BRAINIEST!

A 0.2722 g sample of a pure carbonate, X n CO 3 ( s ) , was dissolved in 50.0 mL of 0.1200 M HCl ( aq ) . The excess HCl ( aq )

Alex73 [517]

Answer:

0.00369 moles of HCl react with carbonate.

Explanation:

Number of moles of HCl present initially = $\frac{0.1200}{1000}\times 50.0$ moles = 0.00600 moles

Neutralization reaction (back titration): $NaOH+HCl\rightarrow NaCl+H_{2}O$

According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.

So, excess number of moles of HCl present = number of NaOH added for back titration = $\frac{0.0980}{1000}\times 23.60$ moles = 0.00231 moles

So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles

Hence, 0.00369 moles of HCl react with carbonate.

3 0

3 years ago

A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH

Yuki888 [10]

Answer:

1) pH = 5.05

2) pH = 5.13

3) pH = 4.97

Explanation:

Step 1: Data given

Number of moles of propionic acid = 0.18 moles

Number of moles sodium propionate = 0.26 moles

Volume = 1.20 L

Ka = 1.3 * 10^-5 → pKa = 4.989

Step 2: Calculate concentrations

Concentration = moles / volume

[acid]= 0.18/ 1.2 =0.150 M

[salt]= 0.26/ 1.3 = 0.217 M

pH = 4.89 + log(0.217/0.150)=5.05

What is the pH of the buffer after the addition of 0.02 mol of NaOH?

moles acid = 0.18 - 0.02 = 0.16

[acid]= 0.16/ 1.2=0.133 M

moles salt = 0.26 + 0.02 = 0.28

[salt]= 0.28/ 12=0.233

pH = 4.89 + log 0.233/ 0.133 = 5.13

What is the pH of the buffer after the addition of 0.02 mol of HI?

moles acid = 0.18+ 0.02 = 0.20 moles

[acid]= 0.20/ 1.2 = 0.167 M

[salt]= 0.26 - 0.02= 0.24 moles

[salt]= 0.24/ 1.2 = 0.20 M

pH = 4.89 + log 0.20/ 0.167= 4.97

8 0

3 years ago

Briefly explain the difference between hard water and soft water.

melisa1 [442]

Answer:

I HOPE THE ABOVE INFORMATION WILL HELP YOU A LOT.

HAVE A NICE DAY.

3 0

3 years ago

What is the percent composition of carbon in heptane c7h16

Gnom [1K]

Answer: the percent composition of carbon in heptane is 83.9%

Explanation:

1) Atomic masses of the atoms:

C: 12.01 g/mol

H: 1.008 g/mol

2) Molar mass of heptane:

C₇H₁₆: 7 × 12.01 g/mol + 16×1.008 g/mol = 100.2 g/mol

3) Mass of carbon in one mole of heptane:

C₇: 7 × 12.01 g/mol = 84.07 g/mol

3) Percent composition of carbon:

% = (mass in grams of C) / (mass in grams of C₇H₁₆) × 100 =

= (84.07 g/ 100.2 g) × 100 = 83.9% ← answer

8 0

3 years ago

Lauryl alcohol is a nonelectrolyte obtained from coconut oil and is used to make detergents. A solution of 6.80 g of lauryl alco

natka813 [3]

Answer:

The approximate molar mass of lauryl alcohol is 174.08 g/m

Explanation:

An excersise to apply the colligative property of Freezing-point depression.

This is the formula: ΔT = Kf . m

First of all, think the T° of fusion of benzene → 5.5°C

ΔT = T° pure solvent - T° fusion solution

Kf for benzene: 5.12 °C/m

5.5°C - 4.5°C = 5.12 °C /m . m

1°C / 5.12 m /°C = m

0.195 m = molality

This moles of lauryl alcohol, solute, are in 1 kg of benzene, solvent.

I have to find out in 0.2 kg.

1 kg sv ____ 0.195 moles solute

0.2 kg sv ____ (0.195 . 0.2)/1 = 0.039 moles solute

The mass for these moles is 6.80 g, so if I want to know the molar mass, I have to divide mass / moles

6.80 g/ 0.039 moles = 174.08 g/m

5 0

3 years ago