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3 years ago

Which method would increase the solubility of a gas?

Chemistry

2 answers:

Irina18 [472]3 years ago

4 0

Answer:

increasing the pressure

Explanation:

UNO [17]3 years ago

3 0

Temperature and Pressure One way to increase the solubility of a gas is to decrease the temperature of the liquid. The solubility of a gas in a liquid is usually temperature dependent, although it depends on the particular combination of which gas and which liquid. Usually the solubility of a gas goes down with increasing temperature (think of warm carbonated beverages going flat).

<span>The other way to increase the solubility is to increase the pressure of the gas. The higher the pressure of the gas above the liquid, the more will dissolve. Again, think of a carbonated beverage: when it is sealed it doesn't go flat because it is under pressure, but when open to air, it will go flat. </span>

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what do the symbols tell you about the conditions of the reaction shown to the right check all boxes that apply

alukav5142 [94]

Answer:

Sorry but erm was their supposed to be a image attach with it

Explanation:

6 0

3 years ago

Read 2 more answers

A solution contains [Ba2+] = 5.0 × 10−5 M, [Zn2+] = 2.0 × 10−7 M, and [Ag+] = 3.0 × 10−5 M. Sodium oxalate (Na2C2O4) is slowly a

Jet001 [13]

Answer:

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

Explanation:

1. Calculate the equilibrium concentrations of oxalate ion

Let [C₂O₄²⁻] = c

(a) Barium oxalate

BaC₂O₄ ⇌ Ba²⁺ + C₂O₄²⁻

E/mol·L⁻¹: 5.0 × 10⁻⁵ c

Ksp = [Ba²⁺][C₂O₄²⁻] = 5.0 × 10⁻⁵c = 1.5 × 10⁻⁸

c = (1.5 × 10⁻⁸)/(5.0 × 10⁻⁵) = 3.0 × 10⁻⁴ mol·L⁻¹

(b) Zinc oxalate

ZnC₂O₄ ⇌ Zn²⁺ + C₂O₄²⁻

E/mol·L⁻¹: 2.0 × 10⁻⁷ c

Ksp = [Zn²⁺][C₂O₄²⁻] = 2.0 × 10⁻⁷c = 1.35 × 10⁻⁹

c = (1.35 × 10⁻⁹)/(2.0 × 10⁻⁷) = 6.8 × 10⁻³ mol·L⁻¹

(c) Silver oxalate

Ag₂C₂O₄ ⇌ 2Ag⁺ + C₂O₄²⁻

E/mol·L⁻¹: 3.0 × 10⁻⁵ c

Ksp = [Ag⁺]²[C₂O₄²⁻] = (3.0× 10⁻⁵)²c = 9.0 × 10⁻¹⁰c = 1.1 × 10⁻¹¹

c = (1.1 × 10⁻¹¹)/(9.0 × 10⁻¹⁰) = 0.012 mol·L⁻¹

2. Decide the order of precipitation

BaC₂O₄ will precipitate when c > 3.0 × 10⁻⁴ mol·L⁻¹

ZnC₂O₄ will precipitate when c > 6.8 × 10⁻³ mol·L⁻¹

Ag₂C₂O₄ will precipitate when c > 0.028 mol·L⁻¹

This happens to be the order of increasing concentration of oxalate ion.

The order of precipitation is

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

4 0

3 years ago

Calculate the number of grams of CO that can react with 0.400 kg of Fe2O3.

aleksandr82 [10.1K]

Answer:

mass of CO = 210.42 g

mass in three significant figures = 210. g

Explanation:

Given data:

mass of Fe2O3 = 0.400 Kg

mass of CO= ?

Solution:

chemical equation:

Fe2O3 + 3CO → 2Fe + 3CO2

Now we will calculate the molar mass of Fe2O3 and CO.

Molar mass of Fe2O3 = (55.845 × 2) + (16 × 3) = 159.69 g/mol

Molar mass of CO = 12+ 16 = 28 g/mol

now we will convert the kg of Fe2O3 in g.

mass of Fe2O3 = 0.400 kg × 1000 = 400 g

number of moles of Fe2O3 = 400 g/ 159.69 g/mol = 2.505 mol

mass of CO = moles of Fe2O3 × 3( molar mass of CO)

mass of CO = 2.505 mol × 84 g/mol

mass of CO = 210.42 g

mass in three significant figures = 210. g

5 0

3 years ago

Read 2 more answers

Find the empirical formula of the following compounds:

Aneli [31]

The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus $\frac{mol phosphorus}{ 30.97 g phosphorus}$ = 0.0293 mol

Moles bromine 6.99 g bromine $\frac{mol bromine}{79.90 g bromine}$ =0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

To learn more about empirical formula visit:

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8 0

1 year ago

A potassium bromide solution is 8.30% potassium bromide by mass and its density is 1.03 g/ml. what mass of potassium bromide is

pshichka [43]

<u>Answer:</u> The mass of potassium bromide present in 41.2 mL of solution will be 3.522 grams.

<u>Explanation:</u>

We are given that KBr is present in 8.3% KBr solution, which means that 8.3 grams of potassium bromide is present in 100 gram of the solution.

To calculate the volume of KBr, we use the formula:

$Density=\frac{Mass}{Volume}$

Mass of the solution = 100 grams

Density of KBr solution = 1.03g/mL

Volume of the solution = ? mL

Putting values in above equation, we get:

$1.03g/mL=\frac{100g}{Volume}\\\\Volume=97.08mL$

Now, to calculate the mass of KBr in 41.2mL of the solution, we use unitary method.

In 97.08 mL of solution, mass of KBr present is 8.3 grams.

So, 41.2 mL of solution will contain = $\frac{8.3g}{97.08mL}\times 41.2mL=3.522g$ of KBr.

Hence, the mass of potassium bromide present in 41.2 mL of solution will be 3.522 grams.

7 0

3 years ago