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Butoxors [25]
4 years ago
13

What is the molality of a solution in which 0.32 moles AlCl3 has been dissolved in 2,200 g water? What mass of water is needed t

o prepare a 1.20 molal solution using 0.60 mol propyleneglycol? What is the molality of a solution in which 0.145 mol CO2 is dissolved in 591 g water?
Chemistry
1 answer:
klemol [59]4 years ago
8 0

Answer:

1) The molality is 0.145 molal

2) We need 0.50 kg water

3) The molality is 0.245 molal

Explanation:

What is the molality of a solution in which 0.32 moles AlCl3 has been dissolved in 2,200 g water?

Molality = moles AlCl3 / mass water (in kg)

Molality = 0.32 moles / 2.2 kg water

Molality = 0.145 moles/kg = 0.145 molal

The molality is 0.145 molal

What mass of water is needed to prepare a 1.20 molal solution using 0.60 mol propyleneglycol?

Molality = moles propyleneglycol / mass water

1.20 = 0.60 / x kg

X = 0.60 / 1.2

X = 0.50 kg water

We need 0.50 kg water

What is the molality of a solution in which 0.145 mol CO2 is dissolved in 591 g water?

Molality = moles CO2 / mass water (in kg)

Molality = 0.145 moles / 0.591 kg

Molality = 0.245 molal

The molality is 0.245 molal

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ead is a toxic metal that affects the central nervous system. A Pb-contaminated water sample contains 0.0012 %Pb by mass. How mu
Flura [38]

Answer:

1.3 × 10⁴ mL

Explanation:

A Pb-contaminated water sample contains 0.0012 %Pb by mass, that is, there are 0.0012 g of Pb in 100 g of solution. The mass of the sample that contains 150 mg (0.150 g) of Pb is:

0.150 g Pb × (100 g sample/0.0012 g Pb) = 1.25 × 10⁴ g sample

The density of the sample is 1.0 g/mL. The volume of the sample is:

1.25 × 10⁴ g × (1 mL/1.0 g) = 1.3 × 10⁴ mL

7 0
3 years ago
Based on the recipe for the Dextrose Calcium Carbonate (DCC) medium, what is the percent concentration (%weight/volume) of dextr
yanalaym [24]

<u>Answer:</u> The percent concentration ( % m/v) of dextrose in the solution is 2 %

<u>Explanation:</u>

We are given:

Mass of Yeast Extract = 5.0 grams

Mass of K_2HPO_4 = 1.0 grams

Mass of MgSO_4 = 0.5 grams

Mass of Dextrose = 20.0 grams

Mass of CaCO_3 = 10. grams

Mass of Agar = 18.0 grams

Volume of solution = 1 L = 1000 mL      (Conversion factor: 1 L = 1000 mL)

% (m/v) is defined as the concentration, which is mass of solute present in 100 mL of solution

To calculate the mass of solute, we apply unitary method:

In 1000 mL of solution, the mass of dextrose present is 20.0 grams

So, in 100 mL of solution, the mass of dextrose present will be = \frac{20.0}{1000}\times 100=2g

Calculating the % (m/v) of dextrose, we get:

\% \text{(m/v) of dextrose}=\frac{\text{Mass of dextrose}}{\text{Volume of solution}}\times 100\\\\\% \text{(m/v) of dextrose}=\frac{2g}{100mL}\times 100=2\%

Hence, the percent concentration ( % m/v) of dextrose in the solution is 2 %

4 0
4 years ago
I need help with this question
dsp73

Answer:

B?

Explanation:

it just makes the most sense in my head

7 0
3 years ago
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What are the characteristics of a good (effective) window cleaner?
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Removes smudges, leaves no after streaks

8 0
3 years ago
At what temperature will .07 moles of CL2 exert a pressure of 1.18 atm at a volume of 750 mL? ______K
Ymorist [56]

Answer:

T = 154 K

Explanation:

Given data:

Number of moles of Cl₂ = 0.07 mol

Pressure = 1.18 atm

Volume = 750 mL

Temperature = ?

Solution:

The given problem will be solve by using general gas equation, which is,

PV = nRT

R = general gas constant (0.0821 atm.L/mol.K)

Now we will convert the mL to L.

Volume = 750 mL × 1 L/1000 mL

Volume = 0.75 L

Now we will put the values in formula.

T = PV/nR

T = 1.18 atm × 0.75 L / 0.07 mol ×0.0821 atm.L/mol.K

T = 0.885/0.00575 /K

T = 154 K

5 0
3 years ago
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