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sasho [114]
3 years ago
10

Write the net ionic reaction that occurs upon the addition of hno3 to a solution which contains methylamine (ch3nh2) and methyla

mmonium chloride (ch3nh3cl). (be sure to include states of matter.)

Chemistry
2 answers:
pychu [463]3 years ago
5 0

equation solution is attached

ra1l [238]3 years ago
4 0
Hello!

First, HNO₃ dissociates in the following way:

HNO₃(aq) → H⁺(aq) + NO₃⁻(aq)

Next, there is a neutralization reaction with Methylamine:

H⁺(aq) + CH₃NH₂(aq) → CH₃NH₃⁺(aq)

Finally, the formed methylammonium ion weakly dissociates in the following way:

CH₃NH₃⁺(aq) + H₂O(l)  ⇄ H₃O⁺(aq) + CH₃NH₂ (aq)

From this sequence of reactions, we can see that the resulting solution would be slightly more acidic than the initial one, as the buffer solution resists the addition of strong acid.

Have a nice day!
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3 years ago
A gas has a volume of 300 mL in a rigid container at 50oC and 1.75 atm. What will be its pressure at 100K?
eduard

Answer:

Its pressure will be 0.54 atm at 100 K.

Explanation:

Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.

Gay-Lussac's law can be expressed mathematically as the quotient between pressure and temperature equal to a constant:

\frac{P}{T} =k

Studying two different states, an initial state 1 and a final state 2, it is satisfied:

\frac{P1}{T1} =\frac{P2}{T2}

In this case:

  • P1= 1.75 atm
  • T1= 50 °C= 323 K (being 0 C=273 K)
  • P2= ?
  • T2= 100 K

Replacing:

\frac{1.75 atm}{323 K} =\frac{P2}{100 K}

Solving:

P2= 100 k*\frac{1.75 atm}{323 K}

P2= 0.54 atm

<u><em>Its pressure will be 0.54 atm at 100 K.</em></u>

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7 0
3 years ago
I need help on this one for an exam real quick please​
gogolik [260]

Answer:

See Explanation

Explanation:

In the first case, when potassium iodide is added to an aqueous bromine solution, a chemical reaction occurs as follows;

Br2(l) + 2 KI(aq) = 2 KBr(aq) + I2(l)

This reaction produces iodine solution which is brown in colour.

Secondly, when potassium iodide is added to aqueous chlorine solution, the following reaction occurs;

2KI(aq) + Cl2(l)→ 2KCl(aq) + I2 (l)

This reaction also yields iodine solution which is brown in colour.

KI(aq) + I2(l) -------->K^+(aq) + I3^-(aq)

The  I3^-(aq) solution appears brown at high concentrations.

8 0
3 years ago
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