At 25.0∘c, the molar solubility of barium chromate in water is 1.10×10−5 m . calculate the solubility in grams per liter.

1 answer:

6 0

When in water, barium chromate dissociates into ions. The dissociation equation would be as follows:

<span>BaCrO4 = Ba^2+ + CrO4^2+
</span>
For this case, we are to convert only from moles per kg to grams per liter. We do as follows:

<span>1.10×10^−5 mol / kg ( 253.37 g /mol) ( 1 kg/L ) = 2.79x10^-3 kg / L = 2.79 g/L</span>

You might be interested in

Concentrated HCl is 15 M (15 molar). A 500 mL quantity of this solution is diluted by adding water to give a final volume of 2.5

Murrr4er [49]

The molarity of the resulting solution obtained by diluting the stock solution is 3 M

<h3>Data obtained from the question </h3>

Molarity of stock solution (M₁) = 15 M
Volume of stock solution (V₁) = 500 mL
Volume of diluted solution (V₂) = 2.5 L = 2.5 × 1000 = 2500 mL
Molarity of diluted solution (M₂) =?

<h3>How to determine the molarity of diluted solution </h3>

M₁V₁ = M₂V₂

15 × 500 = M₂ × 2500

7500 = M₂ × 2500

Divide both side by 2500

M₂ = 7500 / 2500

M₂ = 3 M

Thus, the volume of the resulting solution is 3 M

Learn more about dilution:

brainly.com/question/15022582

#SPJ1

7 0

2 years ago

How do you solve this ?NH4+NO4->N2+O2+H2O2

ololo11 [35]

I'm assuming you wanted this equation balanced? If you typed it correctly, it's already balanced for you. :)

8 0

3 years ago

A 7.5 L cylinder contains 5 moles of gas at a temperature of 274°C. What is its pressure in kiloPascals (kPa)?

AleksAgata [21]

<h3><u>Answer;</u></h3>

= 3032.15 kPa

<h3><u>Explanation;</u></h3>

Using the equation;

PV = nRT , where P is the pressure,. V is the volume, n is the number of moles and T is the temperature and R is the gas constant, 0.08206 L. atm. mol−1.

Volume = 7.5 L, T = 274 +273 = 547 K, N = 5 moles

Therefore;

Pressure = nRT/V

= (5 × 0.08206 × 547)/7.5 L

= 29.925 atm

But; 1 atm = 101325 pascals

Hence; Pressure = 3032150.63 pascals

<u>= 3032.15 kPa</u>

3 0

3 years ago

How many grams are there in 7.5x10^23 molecules of H2SO4

Xelga [282]