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3 years ago

Describe a method to separate the substances in a solution.

Chemistry

2 answers:

AveGali [126]3 years ago

7 0

Answer:

1. Chromatography involves solvent separation on a solid medium.

2. Distillation takes advantage of differences in boiling points.

3. Evaporation removes a liquid from a solution to leave a solid material.

4. Filtration separates solids of different sizes.

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zhenek [66]3 years ago

6 0

Answer:

By using different techniques.

Explanation:

For example: Chromatography involves solvent separation on a solid medium. Distillation takes advantage of differences in boiling points. Evaporation removes a liquid from a solution to leave a solid material. Filtration separates solids of different sizes.

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An ostrich runs north in a straight line for 21 meters. She then turns around and runs 25 meters back toward her nest (south) on

drek231 [11]

The ostrich runs 21 N then goes back by 25 S

The displacement is -4 S

21 - 25 = -4

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3 years ago

Identify the control group the experimental group the independent variable and the dependent variable

denis-greek [22]

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Independent variable: dog food
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3 years ago

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

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3 years ago

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VARVARA [1.3K]

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3 years ago

2N2H4+ N2O4———3N2+4H2O SalmaKhan99 avatar How many grams of N2 gas will be formed by reacting 100g of N2H4 and 200g of N2 Kindly

Alona [7]

Answer:

131.26 g

Explanation:

From the balanced equation,

2 moles of N₂H₄reacts with 1 mole of N₂O₄ to give 3 moles of N₂

Now number of moles of N₂H₄ present in 100 g N₂H₄ is n = 100 g/molar mass N₂H₄.

Molar mass N₂H₄ = 2 × 14.01 g/mol + 1 × 4 g/mol = 28.02 g/mol + 4 g/mol = 32.02 g/mol

n₁ = 100/32.02 = 3.123 mol

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Now number of moles of N₂O₄ present in 200 g N₂O₄ is n = 200 g/molar mass N₂O₄.

Molar mass N₂O₄ = 2 × 14.01 g/mol + 16 × 4 g/mol = 28.02 g/mol + 64 g/mol = 92.02 g/mol

n₂ = 200/92.02 = 2.173 mol

Since the mole ratio of N₂H₄ to N₂O₄ is 2 : 1, We require 2 × 2.173 mol N₂H₄ to react with 2.173 mole N₂O₄

Number of moles of N₂H₄ required is 4.346. But the number of moles of N₂H₄ present is 3.123 so N₂H₄ is the limiting reagent.

So, from the equation, 2 moles of N₂H₄ produces 3 moles of N₂

Therefore number of mole N₂ = 3/2 moles of N₂H₄ = 3/2 × 3.123 mol = 4.6845 mol

From n = m/M where n = number of moles of nitrogen gas = 4.6845 mol and M = molar mass of nitrogen gas = 28.02 g/mol and m = mass of nitogen gas.

m = nM = 4.6845 mol × 28.02 g/mol = 131.26 g

So the mas of nitrogen gas produced is 131.26 g

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3 years ago