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3 years ago

What must happen to uranium before it can be used as a fuel source?

Chemistry

1 answer:

Otrada [13]3 years ago

5 0

Answer:

Uranium must be purified before it is used as a fuel source

Explanation:

The purer the uranium sample, the more the concentration of uranium in the fuel is.

Whenever uranium is extracted from nature, it contains a lot of impurities. Only a few special nuclear reactors can utilize uranium in this raw state. most of the others have to get uranium to become about 3% pure before they begin using it.

To do this, uranium has to be passed through a series of chemical reactions all with the aim of extracting the other compounds that may be present in the fuel.

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A solution has a pOH of 8.7 so what is the pH of the solution? Is the solution acidic, basic, or neutral?

RSB [31]

PH + pOH = 14

pH = 14 - pOH

pH = 14 - 8.7

pH = 5,3

This solution is acidic.

If pH<7 - acidic
If pH=7 - neutral
If pH>7 - basic

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3 years ago

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A solution with a ph of 4 has _________ the concentration of h+ present compared to a solution with a ph of 5.

Marina CMI [18]

A solution with a pH of 4 has ten times the concentration of H⁺ present compared to a solution with a pH of 5.
pH is a numeric scale for the acidity or basicity of an aqueous solution. It is the negative of the base 10 logarithm of the molar concentration of hydrogen ions.
[H⁺] = 10∧-pH.
pH = 4 → [H⁺]₁ = 10⁻⁴ M = 0,0001 M.
pH = 5 → [H⁺]₂ = 10⁻⁵ M = 0,00001 M.
[H⁺]₁ / [H⁺]₂ = 0,0001 M / 0,00001 M.
[H⁺]₁ / [H⁺]₂ = 10.

7 0

3 years ago

How many atoms are in 1.00 molecules of he

irina1246 [14]

Answer:

Explanation:

6 0

3 years ago

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Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

Explanation:

According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

⇒ $-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}$ -----------------------------(1)

Given that the rate of disappearance of oxygen = $-\frac{d[O_{2} ]}{dt}$ = 3.64 x 10⁻³ M/s

So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = ?

from equation (1) we can write

$\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]$

⇒ $\frac{d[SO_{3}] }{dt}$ = 2 x 3.64 x 10⁻³ M/s

⇒ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

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3 years ago

Select the pair that consists of a base and its conjugate acid in that order. CO32−/CO22−

lana [24]

Answer: The pair that consists of a base and its conjugate acid in that order. $NH_3/NH_4^+$

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

$H_3PO_4\rightarrow H_PO_4{2^-}+2H^+$

$H_2CO_3\rightarrow HCO_3^-+H^+$

$NH_3+H^+\rightarrow NH_4^+$

$HCO_3^-\rightarrow CO_3^{2-}+H^+$

$NH_3$ is gaining a proton, thus it is considered as a brønsted-lowry base and after gaining a proton, it forms $NH_4^+$ which is a conjugate acid.

3 0

3 years ago