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zhannawk [14.2K]
2 years ago
15

Calculate the pH during the titration of 20.00 mL of 0.1000 M dimethylamine, (CH3)2NH(aq), with 0.1000 M HCl(aq) after 21.23 mL

of the acid have been added.
Chemistry
1 answer:
Evgesh-ka [11]2 years ago
8 0

The pH value of the solution is mathematically given as

pH=2.35

<h3>What pH value of the solution?</h3>

Question Parameters:

pH during the titration of 20.00 mL of 0.1000 M dimethylamine,

with 0.1000 M HCl(aq) after 21.23 mL of the acid

Generally, the equation for the  Chemical Reaction  is mathematically given as

(CH3)2NH(aq), +Hcl   ---> <---- (CH3)2NH2Cl(aq)

Therefore

HCl=\frac{0.186mol}{41.86}

HCL=0.00444M

WHere

HClaq--->H+(aq)+Cl-(aq)

Hence

H+=0.00444M

pH= -log{H+}

pH=log(0.00444)

pH=2.35

For more information on Chemical Reaction

brainly.com/question/11231920

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1 year ago
Entropy is greater at higher trophic levels than at lower levels.. True of False
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8 0
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Read 2 more answers
The air inside a bicycle tire pump has 27 joules of heat conducted away from while 77.9 joules of work done are being done on it
tatyana61 [14]

Answer:

51 J

Explanation:

The air inside a bicycle tire pump has 27 joules of heat conducted away. By convention, when heat is released, it takes the negative sign, so Q = -27 J.

77.9 joules of work done are being done on the air inside a bicycle tire pump. By convention, when work is being done on the system, it takes the positive sign, so W = 77.9 J

We can calculate the change in the internal energy (ΔU) using the following expression.

ΔU = Q + W

ΔU = (-27 J) + 77.9 J

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7 0
3 years ago
A compound is found to be 30.45% n and 69.55 % o by mass. if 1.63 g of this compound occupy 389 ml at 0.00°c and 775 mm hg, what
Charra [1.4K]
1) mass composition

N: 30.45%
O: 69.55%
   -----------
   100.00%

2) molar composition

Divide each element by its atomic mass

N: 30.45 / 14.00 = 2.175 mol

O: 69.55 / 16.00 = 4.346875

4) Find the smallest molar proportion

Divide both by the smaller number

N: 2.175 / 2.175 = 1

O: 4.346875 / 2.175 = 1.999 = 2

5) Empirical formula: NO2

6) mass of the empirical formula

14.00 + 2 * 16.00 = 46.00 g

7) Find the number of moles of the gas using the equation pV = nRT

=> n = pV / RT = (775/760) atm * 0.389 l / (0.0821 atm*l /K*mol * 273.15K)

=> n = 0.01769 moles

8) Find molar mass

molar mass = mass in grams / number of moles = 1.63 g / 0.01769 mol = 92.14 g / mol

9) Find how many times the mass of the empirical formula is contained in the molar mass

92.14 / 46.00 = 2.00

10) Multiply the subscripts of the empirical formula by the number found in the previous step

=> N2O4

Answer: N2O4
3 0
3 years ago
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