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3 years ago

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The titration of a 25.00 mL sample of NaOH required 28.25 mL of a 0.200 M HCl solution to reach the end point. What was the mola

r concentration of the sodium hydroxide solution?

1 answer:

Ksju [112]3 years ago

5 0

Answer:

Mb = 0.226 M

Explanation:

This is a acid base titration between a strong acid and a strong base, therefore, this reactions is completely done, and the reaction is the following:

NaOH + HCl -------> NaCl + H2O

We have here a 1:1 mole ratio, this means that:

1 mole NaOH = 1 mole HCl

Expressing in concentration and volume:

Mb * Vb = Ma * Va

We have concentration and volume of the acid, and the volume of the base, so we can solve for the concentration:

Mb = Ma * Va / Vb

Mb = 28.25 * 0.2 / 25

Mb = 0.226 M

This is the concentration of the NaOH solution

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What is the mass in grams of Al that were reacted with excess HCl if 2.12 L of hydrogen gas were collected at STP in the followi

Inessa [10]

4.176 g of Al that was reacted with excess HCl if 2.12 L of hydrogen gas were collected at STP in the following reaction.

2 Al (s) + 6 HCl (aq) → 2 AlCl₃ (aq) + 3 H₂ (g)

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

2 Al (s) + 6 HCl (aq) → 2 AlCl₃ (aq) + 3 H₂ (g)

Given data:

Volume of $H_2$ (g) = 5.20L

At STP

Pressure= 1 atm

Temperature =273K

R = 0.0821 L.atm/mol K

PV=nRT

n= $\frac{PV}{RT}$

n= $\frac{1 atm X 5.20 L}{0.0821L.atm/mol K}$

n= 0.2324 mol $H_2$

2 Al (s) + 6 HCl (aq) → 2 AlCl₃ (aq) + 3 H₂ (g)

= $0.2324 mol H_2 X$ $\frac{2 mol Al}{3 mol H_2} X\frac{27 g Al}{1 mol Al}$

=4.176 g of Al

Hence, 4.176 g of Al that were reacted with excess HCl if 2.12 L of hydrogen gas were collected at STP in the following reaction.

Learn more about the ideal gas here:

brainly.com/question/27691721

#SPJ1

8 0

2 years ago

Read 2 more answers

A mixture containing nitrogen and hydrogen weighs 3.49 g and occupies a volume of 7.45 L at 305 K and 1.03 atm. Calculate the ma

-BARSIC- [3]

Answer:

Mass percent N₂ = 89%

Mass percent H₂ = 11%

Explanation:

First we <u>use PV=nRT to calculate n</u>, which is the total number of moles of nitrogen and hydrogen:

1.03 atm * 7.45 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305 K

n = 0.307 mol

So now we know that

MolH₂ + MolN₂ = 0.307 mol

and

MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49 g

So we have a <u>system of two equations and two unknowns</u>. We use algebra to solve it:

Express MolH₂ in terms of MolN₂:

MolH₂ + MolN₂ = 0.307 mol

MolH₂ = 0.307 - MolN₂

Replace that value in the second equation:

MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49

(0.307-MolN₂) * 2 + MolN₂ * 28 = 3.49

0.614 - 2MolN₂ + 28molN₂ = 3.49

0.614 + 26MolN₂ = 3.49

MolN₂ = 0.111 mol

Now we calculate MolH₂:

MolH₂ + MolN₂ = 0.307 mol

MolH₂ + 0.111 = 0.307

MolH₂ = 0.196 mol

Finally, we convert each of those mol numbers to mass, to <u>calculate the mass percent</u>:

N₂ ⇒ 0.111 mol * 28 g/mol = 3.108 g N₂

H₂ ⇒ 0.196 mol * 2 g/mol = 0.392 g H₂

Mass % N₂ = 3.108/3.49 * 100% = 89.05% ≅ 89%

Mass % H₂ = 0.392/3.49 * 100% = 11.15% ≅ 11%

5 0

3 years ago

If the bag was made from a metal (like sturdy aluminum foil), do you think it would lose its heat faster or slower than the plas

Iteru [2.4K]

Answer:

Fish

Explanation:

Because fish and why fish? because fish was one fish so i know this answer is 11

3 0

2 years ago

What will happened to the rock cycle without erosion

mr_godi [17]

Wind ,ice,and water eroded it over years

4 0

3 years ago

Read 2 more answers

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

sammy [17]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

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

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

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

<u>Calculation for freezing point of solution:</u>

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

<u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

5 0

3 years ago