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lozanna [386]
4 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]4 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]4 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Control rods may be inserted into a nuclear reactor core to reduce the reaction rate or withdrawn to increase it depending on the desired rate of reaction.

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How long does it take the Earth to move from Point A to Point C?
e-lub [12.9K]

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Water is a pure substance. Which of the following is true about water? (4 points) Its compounds can only be physically separated
jonny [76]

Answer:

Water/H20 is a compound made of hydrogen and oxygen. Although water is the most abundant substance on earth, it is rarely found naturally in its pure form. Most of the time, pure water has to be created. Pure water is called distilled water or deionized water.

Explanation:

7 0
3 years ago
Given the chemical equation: 2 Pb + O2 → 2 PbO, if 51.8 grams of Pb are formed in this reaction, then 8.00 grams of O2 must have
Nutka1998 [239]

Answer:

If 51.8 of Pb is reacting, it will require 4.00 g of O2

If 51.8 g of PbO is formed, it will require 3.47 g of O2.

Explanation:

Equation of the reaction:

2 Pb + O2 → 2 PbO

From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO

Molar mass of Pb = 207 g

Molar mass of O2 = 32 g

Molar mass of PbO = 207 + 32 = 239 g

Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO

= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO

Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.

If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2

7 0
3 years ago
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