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STALIN [3.7K]
2 years ago
12

What volume of a 15.0% by mass NaOH solution, which has a density of 1.116 g/mL , should be used to make 5.05 L of an NaOH solut

ion with a pH of 10.6
Chemistry
1 answer:
Vilka [71]2 years ago
7 0

Answer:

NaOH = 40 g/mole

M = 10*d*m% /MW

M = 10*1.116*15/ 40

M = 4.185

––––––––––––––––––––

pH = 11 –––> pOH = 3 –––> [OH–] = 10^–3 M

M1*V1 = M2*V2

4.185* V1 = 10^–3 * 5.3

V1 = 1.27×10^–3 L = 1.27 ml

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The relative volumes of chloroform and water that should be used is 9:10

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Meaning;

K_D = \frac{\frac{mole\ of\ chloroform}{volume\ of\ chloroform} }{\frac{mole\ of\ water}{volume\ of\ water} }

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8 0
3 years ago
a compound has 15.39 g of gold for every 2.77 g of chlorine. simplified there is _____ g of gold for every 1 g of chlorine
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Answer:

There is 5.56 g of gold for every 1 g of chlorine

Explanation:

The ratio is the relationship between two numbers, defined as the ratio of one number to the other. So, the ratio between two numbers a and b is the fraction \frac{a}{b}

You know that a compound has 15.39 g of gold for every 2.77 g of chlorine. This can be expressed by the ratio:

\frac{15.39 g  of gold}{2.77 g of chlorine}

The proportion is the equal relationship that exists between two reasons and is represented by:    \frac{a}{b}=\frac{c}{d}

This reads a is a b as c is a d.

To calculate the amount of gold per 1 g of chlorine, the following proportion is expressed:

\frac{15.39 g  of gold}{2.77 g of chlorine}=\frac{mass of gold}{1 g of chlorine}

Solving for the mass of gold gives:

mass of gold=1 g of chlorine*\frac{15.39 g  of gold}{2.77 g of chlorine}

mass of gold= 5.56 grams

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