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3 years ago

What is the concentration of water in liquid (Percent)

Chemistry

1 answer:

scZoUnD [109]3 years ago

6 0

Hello!

Reading the worksheet shows you how to find the concentration of liquid water in a substance, but when converting a decimal to a percentage, remember that ALL percentages are out of 100, so, we multiply the decimal by 100 to convert it into a percentage.

Liquid A: total amount: 10 millimeters & amount of water: 7 millimeters

7/10 = 0.7 × 100 = 70%

Liquid B: total amount: 100 millimeters & amount of water: 92 millimeters

92/100 = 0.92 × 100 = 92%

Liquid C: total amount: 15 millimeters & amount of water: 13 millimeters

13/15 = 0.867 × 100 = 86.7%

Liquid D: total amount: 28 millimeters & amount of water: 22 millimeters

22/28 = 0.786 × 100 = 78.6%

<u>Final answers</u>:

Liquid A: 70%
Liquid B: 92%
Liquid C: 86.7%
Liquid D: 78.6%

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Does changing the number of neutrons change the identity of the element you have built ?

worty [1.4K]

Answer:

If you change the number of neutrons somehow, nothing will happen because it carry's no charge at all.

Explanation:

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3 years ago

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bija089 [108]

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3 years ago

Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.

Colt1911 [192]

Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

$\%C_E = \frac{E}{T} * 100\%$

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In $(Ca(OH)_2)$

The amount of all elements is: (i.e formula mass of $(Ca(OH)_2)$ )

$T = 1 * Ca + 2 * H + 2 * O$

$T = 1 * 40 + 2 * 1 + 2 * 16$

$T = 74$

The amount of calcium is: (i.e formula mass of calcium)

$E = 1 * Ca$

$E = 1 * 40$

$E = 40$

So, the percentage component of calcium is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{40}{74} * 100\%$

$\%C_E = \frac{4000}{74}\%$

$\%C_E = 54.05\%$

The amount of hydrogen is:

$E = 2 * H$

$E = 2 * 1$

$E = 2$

So, the percentage component of hydrogen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{2}{74} * 100\%$

$\%C_E = \frac{200}{74}\%$

$\%C_E = 2.70\%$

Similarly, for oxygen:

The amount of oxygen is:

$E = 2 * O$

$E = 2 * 16$

$E = 32$

So, the percentage component of oxygen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{32}{74} * 100\%$

$\%C_E = \frac{3200}{74}\%$

$\%C_E = 43.24\%$

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2 years ago

A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4

Harman [31]

Na₃PO₄ -----> 3Na(+) + PO₄(3-)
y-y.......................3y...........y

3y = 1.2
y = 0,4M

Na₃PO₄ -----> 3Na(+) + PO₄(3-)
0,4-0,4..............1,2..........0,4
0.........................1,2..........0,4

C = n/V
n = C×V
n = 0,4×0,65L
n = 0,26 mol Na₃PO₄

mNa₃PO₄: (23×3)+31+(16×4) = 164 g/mol

164g ----- 1 mol
Xg -------- 0,26 mol
X = 164×0,26
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3 years ago

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Nikitich [7]

Answer:

conserve mass.

Explanation:

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3 years ago