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Vika [28.1K]
3 years ago
13

A solution is prepared by dissolving 0.23 mol of hydrazoic acid and 0.27 mol of sodium azide in water sufficient to yield 1.00 L

of solution.The addition of 0.05 mol of NaOH to this buffer solution causes the pH to increase slightly. The pH does not increase drastically because the NaOH reacts with the ________ present in the buffer solution. The Ka of hydrazoic acid is 1.9 ⋅ 10-5.
Chemistry
1 answer:
GarryVolchara [31]3 years ago
7 0

Answer:

The pH does not increase drastically because the NaOH reacts with the _HYDRAZOIC ACID__ present in the buffer solution.

Explanation:

NaOH reacts with the hydrazoic acid in the buffer

HN3 + OH? --> H2O + N3-

If you put acid in, it will react with N3-. Here is the equation below:

N3- + H+ --> HN3

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Olenka [21]
<span>To find the amount of heat energy required to raise a substance by a certain number of degrees, we simply multiply: (specific heat * change in temperature * mass) Heat = 2.22 * 35 * 3 = 233.1 joules Therefore, we would need 233.1 joules of heat to raise the temperature by 3 degrees Celsius.</span>
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1144 grams of Chromium =<br> number.<br> moles of Chromium, rounded to the nearest whole
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Explanation:

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A 2.600×10−2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created by dissolving a sample of C3H8O3 in w
sesenic [268]

Answer:

A. 0.2395 w/w %

B. 2394ppm

Explanation:

A. To find concentrationin percent by mass of the solution we need to calculate mass of glycerol and mass of water. The formula is:

Mass glycerol / Total mass * 100

<em>Mass glycerol:</em>

The solution is 2.6x10⁻²moles / L. As there is 1L of solution there are 2.6x10⁻² moles of glycerol. In mass (Using molar mass glycerol: 92.09g/mol):

2.6x10⁻² moles of glycerol * (92.09g / mol) = 2.394g glycerol

<em>Mass of water:</em>

998.9mL and density = 0.9982g/mL:

998.9mL * (0.9982g/mL) = 997.1g of water.

That means percent by mass is:

% by mass: 2.394g / (997.1g + 2.394g) * 100 = 0.2395 w/w %

B. Parts per million are mg of glycerol per L of solution. As in 1L there are 2.394g. In mg:

2.394g * (1000mg / 1g) = 2394mg:

Parts per million: 2394mg / L = 2394ppm

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3 years ago
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4 0
4 years ago
At 35°C, K = 1.6 × 10^-5 for the reaction
TEA [102]

Answer:

a) [NOCl] = 0.968 M

[NO] = 0.032M

[Cl²] = 0.016M

b) [NOCl] = 1.992M

[NO] = 0.008 M

[Cl2]  = 1.004 M

Explanation:

Step 1: Data given

Temperature = 35°C = 308K

K = 1.6 × 10^-5

Step 2: The reaction

2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)

For 2 moles NOCl we'll have 2 moles NO and 1 mol Cl2

Step 3

a. 2.0 mol pure NOCl in a 2.0 L flask

Concentration at the start:

Concentration = mol / volume

[NOCl] = mol / volume

[NOCl] = 2.0 / 2.0 L

[NOCl] = 1.0 M

[NO] = 0 M

[Cl] = 0M

Concentration at the equillibrium

[NOCl] = 1.0M - 2x

[NO] = 2x

[Cl2]= x

K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5

1.6*10^-5 = ((2x)² * x) / (1.0-2x)²

x = 0.016

[NOCl] = 1.0 -  2*0.016 = 0.968 M

[NO] = 2*0.016 = 0.032M

[Cl²] = 0.016M

b. 2.0 mol NOCl and 1.0 mol Cl2 in a 1.0 L flask

Concentration at the equillibrium

[NOCl] = 2.0 mol / 1.0 L = 2.0 M

[NO] = 0 M

[Cl2]= 1.0 mol / 1.0 L = 1.0 M

Concentration at the equillibrium

[NOCl] = 2.0M - 2x

[NO] = 2x

[Cl2]= 1.0 + x

K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5

1.6 *10^-5 = (2x)²*(1.0+x) / ((2.0-2x)²)

1.6 *10^-5= (2x)² * 1 )/2.0²

1.6 *10^-5= 4x² / 4 = x²

x = \sqrt{1.6 *10^-5} = 4.0*10^-3

[NOCl] = 2.0 - 2*0.004 = 1.992M

[NO] = 2*0.004 = 0.008 M

[Cl2] = 1+ 0.004M = 1.004 M

5 0
3 years ago
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