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olga55 [171]
3 years ago
9

Which of the following is true about homogeneous mixtures? A. They are known as solutions

Chemistry
2 answers:
Natali [406]3 years ago
8 0

Answer: Option (A) is the correct answer.

Explanation:

A homogeneous mixture is defined as the mixture in which solute particles are evenly distributed in a solvent.

A homogeneous mixture is a clear solution.

For example, salt dissolved in water is a homogeneous mixture.

A homogeneous mixture is also known as solution as a homogeneous mixture is basically a clear solution.

Composition of the solute and solvent particles can vary with time or set of conditions.

A homogeneous mixture can be present any state.

Thus, we can conclude that the statement they are known as solutions is true about homogeneous mixtures.

Helga [31]3 years ago
7 0

Answer:

A is your answer..... !!!!!!!!!!!!!

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Which of the following transfers carbon from the nonliving environment to the living environment?
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What is the relative molecular mass (mr) of ammonia, formula NH3?
egoroff_w [7]

Answer:

APROXIMENTLY 17

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7 0
3 years ago
What is the pH of this solution?
Vesnalui [34]

Answer:

pH = 11.216.

Explanation:

Hello there!

In this case, according to the ionization of ammonia in aqueous solution:

NH_3+H_2O\rightleftharpoons NH_4^++OH^-

We can set up its equilibrium expression in terms of x as the reaction extent equal to the concentration of each product at equilibrium:

Kb=\frac{[NH_4^+][OH^-]}{[NH_3]} \\\\1.80x10^{-5}=\frac{x*x}{0.150-x}

However, since Kb<<<1 we can neglect the x on bottom and easily compute it via:

1.80x10^{-5}=\frac{x*x}{0.150}\\\\x=\sqrt{1.80x10^{-5}*0.150}=1.643x10^{-3}M

Which is also:

[OH^-]=1.643x10^{-3}M

Thereafter we can compute the pOH first:

pOH=-log(1.643x10^{-3}M)\\\\pOH=2.784

Finally, the pH turns out:

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5 0
3 years ago
A 104 m3 thoroughly mixed pond has a water inflow and outflow of 5 m3/h. The inflow water contains 0.01 mol/m3 of chemical. Chem
Naily [24]

Answer:

C ≈ 1.44 × 10⁻³ mol/m³

Explanation:

The given information are;

The liquid volume the pond can hold = 104 m³

The volume of inflow into the pond = 5 m³/h

The volume of outflow into the pond = 5 m³/h

The concentration of the chemical in the inflow water = 0.01 mol/m³

The concentration of the chemical discharged directly into the water = 0.1 mol/h

The concentration, c_{(inflow)}, of chemical that enters the water through inflow per hour is given as follows;

c_{(inflow)} = 0.01 mol/m³ × 5 m³/h = 0.05 mol/h

The concentration, c_{(discharge)}, of chemical that enters the water through direct discharge per hour is given as follows;

c_{(discharge)} = 0.1 mol/h

The total concentration that enters the pond per hour is given as follows;

c_{(inflow)} + c_{(discharge)} = 0.1 mol/h + 0.05 mol/h = 0.15 mol/h

Whereby the water in the pond properly mixes with the pond, we have;

The concentration of chemicals (C) in the outflow water = 0.15 mol/(104 m³) ≈ 0.00144 mol/m³

C ≈ 1.44 × 10⁻³ mol/m³.

4 0
3 years ago
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