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Aneli [31]
1 year ago
7

Recall that you're hypothesis is that these values are the fractions of atoms that are still radioactive after n half life cycle

s record it in the appropriate blanks
Chemistry
1 answer:
Sergeu [11.5K]1 year ago
4 0

A= 0.5, B = 0.25 , C = 0.125, D = 0.015625 and E = 0.00390625

<h3>What is a hypothesis?</h3>

A research hypothesis is a statement of expectation or prediction that will be tested by research. Before formulating your research hypothesis, read about the topic of interest to you.

Half life of a substance is defined as the amount of time taken by the substance to reduce to half of its original amount.

Here n represents the number of half lives.

The amount of substance that remains after n half lives can be calculated using the given formula,

So when we have n =1,

Fraction of substance that remains = 0.5¹ = 0.5.

That means after first half life over, the amount of substance that remains is 0.5 times that of original.

Therefore we have A = 0.5

When n = 2, we have 0.5² = 0.25

So when 2 half lives are over, the amount of substance that remains is 0.25 times that of original

Therefore B = 0.25

When n = 3, we have 0.5³ = 0.125

So when 3 half lives are over, the amount of substance that remains is 0.125 times that of original.

Therefore we have C = 0.125

When n = 6 , we have 0.5⁶ = 0.015625

So D = 0.015625

When n = 8, we have 0.5⁸ = 0.00390625

Therefore E = 0.00390625

The values for A, B, C, D and E are 0.5, 0.25, 0.125, 0.015625 and 0.00390625 respectively.

Learn more about the hypothesis here:

brainly.com/question/17173491

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Sodium phosphate is added to a solution that contains 0.0070 M aluminum nitrate and 0.052 M calcium chloride. The concentration
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Answer:

The answer to the question is;

The first ion to precipitate out is the Al³⁺ ion and the concentration of the Al³⁺ ion when the Ca²⁺ ion begins to precipitate is 1.12 × 10⁻⁵ M.

Explanation:

To solve the question, we note that

aluminum nitrate, Al(NO₃)₃ will dissociate as follows

Al(NO₃)₃ → Al³⁺ (aq) + 3NO₃⁻ (aq)

Therefore when sodium phosphate is added to a solution that contains aluminum nitrate  we have the following system  of aluminium phosphate which is

AlPO₄(s) ⇄ Al³⁺(aq) + PO₄³⁻(aq)

The solubility product for the above reaction is

Ksp = [Al³⁺][PO₄³⁻] = 9.84×10⁻²¹

The solubility product for calcium phosphate is expressed as

Ca₃(PO₄)₂(s) ⇄ 3 Ca²⁺(aq) + 2 PO₄³⁻(aq)

With Ksp =  [Ca²⁺]³[PO₄³⁻]² = 2.07×10⁻³³

From the solubility product, we can find the concentration of [PO₄³⁻] at which precipitation starts as follows

The phosphate concretion for Al³⁺ when precipitation starts is

[PO₄³⁻] = \frac{K_{sp}}{[Al^{3+}]}= 9.84×10⁻²¹ / 0.007 = 1.406×10⁻¹⁸ M

The phosphate concretion for Ca²⁺ when precipitation starts is

[PO₄³⁻] =\sqrt{\frac{K_{sp}}{[Ca^{2+}]^2}}  = \sqrt{\frac{2.07\times10^{-33}}{[0.052]^2}} = 8.75×10⁻¹⁶ M

(Aluminium phosphate precipitates out first)

The reaction favors the precipitation of the aluminum phosphate first due to the lower concentration of the [PO₄³⁻]  ions in the [Al³⁺][PO₄³⁻] system which  is lower than the relative [PO₄³⁻] in the [Ca²⁺]³[PO₄³⁻]².

Therefore, the more sodium phosphate added serves to precipitate the remaining aluminium phosphate.

The process continues and the concentration of Al³⁺ decreases as more precipitates form. The process continues until the equilibrium conditions satisfies the precipitation threshold level for the calcium phosphate system concentration whereby the concentration of the Al³⁺ in the solution is given by.

[Al³⁺] = \frac{K_{sp}}{[PO_4^{3-}]} = \frac{9.84\times 10^{-21}}{8.75\times 10^{-16}}  = 1.12 × 10⁻⁵ M

Therefore the concentration of this aluminium ion when the calcium ion begins to precipitate =  1.12 × 10⁻⁵ M.

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