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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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Which of the following laboratory procedures best illustrates the law of conservation of mass?
fomenos

The  laboratory  procedure that best  illustrate the law of conservation  is

heating 100 g of CaCo3  to produce  56 g of CaO  (answer C)

<u><em>explanation</em></u>

According to the law  of mass conservation ,  the mass of the  reactant  must   be equal  to the mass  of the product.

 According  to  option c Heating  100 g CaCO3  to produces  56 g CaO  (  40 +16=56)

The remaining mass  = 100-56  =  44  which  would the mass of CO2  [  12 + (16 x2)]= 44   since  CaCO3  decomposes to produce CaO  and  CO2


Therefore  the  mass  of reactant=   100g

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Therefore the laboratory    procedure for  decomposition  of CaCO<em>3</em>  illustrate the law of mass  conservation since the mass of reactant = mass of product.

   


4 0
3 years ago
Read 2 more answers
A sample of nitrogen gas occupies 1.55 L at
34kurt

Answer:

The new volume after the temperature reduced to -100 °C is 0.894 L

Explanation:

Step 1: Data given

Volume of nitrogen gas = 1.55 L

Temperature = 27.0 °C = 300 K

The temperature reduces to -100 °C = 173 K

The pressure stays constant

Step 2: Calculate the new volume

V1/T1 = V2/T2

⇒with V1 = the initial volume of the gas = 1.55 L

⇒with T1 = the initial temperature = 300 K

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the reduced temperature = 173 K

1.55 L / 300 K = V2 / 173 K

V2 = (1.55L /300K) * 173 K

V2 = 0.894 L

The new volume after the temperature reduced to -100 °C is 0.894 L

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Answer:

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