2. How many atoms of carbon are in a 0.150 mole sample?

Determine the number of moles present in 32.5 g aluminum chloride.

Artyom0805 [142]

The number of moles in 32.5g of aluminum chloride is approximately 0.250 moles.

3 0

3 years ago

A sample of nitrogen gas occupies a volume of 2.55 L when it is at 755 mm Hg and 23 degrees Celsius. Use this information to det

ivolga24 [154]

<u>Answer:</u> The number of moles of nitrogen gas are 0.1043 moles and the pressure when volume and temperature has changed is 461.6 mmHg

<u>Explanation:</u>

To calculate the amount of nitrogen gas, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 755 mmHg

V = Volume of the gas = 2.55 L

T = Temperature of the gas = $23^oC=[23+273]K=296K$

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of nitrogen gas = ?

Putting values in above equation, we get:

$755mmHg\times 2.55L=n\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 296K\\\\n=\frac{755\times 2.55}{62.364\times 296}=0.1043mol$

To calculate the pressure when temperature and volume has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=755mmHg\\V_1=2.55mL\\T_1=23^oC=[23+273]K=296K\\P_2=?\\V_2=4.10L\\T_2=18^oC=[18+273]K=291K$

Putting values in above equation, we get:

$\frac{755mmHg\times 2.55L}{296K}=\frac{P_2\times 4.10L}{291K}\\\\P_2=\frac{755\times 2.55\times 291}{4.10\times 296}=461.6mmHg$

Hence, the number of moles of nitrogen gas are 0.1043 moles and the pressure when volume and temperature has changed is 461.6 mmHg

6 0

3 years ago

Read 2 more answers

Use Dot and cross method to show the bonding of the atoms and state the number of atoms of each element present

goldfiish [28.3K]

Answer:

Check the attached one..I only draw the Dot method of Magnesium and Fluorine.. since i love chemistry ..Hope this helps you...Balance you have to do it yourself..If you didn't get the answer let me know in the comments..But you have try okay..

5 0

2 years ago

A saturated solution of calcium hydroxide contains 1.85 g of solute in 100. mL of solution. What is its molarity

Leona [35]

Answer:

$M=0.250M$

Explanation:

Hello!

In this case, since the molarity of a solution is computed by dividing the moles of solute by the volume of solution in liters, we first need to compute the moles of solute knowing that the molar mass of calcium hydroxide is 74.1 g/mol as follows:

$n=1.85g*\frac{1mol}{74.1g}=0.0250mol$

Next, since the 100-mL solution is also expressed in liters by 0.100 L, we directly compute the molarity as shown below:

$M=\frac{0.0250mol}{0.100L}\\\\M=0.250M$

Which is expressed in molar units that are mol/L.

Best regards!

8 0

3 years ago

A 75.0 mL sample of 0.020 M acetic acid (HC2H3O2) is titrated with 0.020 M NaOH. ? Determine the pH of the solution before the a

Bad White [126]

Answer:

pH = 3.23

Explanation:

Before the addition of any NaOH, the only you have is a 0.020M acetic acid solution. That is in equilibrium with water as follows:

HC₂H₃O₂(aq) + H₂O(l) ⇄ C₂H₃O₂⁻(aq) + H₃O⁺(aq)

The Ka of this reaction is:

Ka = 1.8x10⁻⁵ = [C₂H₃O₂⁻] [H₃O⁺] / [HC₂H₃O₂]

<em>Where [] are concentrations in equilibrium of each species</em>

As you have in solution just HC₂H₃O₂, the equilibrium concentrations will be:

[HC₂H₃O₂] = 0.020M - X

[C₂H₃O₂⁻] = X

[H₃O⁺] = X

<em>Where X is reaction coordinate.</em>

<em />

Repalcing in Ka expression:

1.8x10⁻⁵ = [C₂H₃O₂⁻] [H₃O⁺] / [HC₂H₃O₂]

1.8x10⁻⁵ = [X] [X] / [0.020M - X]

3.6x10⁻⁷ - 1.8x10⁻⁵X = X²

3.6x10⁻⁷ - 1.8x10⁻⁵X - X² = 0

Solving for X:

X = -0.0006M → False solution. There is no negative concentrations

X = 0.000591M → Right solution.

As:

[H₃O⁺] = X

[H₃O⁺] = 0.000591M

As pH = -log[H₃O⁺]

6 0

3 years ago