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2 years ago

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?

1 answer:

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

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yes

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3 years ago

A cork has a mass of 3 grams and a volume of 16 cms calculate the density

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hey mate here is ur answer

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3 years ago

A substance that does not allow x-rays to pass through is described as being

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A substance that is impenetrable by x-rays is described as being radiopaque.

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2 years ago

Using the black numbers on the stopwatch to answer the questions.

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4 years ago

Read 2 more answers

Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. How many grams of di

pychu [463]

Answer:

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant. Explanation:

$Ca(NO_3)_2(aq)+2NH_4F(aq)\rightarrow CaF_2(aq)+2N_2O(g)+4H_2O(g)$

Moles of calcium nitrate = $\frac{31.3 g}{164 g/mol}=0.1908 mol$

Moles of ammonium fluoride = $\frac{38.7 g}{37 g/mol}=1.046 mol$

According to reaction , 2 moles of ammonium fluoride reacts with 1 mole of calcium nitrate.

Then 1.046 moles of ammonium fluoride will react with :

$\frac{1}{2}\times 1.046 mol=0.523 mol$ calcium nitarte .

This means that ammonium fluoride is in excess amount and calcium nitrate is in limiting amount.

Hence, calcium nitrate is a limiting reactant.

So, amount of dinitrogen monoxide will depend upon moles of calcium nitrate.

So, according to reaction , 1 mole of calcium nitarte gives 2 moles of dinitrogen monoxide gas .

Then 0.1908 moles of calcium nitrate will give:

$\frac{2}{1}\times 0.1908 mol=0.3816 mol$ of dinitrogen monoxide gas.

Mass of 0.03816 moles of dinitrogen monoxide gas:

0.03816 mol × 44 g/mol = 16.79 g

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant.

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3 years ago