hydrogen and carbon, hope that helped
An example of a reaction that occurs within the core of a nuclear reactor is the nuclear fission reaction given:
- ²³⁵₉₂U + ¹₀n ---> ⁹⁰₃₈Sr + ¹⁴³₅₄ + 3 ¹₀n
<h3>What is a nuclear reactor?</h3>
A nuclear reactor is a device which produces electrical energy as a result of the nuclear reactions that take place within it.
In a nuclear reactor, the reaction that takes place within the core is a nuclear fission chain reaction.
In a nuclear fission reaction, the nucleus of larger atoms are split into the nucleus of smaller atoms when fast moving neutrons are used to bombard the nucleus of the large atom. The fission of the nucleus of the large atom results in the formation of atoms of lighter nucleus as well as more protons which then bombard more nucleus of the large atoms resulting in a chain reaction.
The chain reaction occurring within the nuclear reactor core is controlled by the insertion of boron rods which absorbs the excess neutrons produced.
An example of a reaction that occurs within the core of a nuclear reactor is given below:
²³⁵₉₂U + ¹₀n ---> ⁹⁰₃₈Sr + ¹⁴³₅₄ + 3 ¹₀n
Learn more about nuclear fission at: brainly.com/question/913303
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Answer:
4.5 kilograms of silicon dioxide is required to produce 3.00 kg of SiC.
Explanation:
The balanced equation for the reaction between silicon dioxide and carbon at high temperature is given as:
1 mole silicon dioxide reacts with 3 moles of carbon to give 1 moles of silicon carbide and 2 moles of carbon monoxide.
Mass of SiC = 3.00kg = 3000.00 g
1 kg = 1000 g
Molecular mass of SiC = 40 g/mol
Moles of SiC = 
According to reaction, 1 mole of SiC is produced from 1 mole of silicon dioxide.
Then 75 moles of SiC will be produce from:
of silicon dioxide.
mass of 75 moles of silicon dioxde:
4.5 kilograms of silicon dioxide is required to produce 3.00 kg of SiC.
Answer:
The outside temperature is -45.8°C
Explanation:
When a gas keeps on constant its moles and its pressure, we can assume that volume will be increased or decreased as the T° (absolute T° in K).
V1 / T1 = V2 / T2
2.95L/298K = 2.25L / T2
(2.95L/298K ) . T2 = 2.25L
T2 = 2.25L . 298K / 2.95L
T2 = 227.2K
T°K - 273 = T°C
227.2K - 273 = -45.8°C
