All categories

3 years ago

Which of the following statements is TRUE?

Chemistry

1 answer:

borishaifa [10]3 years ago

6 0

Answer:

Dynamic equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction:

Explanation:

Reaction quotient is the ratio of product of concentrations of products to product of concentrations of reactants at any time.

The same ratio at equilibrium (when rate of forward reaction becomes equal to rate of backward reaction) is equilibrium constant.

when Q < Kc, a forward reaction is favored.

When when Q > Kc, a backward or reverse reaction is favored

So the first statement that

a) A reaction quotient (Q) larger than the equilibrium constant (K) means that the reaction will favor the production of more products: false

b) No the rate of forward and backward reaction are equal.

c) c. Dynamic equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction: True

d) Dynamic equilibrium indicates that the amount of reactants and products are equal: This could be static equilibrium but not dynamic.

You might be interested in

Please help! 20 Points!

Pani-rosa [81]

Answer: YOU MAY FIND THE ANSWER IF YOU SEARCH THE WEB.

Explanation:

7 0

3 years ago

The energy required to break existing chemical bonds in reactants is called the ______ energy.

mixer [17]

The energy required to break existing chemical bonds in reactants is called the activation energy.

<h3>What is activation energy?</h3>

Activation energy in chemistry is the energy required to initiate a chemical reaction.

Chemical reactions involve the breaking of chemical bonds in substances called reactants to form new substances called products.

The energy required to break the bond in the existing reactants thus elevating these substances to a state of high activation is known as activation energy.

Therefore, it can be said that energy required to break existing chemical bonds in reactants is called the activation energy.

Learn more about activation energy at: brainly.com/question/11334504

#SPJ1

8 0

1 year ago

What is the value for (delta)G at 1000 K if (delta)H = -220 kJ/mol and (delta)S = -0.05 kJ/(molK)?

tankabanditka [31]

The system is isothermal, so we use the formula:
(delta)G = (delta)H - T (delta) S

Plugging in the given values:
(delta)G = -220 kJ/ mol - (1000K) (-0.05 kJ/mol K)
(delta)G = -170 kJ/mol

If we take a basis of 1 mol, the answer is
D. -170 kJ

6 0

3 years ago

Read 2 more answers

At which part of a river would you have the best luck growing food?

krek1111 [17]

Answer:

B. Delta

Explanation:

3 0

2 years ago

Read 2 more answers

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T

Eduardwww [97]

Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 730.5 mmHg

V = Volume of the gas = 9.49 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol$

According to the reaction shown below as:-

$2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)$

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when $\frac{2}{3}\times 0.3794$ moles of potassium chlorate undergoes reaction.

<u>Moles of potassium chlorate reacted = 0.2529 moles</u>

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 735.5 mmHg

V = Volume of the gas = 9.99 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol$

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g$

<u>Hence, the amount of oxygen gas collected will be 12.8675 g</u>

5 0

2 years ago