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3 years ago

Predict the moon phases

Chemistry

1 answer:

Afina-wow [57]3 years ago

7 0

Moon phases of the world is the best way to get to the next level

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High concentrations of ammonia (NH3), nitrite ion, and nitrate ion in water can kill fish. Lethal concentrations of these specie

kykrilka [37]

Explanation:

It is known that molality is the number of moles present in kg of solution.

Mathematically, Molality = $\frac{\text{no. of moles of solute}}{\text{mass of solvent in Kg}}$

The given data is as follows.

Molar mass of ammonia = 17 g/mol

Concentration = 1.002 mg/L = $\frac{0.001002 g/L}{17 g/mol}$

= $5.89 \times 10^{-4}$ mol/L

Also, density = $\frac{1 g}{mL}$ = 1 kg/L

Therefore, molality will be calculated as follows.

Molality = $\frac{5.89 \times 10^{-4} mol/L}{1 kg/L}$

= $5.89 \times 10^{-4} mol/kg$

And,

Molar mass of nitrite = 46 g/mol

Concentration = 0.387 mg/L = $\frac{0.000412 g/L}{46 g/mol}$

= $8.956 \times 10^{-6}$ mol/L

And, density = $\frac{1 g}{mL}$ = 1 kg/L

Hence, molality = $\frac{8.956 \times 10^{-6} mol/L}{1 kg/L}$

= $8.956 \times 10^{-6}$ mol/kg

Now, Molar mass of nitarte = 62 g/mol

Concentration = 1352.2 mg/L

= $\frac{1.3522 g/L}{62 g/mol}$

= 0.02181 mol/L

Also, density = $\frac{1 g}{mL}$ = 1 kg/L

Hence, molality will be calculated as follows.

Molality = $\frac{0.02181 mol/L}{1 kg/L}$

= 0.02181 mol/kg

Therefore, molality of given species is $5.89 \times 10^{-4} mol/kg$ for ammonia, $8.956 \times 10^{-6}$ mol/kg for nitrite, and 0.02181 mol/kg for nitrate ion.

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4 years ago

How does energy change form when friction causes the temperature of an object to increase?

Oksanka [162]

I believe the answer would be through the movement of the molecules, the higher the temperature the more the molecules move around and vibrate.

7 0

3 years ago

Which is an example of flow of heat through conduction

JulsSmile [24]

Answer:

a metal spoon left in boiling water

Explanation:

5 0

3 years ago

Save the ________ used to extract metals.

MrRa [10]

Answer:

Save the *minerals* used to extract metals.

6 0

2 years ago

The rate constant for the reaction 3A equals 4b is 6.00×10 how long will it take the concentration of a to drop from 0.75 to 0.2

Lelu [443]

This question is incomplete, the complete question is;

The rate constant for the reaction 3A equals 4B is 6.00 × 10⁻³ L.mol⁻¹min⁻¹.

how long will it take the concentration of A to drop from 0.75 to 0.25M ?

from the unit of the rate constant we know it is a second reaction order

OPTIONS

a) 2.2×10^−3 min

b) 5.5×10^−3 min

c) 180 min

d) 440 min

e) 5.0×10^2 min

Answer:

it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

Explanation:

Given that;

Rate constant K = 6.00 × 10⁻³ L.mol⁻¹min⁻¹

3A → 4B

given that it is a second reaction order;

k = 1/t [ 1/A - 1/A₀]

kt = [ 1/A - 1/A₀]

t = [ 1/A - 1/A₀] / k

K is the rate constant(6.00 × 10⁻³)

A₀ is initial concentration( 0.75 )

A is final concentration(0.25)

t is time required = ?

so we substitute our values into the equation

t = [ (1/0.25) - (1/0.75)] / (6.00 × 10⁻³)

t = 2.6666 / (6.00 × 10⁻³)

t = 444.34 ≈ 440 min {significant figures}

Therefore it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

6 0

3 years ago