All categories

1 year ago

What volume of 0.210 M sulfuric acid is required to completely react with 2.14 g aluminium hydroxide?

1 answer:

3 0

3 H2SO4 + 2 Al(OH)3 → Al2(SO4)3 + 6 H2O

(2.14 g Al(OH)3) / (78.0036 g Al(OH)3/mol) x (3 mol H2SO4 / 2 mol Al(OH)3) / (0.210 mol/L H2SO4) =
0.19596 L = 196 mL H2SO4

You might be interested in

The diagram below shows the different phase transitions that occur in matter. Which arrow represents the transition in which dew

oksian1 [2.3K]

Answer:

The arrow labeled 4: from gas to liquid.

Explanation:

Dew is a manifestation of water condensation.

The air that surrounds us contains water vapor (humidity) from the evaporatoin of the water in the rivers, lakes, and the water with which you water the plants of your garden.

During the night, and specially in the early morning, before dawn, the temperature of the air descends, and part of the vapor in the air condensates in tiny droplets that accumulate over the surface of the plant's leaves, and other solid surfaces like the winshields and hoods of the cars.

Then, the phase transition that occurs is from gas (vapor) to liquid, which is called condensation and represented with the arrow labeled 4 on the diagram.

3 0

1 year ago

Read 2 more answers

Which aqueous solution would exhibit a lower freezing point, 0.35 m k2so4 or 0.50 m kcl? explain and show necessary calculations

Umnica [9.8K]

Answer is: a lower freezing point has solution of K₂SO₄.

Change in freezing point from pure solvent to solution: ΔT =i · Kf · b.
Kf - molal freezing-point depression constant for water is 1.86°C/m.
b - molality, moles of solute per kilogram of solvent.
i - Van't Hoff factor.
b(K₂SO₄) = 0.35 m.
b(KCl) = 0.5 m.
i(K₂SO₄) = 3.
i(KCl) = 2.
ΔT(K₂SO₄) = 3 · 0.35 m · 1.86°C/m.
ΔT(K₂SO₄) = 1.953°C.
ΔT(KCl) = 2 · 0.5 m · 1.86°C/m.
ΔT(KCl) = 1.86°C.

8 0

2 years ago

Read 2 more answers

How many atoms are in a molecule of RSq?

Wewaii [24]

2 atoms in a molecule of RSq, TSq contains 3 different types of atoms (Sq, R, and T.)

6 0

2 years ago

A 600.0 mL sample of 0.20 MHF is titrated with 0.10 MNaOH. Determine the pH of the solution after the addition of 600.0 mL of Na

Leona [35]

Answer: pH=12.69

Explanation:

${\text{Moles of HF}=Molarity\times {\text{Volume of solution in liters}}$

${\text{Moles of HF}=0.20M\times 0.6L=0.12 moles$

$HF\rightarrow H^++F^-$

Initial 0.12 0 0

Eqm 0.12-x x x

$K_a=\frac{[H^+][F^-]}{[HF]}$

$3.5\times 10^{-4}=\frac{x^2}{0.12-x}$

(neglecting small value of x in comparison to 0.12)

$x=4.2\times 10^{-5}$

Moles of $H^+=4.2\times 10^{-5}$

$NaOH\rightarrow Na^++OH^-$

${\text{Molesof NaOH}}=Molarity\times {\text{Volume of solution in liters}}$

${\text{Moles of NaOH}}=0.10M\times 0.6L=0.06 moles$

0.06 moles of NaOH will give 0.06 moles of $[OH^-]$

Now $4.2\times 10^{-5}$ moles of $OH^-$ will be neutralized by $4.2\times 10^{-5}$ moles of $H^+$ and $(0.06-4.2\times 10^{-5})=0.059$ moles of $OH^-$ will be left.

Molarity of $OH^-=\frac{0.059moles}{1.2L}=0.049M$

$pOH=-\log[OH^-]=-\log[0.049]=1.31$

pH = 14 - pOH= 14 - 1.31 = 12.69

5 0

2 years ago

Which is the stronger acid in each of the following pairs? Explain your reasoning.

Makovka662 [10]

Answer:p-hydroxybenzaldehyde is stronger acid to phenol

para-cyanophenol is stronger acid to meta-cyanophenol

o-fluorophenol is stronger acid to p-fluorophenol.

Explanation:

The PKa tool relative to Ph are used to contrast the pairs.

The pKa of phenol is 10. The pKa of p-hydroxybenzaldehyde is 9.24

The pKa for meta-cyanophenol is 8.61 and the pKa for para-cyanophenol is 7.95.

The pKa value of o-fluorophenol is 8.7, while that of the p-fluorophenol is 9.9. It's obvious that the inductive effect is more dominant at ortho-position, which results in a more acidic nature

The pKa is the pH value at which a chemical species will accept or donate a proton. The lower the pKa, the stronger the acid and the greater the ability to donate a proton in aqueous solution.

6 0

2 years ago