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lozanna [386]
3 years ago
9

in a nuclear power plant control rods are used to a. speed up the reaction. b. slow down the reaction. c. feed material to the r

eaction. d. stabilize the walls of the nuclear reactor.
Chemistry
2 answers:
Natali [406]3 years ago
6 0
<span>In a nuclear power plant control rods are used to:

</span><span>b. slow down the reaction.</span>
ale4655 [162]3 years ago
3 0

Answer:

speed up the reaction.

slow down the reaction.

Explanation:

Control rods are used in nuclear reactors to control the fission rate of reactions. Their compositions includes certain elements such as boron, cadmium, silver, or indium, which are capable of absorbing many neutrons without themselves fissioning.

Control rods may be inserted into a nuclear reactor core to reduce the reaction rate or withdrawn to increase it depending on the desired rate of reaction.

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What will the concentration of PCl5 be when equilibrium is reestablished after addition of 1.31 g Cl2? PCl5(g) ⇆ PCl3(g) + Cl2(g
Masteriza [31]

Answer:

The new concentration of PCl5 will be 0.01953 M

Explanation:

Step 1: Data given

Mass of Cl2 added = 1.31 grams

Molar mass Cl2 = 70.9 g/mol

Original Equilibrium Mixture:

3.42 g PCl5

4.86 g PCl3

3.59 g Cl2

Volume = 1.0 L

Step 2: The balanced equation

PCl5(g) ⇆ PCl3(g) + Cl2(g)

Step 3: Calculate the original moles and molarity

Moles = mass / molar mass

Moles PCL5 = 3.42 grams / 208.24 g/mol

Moles PCl5 = 0.0164 moles

[PCl5] = 0.0164 M

moles PCl3 = 4.86 grams / 137.33 g/mol

moles PCl3 = 0.0354 moles

[PCl3] = 0.0354 M

moles Cl2 = 3.59 grams / 70.9 g/mol

moles Cl2 = 0.0506 moles

[Cl2] = 0.0506 M

the new mass Cl2 = 3.59 + 1.31 = 4.9 grams

moles Cl2 = 0.0691 moles

[Cl2]= 0.0691 M

The new concentration at the equilibrium

[PCl5] = 0.0164 + X M

[PCl3 ] =  0.0354 - X M

[Cl2] = 0.0691 - X M

Step 4: Calculate Kc

Kc = [Cl2][PCl3] / [PCl5]

Kc = (0.0506*0.0354)/0.0164

Kc = 0.109

Step 5: Calculate [PCl5]

Kc = 0.109 = ((0.0691 - X)(0.0354 - X)) / (0.0164 + X)

X = 0.00313

[PCl5] = 0.0164 + 0.00313 M = 0.01953 M

[PCl3 ] =  0.0354 - 0.00313 M = 0.03227 M

[Cl2] = 0.0691 - 0.00313 M = 0.06597

The new concentration of PCl5 will be 0.01953 M

6 0
3 years ago
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