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2 years ago

What is the function of control rods in a nuclear reactor?

Chemistry

2 answers:

Helga [31]2 years ago

8 0

C is the correct answer

trapecia [35]2 years ago

4 0

The answer is
C. Control reaction rate
hope it helps :)

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The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for

tester [92]

The given question is incomplete. The complete question is:

The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for each of the reactions and decide if the entropy increases decreases or has little to no change:

A. $K(s)+O_2(g)\rightarrow KO_2(s)$

B. $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$

C. $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

D. $N_2O_2(g)\rightarrow 2NO(g)+O_2(g)$

Answer: A. $K(s)+O_2(g)\rightarrow KO_2(s)$ : decreases

B. $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2(g)$ : decreases

C. $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$ : no change

D. $N_2O_2(g)\rightarrow 2NO(g)+O_2(g)$ : increases

Explanation:

Entropy is defined as the randomness of the system.

Entropy is said to increase when the randomness of the system increase, is said to decrease when the randomness of the system decrease and is said to have no change when the randomness remains same.

In reaction $K(s)+O_2(g)\rightarrow KO_2(s)$ , as gaseous reactant is changed to solid product, entropy decreases.

In reaction $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$ , as 4 moles of gaseous reactants is changed to 2 moles of gaseous product, entropy decreases.

In reaction $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$ , as 3 moles of gaseous reactants is changed to 3 moles of gaseous product, entropy has no change.

In reaction $N_2O_2(g)\rightarrow 2NO(g)+O_2(g)$ , as 1 mole of gaseous reactant is changed to 3 moles of gaseous product, entropy increases.

7 0

3 years ago

A 2.20 mol sample of NO 2 ( g ) is added to a 3.50 L vessel and heated to 500 K. N 2 O 4 ( g ) − ⇀ ↽ − 2 NO 2 ( g ) K c = 0.513

igor_vitrenko [27]

Answer:

[NO₂] = 0.434 M

[N₂O₄] = 0.0971 M

Explanation:

The equilibrum is: N₂O₄(g) ⇆ 2NO₂ (g)

1 moles of nitrogen (IV) oxide is in equilibrium with 2 moles of nitrogen dioxide.

Initally we only have 2.20 moles of NO₂. So let's write the equilibrium again:

2NO₂ (g) ⇆ N₂O₄(g)

Initially 2.20 mol -

React x x/2

X amount has reacted, and the half has been formed, according to stoichiometry.

Eq (2.20-x) / 3.50L (x/2)/ 3.50L

We divide by the volume because we need molar concentrations. Let's make the Kc's expression:

Kc = [N₂O₄] / [NO₂]²

0.513 = ((x/2)/ 3.50L) / [(2.20-x) / 3.50L]

0.513 = ((x/2)/ 3.50L) / [(2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / [2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / (4.84 - 4.40x + x²) / 12.25)

0.513 / 12.25 (4.84 - 4.40x + x²) = x/2 / 3.50

0.203 - 0.184x + 0.0419x² = x/2 / 3.50

3.50(0.203 - 0.184x + 0.0419x²) = x/2

7 (0.203 - 0.184x + 0.0419x²) - x = 0

1.421 - 2.288x + 0.2933x² = 0 → Quadratic formula

a = 0.2933 ; b = -2.288 ; c = 1.421

(-b +- √(b²-4ac)) / (2a)

x₁ = 7.12

x₂ = 0.68 → We consider this value, so we can have a (+) concentration.

Concentrations in the equilibrium are:

[NO₂] = (2.20-0.68) / 3.50 = 0.434 M

[N₂O₄] = (0.68/2) / 3.50 = 0.0971 M

8 0

2 years ago

PLEASE HELP!!

sp2606 [1]

Inbox me for the answer

6 0

2 years ago

Match each element to the number of electrons in its valence shell.

irinina [24]

Answer:

C: 4

H: 1

Mg: 2

O : 6

Explanation:

You can quickly find out the number of valence electrons by looking at where the element is on the periodic table and referring to the table that is attached.

6 0

9 months ago

Which of the following is an example of a chemical change?

Lynna [10]

The answer is burning paper

6 0

3 years ago

Read 2 more answers