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Alex777 [14]
3 years ago
8

Nano3 dinitrogen oxide, also known as laughing gas, is a sweet-tasting gas used in dentistry. when solid ammonium nitrate is hea

ted, the products are gaseous water and gaseous dinitrogen oxide. write the balanced chemical equation for the reaction.
Chemistry
1 answer:
Alexxandr [17]3 years ago
3 0
This is a decomposition reaction, where ammonium nitrate is heated and decomposed to gaseous water and dinitrogen oxide.
Formulas for reactants and products with states are as follows;
ammonium nitrate - NH₄NO₃ (s)
water - H₂O (g)
dinitrogen oxide - N₂O (g)
balanced reactions when the masses of both sides of the equation must be balanced 

Balanced chemical equation 
NH₄NO₃(s) ---> 2H₂O(g) + N₂O(g)
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At a certain concentration of H2 and NH3, the initial rate of reaction is 0.120 M / s. What would the initial rate of the reacti
mel-nik [20]

The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

rate=k[H_2]^2[NH_3]

At a certain concentration of H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]

As we are given that:

Initial rate = 0.120 M/s

Expression for rate law for first observation:

0.120=k[H_2]^2[NH_3] ....(1)

Expression for rate law for second observation:

R=k(\frac{[H_2]}{2})^2[NH_3] ....(2)

Dividing 2 by 1, we get:

\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}

\frac{R}{0.120}=\frac{1}{4}

R=0.03M/s

Therefore, the initial rate of the reaction will be, 0.03 M/s

5 0
3 years ago
Which can associate a suspect's gun with a crime?
Anna71 [15]
FBI Fedral Burea of Investagation
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Convert 26 kg/L into g/mL
Art [367]

Answer:

26 kilograms to grams is 26000 grams.

5 0
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5. What are Opioids?
Alex_Xolod [135]
The answer to your question is the first one!
6 0
3 years ago
If a gas sample has a pressure of 30.7 kPa at 0.00*C, by how much does the temperature have to decrease to lower the pressure to
Scrat [10]

Answer:

                      252.68 K  or   -20.46 °C

Explanation:

                    According to Gay-Lussac's Law, "Pressure and Temperature at given volume are directly proportional to each other".

Mathematically,

                                              P₁ / T₁  =  P₂ / T₂   ---- (1)

Data Given:

                  P₁  =  30.7 kPa

                  T₁  =  0.00 °C  =  273.15 K

                  P₂  =  28.4 kPa

                  T₂  =  <u>???</u>

Solving equation for T₂,

                  T₂  =  P₂ T₁ / P₁

Putting values,

                  T₂  =  28.4 kPa × 273.15 K / 30.7 kPa

                  T₂  =  252.68 K  or   -20.46 °C

8 0
3 years ago
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