Answer: 6.2 grams of the sodium acetate can dissolve in 5 milliliters of water. if 124 grams of the sodium acetate dissolves in 100 milliliters of water, then 6.2 grams of the sodium acetate can dissolve in 5 milliliters of water.

4 0

3 years ago

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A 15.0 L tank of gas is contained at a high pressure of 8.20 x 10^4 torr. The tank is opened and the gas expands into an empty c

Inga [223]

Answer:

20.5torr

Explanation:

Given parameters:

V₁ = 15L

P₁ = 8.2 x 10⁴torr

V₂ = 6 x 10⁴L

Unknown:

P₂ = ?

Solution:

To solve this problem we have to apply the claims of Boyle's law.

Boyle's law is given mathematically as;

P₁ V₁ = P₂V₂

where P₁ is the initial pressure

V₁ is the initial volume

P₂ is final pressure

V₂ is final volume

8.2 x 10⁴ x 15 = P₂ x 6 x 10⁴

P₂ = 20.5torr

5 0

3 years ago

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.

8 0

3 years ago

What is the probability that an offspring will have a<br> heterozygous genotype? |

Anastasy [175]

Answer:

25,50,25

Explanation:

5 0

2 years ago