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3 years ago

A 25.00 mL sample of the ammonia solution

Chemistry

1 answer:

musickatia [10]3 years ago

3 0

Answer:

1.634 molL-1

Explanation:

The mol ration between NH3 and HCl is 1 : 1

Using Ca Va / Cb Vb = Na / Nb where a = acid and b = base

Na = 1

Nb = 1

Ca = 0.208 molL-1

Cb = ?

Va = 19.64 mL

Vb = 25.00mL

Solving for Cb

Cb = Ca Va / Vb

Cb = 0.208 * 19.64 / 25.0

Cb = 0.1634 molL-1 (Concentration of diluted ammonia solution)

Using the dilution equation;

C1V1 = C2V2

Initial Concentration, C1 = ?

Initial Volume, V1 = 25.00 mL

Final Volume, V2 = 250 mL

Final Concentration, C2 = 0.1634 molL-1

Solving for C1;

C1 = C2 * V2 / V1

C1 = 0.1634 * 250 / 25.00

C1 = 1.634 molL-1

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Read 2 more answers

Sunflower oil contains 0.080 mol palmitic acid (C16H32O2)/mol, 0.060 mol stearic acid (C18H36O2)/mol, 0.27 mol oleic acid (C18H3

Cerrena [4.2K]

Answer:

$x_H=0.882$

$x_N=0.118$

Explanation:

In this reactor, oleic and linoleic acid react with hydrogen to form stearic acid. This reactions can be represented by:

Oleic: $C_{18}H_{34}O_2 (l) + H_2 (g) \longrightarrow C_{18}H_{36}O_2 (l)$

Linoleic: $C_{18}H_{32}O_2 (l) + 2 H_2 (g) \longrightarrow C_{18}H_{36}O_2 (l)$

Having this reactions in mind, the first thing is to determine the moles of hydrogen required:

<u>Base of caculation: 1 mol of sunflower oil</u>

For oleic acid: $n_{Holeic}=\frac{1 mol H_2}{1 mol oleic}*\frac{0.27 mol oleic}{1 mol oil}*\frac{335 mol oil}{hr}$

$n_{Holeic}=frac{90.45 mol H_2}{hr}$

For linoleic acid: $n_{Hlinoleic}=\frac{2 mol H_2}{1 mol linoleic}*\frac{0.59 mol linoleic}{1 mol oil}*\frac{335 mol oil}{hr}$

$n_{Holeic}=frac{395.3 mol H_2}{hr}$

$n_{Htotal}=\frac{90.45 mol H_2}{hr}+\frac{395.3 mol H_2}{hr}$

$n_{Htotal}=frac{485.75 mol H_2}{hr}$

Applying the excess:

$n_{Htotal}=frac{485.75 mol H_2}{hr}*1.65=801.48 mol$

Nitrogen: $n_N= 801.48 mol*\frac{0.05 mol N}{0.95 mol}$

$n_N= 42.2 mol N$

<u>After the reactions</u>:

$n_H=801.48 mol-485.75mol=315.73 mol$

and the nitrogen is inert.

Purge stream:

$n_total=42.2+315.73 mol=357.93 mol$

$x_H=\frac{315.73mol}{357.93mol}=0.882$

$x_N=\frac{42.2mol}{357.93mol}=0.118$

4 0

3 years ago