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Yakvenalex [24]
2 years ago
13

10 g of a solute is added to 1.00 L of pure water. Half of the added solute fails to dissolve. What can you conclude?

Chemistry
1 answer:
ryzh [129]2 years ago
5 0

If in the solution, half of the added solute fails to dissolve. The solution started out supersaturated. The correct option is b.

<h3>What is supersaturation?</h3>

Supersaturation is the condition where the solutes exceed the amount that can be dissolved in a solution.

Supersaturation occurs when the solute no longer mix in the solution.

Thus, the correct option is b. The solution started out supersaturated.

Learn more about supersaturation

brainly.com/question/16817894

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Molar mass (NH4)2SO4
Hoochie [10]

Answer:

<u>132.15</u>

Explanation:

Molar mass N = 14.00

Molar mass H = 1.01

Molar mass H4 = 1.01 x 4 = 4.04

Molar mass NH4 = 14.00 + 4.04 = 18.04

Molar mass (NH4)2 = 18.04 x 2 = 36.08

Molar mass S = 32.07

Molar mass O = 16.00

Molar mass O4 = 16.00 x 4 = 64.00

Molar mass SO4 = 32.07 + 64.00 = 96.07

Molar mass (NH4)2SO4 = 36.08 + 96.07 = <u>132.14</u>

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Alona [7]

Magnetism, or more specifically electromagnetic repulsion

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Read 2 more answers
_HNO3 + _Al(OH)3
bekas [8.4K]

Answer:

Option D. 3, 1, 3, 1

Explanation:

From the question given above,

HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃

The equation can be balance as follow:

HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃

There are 3 atoms of N on the right side and 1 atom on the left side. It can be balance by 3 in front of HNO₃ as shown below:

3HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃

There are a total of 6 atoms of H on the left side and 2 atoms on the right side. It can be balance by 3 in front of HOH as shown below:

3HNO₃ + Al(OH)₃ —> 3HOH + Al(NO₃)₃

Now, the equation is balanced.

Thus, the coefficients are 3, 1, 3, 1

7 0
3 years ago
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,n
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Answer:

yes

Explanation:

n,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

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