Which describes a way to speed up the collisions between hydrogen and oxygen molecules to produce more water?

Answers

Answer 1

Answer: Equation:2H2(g) + O2(g) → 2H2O(g)

Smaller container means less volume, and the molecules will hit the walls of the container more frequently because there's less space available and the pressure will go up. I guess this would mean that the side with fewer moles would be favored as a result. We count the number of moles on the reactants and products and find that there are fewer moles on the product side, so I guess this would favor the product formation.

Answer 2

Answer:

Equation:2H2(g) + O2(g) → 2H2O(g)

Smaller container means less volume, and the molecules will hit the walls of the container more frequently because there's less space available and the pressure will go up. I guess this would mean that the side with fewer moles would be favored as a result. We count the number of moles on the reactants and products and find that there are fewer moles on the product side, so I guess this would favor the product formation.

Related Questions

Which is true about fossil fuels? A. Coal and natural gas came from swamp environments, while oil came from open oceans. B. Coal came from swampy environments, while oil and natural gas came from open ocean environments. C. Coal, oil, and natural gas came from swamp environments. D. Natural gas and oil came from swamp environments, while coal came from open oceans.
What happend if NH3+ HJ
What is the equivalent of 0 Kelvin on the Celsius scale?
How many grams are there in 7.4x10^23 molecules of Na2SO4?
Electronegativity is a measure of an atoms ability to

Specialized cells are found only in?

Answers

Specialized cells are generally found only in multicellular organisms.

What are multicellular organisms?

A multicellular organism can be described as an organism that contains more than one cell, in contrast to a unicellular organism. Multicellularity has been independently at least 25 times in eukaryotes and in some prokaryotes, like myxobacteria, cyanobacteria, and actinomycetes.

All species of animals, plants, and most fungi are multicellular organisms, whereas a few organisms are partially uni- and partially multicellular, such as slime molds and social amoebae.

Multicellular organisms develop in many ways such as by cell division or by aggregation of many single cells. Colonial organisms can be defined as identical individuals joining together to form a colony.

Unicellular organisms are divided, and the daughter cells failed to separate which results in a conglomeration of identical cells in one organism, which could develop specialized tissues.

Learn more about Multicellular organisms, here:

brainly.com/question/24381583

#SPJ2

Answer:

Specialized cells are found only in multicellular organisms.

Explanation:

Or organisms are made up of more than one cell.

A researcher is using a particle accelerator in an experiment studying isotopes. How can the researcher change one isotope into a different isotope of the same element?by removing valence electrons

by adding valence electrons

by adding or removing protons

by adding or removing neutrons

Answers

The researcher can change one isotope into a different isotope of the same element by adding or removing neutrons. This is because, isotopes are two or more forms of an element which contain equal number of protons and electrons but different number of neutrons. Isotopes differ in relative atomic mass but have same chemical properties.

Answer: by adding or removing neutrons

Explanation:

A student neutralized 16.4 milliliters of HCl by adding 12.7 milliliters of 0.620 M KOH. What was the molarity of the HCl acid?(1) 0.168 M (3) 0.620 M
(2) 0.480 M (4) 0.801 M

Answers

V ( HCl ) = 16.4 mL / 1000 => 0.0164 L

M( HCl) = ?

V( KOH) = 12.7 mL / 1000 => 0.0127 L

M(KOH) = 0.620 M

Number of moles KOH:

n = M x V

n = 0.620 x 0.0127

n = 0.007874 moles of KOH

number of moles HCl :

HCl + KOH = H2O + KCl

1 mole HCl ------ 1 mole KOH
? mole HCl--------0.007874 moles KOH

moles HCl = 0.007874 * 1 / 1

= 0.007874 moles of HCl

M = n / V

M = 0.007874 / 0.0164

= 0.480 M

Answer (2)

hope this helps!

If a man has a mass of 83 kilograms on Earth, what will the force of gravity on his body be on the moon? A. 4,880.4 N B. 135.6 N C. 813.4 N D. 318.5 N

Answers

1/6 of the weight on earth or 20 kgs

814.4 N is the answer

Which change is exothermic?(1) freezing of water(2) melting of iron(3) vaporization of ethanol(4) sublimation of iodine

Answers

The answer is (1) freezing of water. The energy of different status is increase from solid to liquid to gas. So the change of liquid to solid is exothermic. The other three are endothermic.

An equilibrium mixture contains 0.600 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the reactants (carbon monoxide and water vapor) in a 1.00-L container. This is the equation: CO(g)+H2O(g)--><-- CO2(g) + H2(g). How many moles of carbon dioxide would have to be added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol?

Answers

Answer : The moles of $CO_2$ added will be 1.12 mole.

Solution : Given,

Moles of $CO$ and $H_2O$ at equilibrium = 0.200 mol

Moles of $CO_2$ and $H_2$ at equilibrium = 0.600 mol

First we have to calculate the concentration of $CO,H_2O,CO_2\text{ and }H_2$ at equilibrium.

$\text{Concentration of }CO=(Moles)/(Volume)=(0.200mol)/(1L)=0.200M$

$\text{Concentration of }H_2O=(Moles)/(Volume)=(0.200mol)/(1L)=0.200M$

$\text{Concentration of }CO_2=(Moles)/(Volume)=(0.600mol)/(1L)=0.600M$

$\text{Concentration of }H_2=(Moles)/(Volume)=(0.600mol)/(1L)=0.600M$

Now we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

The expression of $K_c$ will be,

$K_c=([H_2][CO_2])/([CO][H_2O])$

$K_c=((0.600)* (0.600))/((0.200)* (0.200))$

$K_c=9$

Now we have to calculate the moles of $CO_2$ added.

