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Fantom [35]
3 years ago
5

How do I find the answer for question 3 and 4?

Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
3 0

Q3) 1.5 mL from the 1.00 M HCl stock solution

Q4) 0.15 mL from the 1.00 × 10⁻¹ M NaOH stock solution

Explanation:

Q3) Determine the amount of 1.00 M HCl stock solution needed to form 100 mL of 1.50 × 10⁻² M HCl solution.

Q4)  Determine the amount of 1.00 × 10⁻¹ M stock NaOH solution needed to form 100 mL of 1.50 × 10⁻² M NaOH solution.

To solve both of the questions we use the following formula:

initial concentration × initial volume =  final concentration × final volume

For Q3 we have:

initial concentration = 1.00 M HCl

initial volume = ?

final concentration =  1.50 × 10⁻² M HCl

final volume = 100 mL

1.00 × initial volume = 1.50 × 10⁻² × 100

initial volume = ( 1.50 × 10⁻² × 100) / 1.00

initial volume = 1.5 mL from the 1.00 M HCl stock solution

For Q4 we have:

initial concentration = 1.00 × 10⁻¹ M NaOH

initial volume = ?

final concentration =  1.50 × 10⁻² M NaOH

final volume = 100 mL

1.00 × 10⁻¹ × initial volume = 1.50 × 10⁻² × 100

initial volume = ( 1.50 × 10⁻² × 100) / 1.00 × 10⁻¹

initial volume = 0.15 mL from the 1.00 × 10⁻¹ M NaOH stock solution

Learn more about:

determining molar concentrations

brainly.com/question/3998372

#learnwithBrainly

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What will be the charge of an atom containing eight protons, nine neutrons, and 10 electrons?
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If 3.31 moles of argon gas occupies a volume of 100 L what volume does 13.15 moles of argon occupy under the same temperature an
kumpel [21]

Answer:

397 L

Explanation:

Recall the ideal gas law:

\displaystyle PV = nRT

If temperature and pressure stays constant, we can rearrange all constant variables onto one side of the equation:

\displaystyle \frac{P}{RT} = \frac{n}{V}

The left-hand side is simply some constant. Hence, we can write that:

\displaystyle \frac{n_1}{V_1} = \frac{n_2}{V_2}

Substitute in known values:

\displaystyle \frac{(3.31 \text{ mol})}{(100 \text{ L})}  = \frac{(13.15\text{ mol })}{V_2}

Solving for <em>V</em>₂ yields:

\displaystyle V_2 = \frac{(100 \text{ L})(13.15)}{3.31} = 397 \text{ L}

In conclusion, 13.15 moles of argon will occupy 397* L under the same temperature and pressure.

(Assuming 100 L has three significant figures.)

3 0
2 years ago
Please help with this chemistry question it’s due by midnight
Scilla [17]

Answer:

HF is the limiting reactant

Explanation:

The balanced equation for the reaction is given below:

SiO₂ + 4HF —> SiF₄ + 2H₂O

From the balanced equation above,

1 mole of SiO₂ reacted with 4 moles of HF.

Finally, we shall determine the limiting reactant. This can be obtained as illustrated below:

From the balanced equation above,

1 mole of SiO₂ reacted with 4 moles of HF.

Therefore, 7.5 moles of SiO₂ will react with = 7.5 × 4 = 30 moles of HF.

From the calculation made above, we can see clearly that it will take a higher amount (i.e 30 moles) of HF than what was given from the question (i.e 5 moles) to react completely with 7.5 moles of SiO₂.

Therefore, HF is the limiting reactant and SiO₂ is the excess reactant.

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3 years ago
Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 (
Alex

Answer:

ΔHorxn = - 11.79 KJ

Explanation:

2 SO 2 ( g ) + O 2 ( g ) ⟶ 2 SO 3 ( g )

The standard enthalpies of formation for SO 2 ( g ) and SO 3 ( g ) are Δ H ∘ f [ SO 2 ( g ) ] = − 296.8 kJ / mol Δ H ∘ f [ SO 3 ( g ) ] = − 395.7 kJ / mol

From the reaction above, 2 mol of SO2 reacts to produce 2 mol of SO3. Assuming ideal gas behaviour,

1 mol = 22.4l

x mol = 2.67l

Upon cross multiplication and solving for x;

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0.1192 mol of SO2 would react to produce 0.1192 mol of SO3.

Amount of heat is given as;

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Because O2(g) is a pure element in its standard state, ΔHοf [O2(g)] = 0 kJ/mol.

ΔHorxn = 0.1192 mol * (− 395.7 kJ / mol) - 0.1192 mol * ( − 296.8 kJ / mol)

ΔHorxn = - 47.17kj + 35.38kj

ΔHorxn = - 11.79 KJ

5 0
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