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3 years ago

If the rate law for the reaction A LaTeX: \longrightarrow

Chemistry

1 answer:

Nostrana [21]3 years ago

6 0

k=.240/(.4)^2=1.5 assuming [A]2 means [A] squared

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A 3.301 mass % aqueous solution of potassium hydroxide has a density of 1.05 g/mL. Calculate the molality of the solution. Give

8_murik_8 [283]

<u>Answer:</u> The molality of potassium hydroxide solution is 0.608 m

<u>Explanation:</u>

We are given:

3.301 mass % of potassium hydroxide solution.

This means that 3.301 grams of potassium hydroxide is present in 100 grams of solution

Mass of solvent = Mass of solution - Mass of solute (KOH)

Mass of solvent = (100 - 3.301) g = 96.699 g

To calculate the molality of solution, we use the equation:

$\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams})}$

Where,

$m_{solute}$ = Given mass of solute (KOH) = 3.301 g

$M_{solute}$ = Molar mass of solute (KOH) = 56.1 g/mol

$W_{solvent}$ = Mass of solvent = 96.699 g

Putting values in above equation, we get:

$\text{Molality of KOH}=\frac{3.301\times 1000}{56.1\times 96.699}\\\\\text{Molality of KOH}=0.608m$

Hence, the molality of potassium hydroxide solution is 0.608 m

4 0

3 years ago

Dissolving 5.28 g of an impure sample of calcium carbonate in hydrochloric acid produced 1.14 L of carbon dioxide at 20.0 °C and

swat32

Answer:

$\%\ mass\ of\ CaCO_3=93.37\ \%$

Explanation:

Given that:

Pressure = 791 mmHg

Temperature = 20.0°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (20 + 273.15) K = 293.15 K

T = 293.15 K

Volume = 100 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 62.3637 L.mmHg/K.mol

Applying the equation as:

791 mmHg × 1.14 L = n × 62.3637 L.mmHg/K.mol × 293.15 K

⇒n of $CO_2$ produced = 0.0493 moles

According to the reaction:-

$CaCO_3 + 2 HCl\rightarrow CaCl_2 + H_2O + CO_2$

1 mole of carbon dioxide is produced 1 mole of calcium carbonate reacts

0.0493 mole of carbon dioxide is produced 0.0493 mole of calcium carbonate reacts

Moles of calcium carbonate reacted = 0.0493 moles

Molar mass of $CaCO_3$ = 100.0869 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0493\ mol= \frac{Mass}{100.0869\ g/mol}$

$Mass_{CaCO_3}=4.93\ g$

Impure sample mass = 5.28 g

Percent mass is percentage by the mass of the compound present in the sample.

$\%\ mass\ of\ CaCO_3=\frac{Mass_{CaCO_3}}{Total\ mass}\times 100$

$\%\ mass\ of\ CaCO_3=\frac{4.93}{5.28}\times 100$

$\%\ mass\ of\ CaCO_3=93.37\ \%$

3 0

3 years ago

The number of protons in an atom is known as its atomic

Fynjy0 [20]

Atomic number should be the answer

4 0

3 years ago

12. What is the name of the compound MnF,? (Chapter 9)

nordsb [41]

Answer: D

Explanation:

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3 years ago

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Who was the first scientist to measure speed as distance over time?

777dan777 [17]

Galileo was the first scientist to measure speed as distance over time.

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3 years ago