Let the moles of $CO_2$ added is 'x'.

The given equilibrium reaction is,

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

Initially 0.200 0.200 0.600 0.600

Added moles 0 0 x 0

Change +0.1 +0.1 -0.1 -0.1

Final 0.3 0.3 (0.5+x) 0.5

The expression of $K_c$ will be,

$K_c=([H_2][CO_2])/([CO][H_2O])$

$9=((0.5)* (0.5+x))/((0.3)* (0.3))$

$x=1.12mol$

Therefore, the moles of $CO_2$ added will be 1.12 mole.

$\boxed{0.{\text{3 mol}}}$ of ${\text{C}}{{\text{O}}_{\text{2}}}$ are added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol.

Further Explanation:

Chemical equilibrium is the state in which the concentration of reactants and products become constant and do not change with time. This is because the rate of forward and backward direction becomes equal. The general equilibrium reaction is as follows:

${\text{P(g)}} + {\text{Q(g)}} \rightleftharpoons {\text{R(g)}} + {\text{S(g)}}$

Equilibrium constant is the constant that relates to the concentration of product and reactant at equilibrium. The formula to calculate the equilibrium constant for the general reaction is as follows:

${\text{K}}=\frac{{\left[ {\text{R}}\right]\left[ {\text{S}}\right]}}{{\left[{\text{P}} \right]\left[ {\text{Q}} \right]}}$

Here,

K is the equilibrium constant.

P and Q are the reactants.

R and S are the products.

The given reaction is as follows:

${\text{CO}}\left(g\right) + {{\text{H}}_2}{\text{O}}\left( g \right)\rightleftharpoons {\text{C}}{{\text{O}}_2}\left( g \right) + {{\text{H}}_2}\left( g \right)$

The expression for the equilibrium constant for the given reaction is as follows:

${\text{K = }}\frac{{\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]\left[ {{{\text{H}}_{\text{2}}}} \right]}}{{\left[ {{\text{CO}}} \right]\left[ {{{\text{H}}_2}{\text{O}}}\right]}}$ ......(1)

Here,

K is the equilibrium constant.

$\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]$ is the concentration of carbon dioxide.

$\left[{{{\text{H}}_{\text{2}}}} \right]$ is the concentration of hydrogen.

$\left[ {{\text{CO}}}\right]$ is the concentration of carbon monoxide.

$\left[ {{{\text{H}}_2}{\text{O}}}\right]$ is the concentration of water.

Substitute 0.600 mol/L for $\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]$ , 0.600 mol/L for $\left[ {{{\text{H}}_{\text{2}}}} \right]$ , 0.200 mol/L for $\left[ {{\text{CO}}}\right]$ and 0.200 mol/L for $\left[ {{{\text{H}}_2}{\text{O}}} \right]$ in equation (1).

$\begin{aligned}{\text{K }}&=\frac{{\left( {{\text{0}}{\text{.600 mol/L}}}\right)\left( {{\text{0}}{\text{.600 mol/L}}}\right)}}{{\left( {{\text{0}}{\text{.200 mol/L}}}\right)\left( {{\text{0}}{\text{.200 mol/L}}}\right)}}\n&= 9\n\end{aligned}$

The value of equilibrium constant comes out to be 9 and it remains the same for the given reaction.

Rearrange equation (1) to calculate .

$\left[{{\text{C}}{{\text{O}}_{\text{2}}}} \right]=\frac{{{\text{K}}\left( {\left[ {{\text{CO}}} \right]\left[ {{{\text{H}}_2}{\text{O}}} \right]} \right)}}{{\left[ {{{\text{H}}_{\text{2}}}} \right]}}$ ......(2)

Substitute 9 for K, 0.300 mol/L for $\left[{{\text{CO}}}\right]$ , 0.200 mol/L for $\left[{{{\text{H}}_2}{\text{O}}}\right]$ and 0.600 mol/L for $\left[ {{{\text{H}}_{\text{2}}}}\right]$ in equation (2).

$\begin{aligned}\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]&=\frac{{{\text{9}}\left( {{\text{0}}{\text{.300 mol/L}}} \right)\left( {{\text{0}}{\text{.200 mol/L}}} \right)}}{{{\text{0}}{\text{.600 mol/L}}}}\n&= 0.{\text{9 mol/L}}\n\end{aligned}$

Initially, 0.6 moles of ${\text{C}}{{\text{O}}_{\text{2}}}$ were present in a 1-L container. But now 0.9 moles of ${\text{C}}{{\text{O}}_{\text{2}}}$ are present in it. So the extra amount of ${\text{C}}{{\text{O}}_{\text{2}}}$ can be calculated as follows:

$\begin{aligned}{\text{Amount of C}}{{\text{O}}_{\text{2}}}{\text{ added}} &= 0.{\text{9 mol}} - 0.{\text{6 mol}}\n&= 0.{\text{3 mol}}\n\end{aligned}$

Therefore 0.3 moles of carbon dioxide are added in a 1-L container.

Learn more:

1. Calculation of equilibrium constant of pure water at 25°c: brainly.com/question/3467841

2. Complete equation for the dissociation of (aq): brainly.com/question/5425813

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Equilibrium

Keywords: CO, H2, CO2, H2O, 0.9 mol/L, 0.2 mol/L, 0.6 mol/L, 0.3 mol/L, K, carbon dioxide, water, hydrogen, carbon monoxide.

Other Questions

The closeness of a measurement to the actual value of what is being measured

What does the nucleus of one atom want to do to the electrons of a nearby atom?

Each period of the periodic table ends with:

A sample composed only of atoms having the same atomic number is classified as(1) a compound (3) an element(2) a solution (4) an isomer