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shutvik [7]
3 years ago
9

A 13.97 g sample of nabr contains 22.34% na by mass. Considering the law of constant composition (definite proportions) how many

grams of sodium does a 5.75 g sample of sodium bromide contain?
Chemistry
1 answer:
Grace [21]3 years ago
3 0

As per the law of constant composition, a given sample will always contain the same number of elements that combine in the same mass proportion.

Therefore if a sample of 13.97 g of NaBr contains 22.39 % of Na by mass then,  a sample of 5.75 g of NaBr would also contain 22.39% Na by mass

Hence:

Mass of Na = 5.75 g * 22.39/100 = 1.287 g

5.75 g of NaBr would contain 1.29 g of Na

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Estimate the solubility of CaCO3
Fudgin [204]

Answer:

CaCO3(s) <==> Ca2+(aq) + CO3 2-(aq)

Let X = the amount of CaCO3 dissolved in Na2CO3 or the amount of each ions formed i.e. Ca2+ and CO3 2–. Then, put this into the formula :

Ksp = [Ca2+] [CO3 2-]

5 × 10^-9 = [X]*[X]

X^2 = 5 × 10^-9

X = √5 × 10^-9

X = 7.07 x 10^-5 mole/L

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4 0
3 years ago
The value of Kc for the reaction between water vapor and dichlorine monoxide, H2O(g) 1 Cl2O(g) 4 2 HOCl(g) is 0.0900 at 25°C. De
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Answer:

[HOCl] = 0.001 127 mol·L⁻¹; [H₂O] = [Cl₂O] = 0.003 76 mol·L⁻¹

Explanation:

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Data:

     Kc = 0.0900

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[Cl₂O] = 0.004 32 mol

1. Set up an ICE table.

\begin{array}{ccccccc}\rm \text{H$_{2}$O}& + & \text{Cl$_{2}$O} & \, \rightleftharpoons \, & \text{2HOCl} & & \\0.00432 & & 0.00432 & & 0 & & \\-x &&-x&&+2x&&\\0.00432-x &&0.00432 - x& & 2x&&\\\end{array}

2. Calculate the equilibrium concentrations

K_{\text{c}} = \dfrac{\text{[HOCl]$^{2}$}}{\text{[H$_{2}$O][Cl$_2$O]}} = \dfrac{(2x)^{2}}{(0.00432 - x)^{2}} = 0.0900\\\\\begin{array}{rcl}\dfrac{4x^{2}}{(0.00432 - x)^{2}} &=& 0.0900\\ \dfrac{2x }{0.00432 - x} & = & 0.300\\2x & = & 0.300(0.00432 - x)\\2x & = & 0.001296 - 0.300x\\2.300x & = & 0.001296\\x & = & \mathbf{5.63\times 10^{-4}}\\\end{array}

[HOCl] = 2x mol·L⁻¹ = 2 × 5.63 × 10⁻⁴ mol·L⁻¹ =0.001 127 mol·L⁻¹

[H₂O] = [Cl₂O] = (0.004 32 - 0.000 563) mol·L⁻¹ = 0.003 76 mol·L⁻¹

Check:

\begin{array}{rcl}\dfrac{0.001127^{2}}{0.00376^{2}} & = & 0.0900\\\\\dfrac{1.270 \times 10^{-6}}{1.411 \times 10^{-5}} & = & 0.0900\\0.0900 & = & 0.0900\\\end{array}

OK.

7 0
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