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Kisachek [45]
3 years ago
12

How many liters of water are required to dissolve 1.00 g of barium chromate? express your answer in liters to three significant

figures?

Chemistry
2 answers:
VashaNatasha [74]3 years ago
8 0
Assuming the water is present at 20 °C, the solubility of barium chromate is 0.2275 mg/100 mL

Converting this to mg/L, we get:

2.275 mg/L

Converting this value again, to grams per liter, we get:

0.002275 g/L

To dissolve one gram, we require:

1/0.002275 = 439.56044

440 liters of water.
Evgen [1.6K]3 years ago
6 0

The solubility  Barium chromate in grams per liter = 2.78 . 10⁻³ grams/L

359 liters of water are required to dissolve 1.00 g of Barium chromate

<h3>Further explanation </h3>

Solubility is the maximum amount of a substance that can dissolve in some solvents. Factors that affect solubility

  • 1. Temperature:
  • 2. Surface area:
  • 3. Solvent type:
  • 4. Stirring process:

Ksp is an ion product in equilibrium

Solubility relationships and solubility constants (Ksp) of the AxBa solution can be stated as follows.

AₓBₐ (s) ← ⎯⎯⎯⎯ → x Aᵃ⁺ (aq) + a Bˣ⁻ (aq)

s                             xs               as

Ksp = [Aᵃ⁺] ˣ [Bˣ⁻] ᵃ

Ksp = (xs) ˣ (as) ᵃ

Solubility units in the form of mol / liter or gram / liter

At 25.°C, the molar solubility of Barium chromate  BaCrO₄ in water is 1.10. 10⁻⁵M.

to change units to grams / liter, we multiply by molar mass:

M BaCrO₄ = Ba + Cr + 4. Ar O

M BaCrO₄ = 137 + 52 + 4.16

M BaCrO₄ = 253

So the solubility is in grams / liter

= 1.10 . 10⁻⁵ mol / liter x 253 grams / mol

=  278.3 .10⁻⁵ = 2.78 . 10⁻³ grams/L

(3 significant numbers, 2.7 and 8)

If we dissolve 1 gram of Barium chromate into the solution, we need water :

= 1 grams / 2.78 . 10⁻³ grams / liter

= 359 liters

(3 significant numbers, 3.5 and 9)

<h3>Learn more </h3>

the rate of solubility

brainly.com/question/9551583

influencing factors are both solubility and the rate of dissolution

brainly.com/question/2393178

The solubility of a substance

brainly.com/question/2847814

Keywords: solubility, silver chromate, a significant number

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A 45 mL sample of nitrogen gas is cooled from 135ºC to 15C in a container that can contract or expand at constant pressure. Wha
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Answer:

V₂ =31.8 mL

Explanation:

Given data:

Initial  volume of gas = 45 mL

Initial temperature = 135°C (135+273 =408 K)

Final temperature = 15°C (15+273 =288 K)

Final volume of gas = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

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V₂ = 12960 mL.K / 408 K

V₂ =31.8 mL

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3 years ago
How many grams of solute are in 1.00 liter of 0.200m na2so4 solution?
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this means there are 0.200 moles in 1 L solution 
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therefore mass of Na₂SO₄ in 1.00 L - 0.200 mol x 142 g/mol = 28.4 g
a mass of 28.4 g of Na₂SO₄ is present in 1.00 L
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