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2 years ago

Explain why uranium must be enriched to be used in a nuclear power plant.

Chemistry

1 answer:

galben [10]2 years ago

5 0

Answer:

The nuclear fuel used in a nuclear reactor needs to have a higher concentration of the U 235 isotope than that which exists in natural uranium ore. U235 when concentrated (or "enriched") is fissionable in light-water reactors (the most common reactor design in the USA).

Explanation:

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How many moles are in 7.2x10^15 atoms of Pb?

fiasKO [112]

<h3>Answer:</h3>

1.2 × 10⁻⁸ mol Pb

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 7.2 × 10¹⁵ atoms Pb

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 7.2 \cdot 10^{15} \ atoms \ Pb(\frac{1 \ mol \ Pb}{6.022 \cdot 10^{23} \ atoms \ Pb})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 1.19562 \cdot 10^{-8} \ mol \ Pb$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.19562 × 10⁻⁸ mol Pb ≈ 1.2 × 10⁻⁸ mol Pb

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2 years ago

Which of the following best describes technology

kotegsom [21]

were are the choices

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2 years ago

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When electrons are added to the outter most shell of carbon atom it forms

GalinKa [24]

Answer:

An anion that has a larger radius.

Explanation:

3 0

3 years ago

HNO3 and H2CO3 are examples of ?

Basile [38]

A) acids because they start with h

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2 years ago

What is the molality of sodium chloride in solution that is 13.0% by mass sodium chloride and that has a density of 1.10 g/ml?

8090 [49]

Answer is: molality od sodium chloride is 2,55 mol/kg.
V(solution) = 100 ml.
m(solution) = d(solution) · V(solution).
m(solution) = 1,10 g/ml · 100 ml.
m(solution) = 110 g.
ω(NaCl) = 13,0% = 0,13.
m(NaCl) = ω(NaCl) · m(solution).
m(NaCl) = 0,13 · 110 g.
m(NaCl) = 14,3 g.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 14,3 g ÷ 58,5 g/mol.
n(NaCl) = 0,244 mol.
m(H₂O) = 110 g - 14,3 g.
m(H₂O) = 95,7 g = 0,0957 kg.
b(NaCl) = n(NaCl) ÷ m(H₂O).
b(NaCl) = 0,244 mol ÷ 0,0957 kg.
b(NaCl) = 2,55 mol/kg.

3 0

3 years ago