poured into a glass cylinder.
A tall cylinder with 3 layers is shown. The top layer is labeled Corn Syrup, the middle layer is labeled Maple Syrup and the bottom layer is labeled Honey.
A ping pong ball released gently into the cylinder floats on top of the corn syrup layer. What best compares the densities of the substances?
Group of answer choices
Honey is denser than corn syrup but less dense than the ball.
All liquids are denser than the ball, and honey is denser than corn syrup and maple syrup.
Corn syrup is denser than the other liquids, and the ball is denser than the liquids.
Corn syrup is denser than the other liquids and the ball.
The correct answer is B. All liquids are denser than the ball, and honey is denser than corn syrup and maple syrup.
Explanation:
Differences in density cause substances such as solids or liquids to float or sink. In general, the substance sinks if it is denser, or floats if it is less dense. In this context, honey is the substance with the most density because it is at the bottom of the container. Also, honey is followed in density by maple syrup, and then by corn syrup. Moreover, if the ping pong ball floats in the most superficial layer, it is because this is less dense than any of the liquids. According to this, it can be concluded honey is denser than the other liquids, and the liquids are denser than the ball (option B.)
Answer:
b is right ( all liquids are denser than the ball and honey is denser than corn syrup.)
Explanation: