Answer:
1+2=2+1 is the answer for the question
Answer:
C₆H₁₂O₆ + 6O₂ —> 6CO₂ + 6H₂O
Explanation:
Glucose (C₆H₁₂O₆) react with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O).
The equation can be written as follow:
C₆H₁₂O₆ + O₂ —> CO₂ + H₂O
The above equation can be balance as illustrated below:
C₆H₁₂O₆ + O₂ —> CO₂ + H₂O
There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 6 in front of CO₂ as shown below:
C₆H₁₂O₆ + O₂ —> 6CO₂ + H₂O
There are 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H₂O as shown below:
C₆H₁₂O₆ + O₂ —> 6CO₂ + 6H₂O
There are a total of 8 atoms of O on the left side and a total of 18 atoms on the right side. It can be balance by 6 in front of O₂ as shown below:
C₆H₁₂O₆ + 6O₂ —> 6CO₂ + 6H₂O
Now, the equation is balanced.
Answer:
0.52 L.
Explanation:
Let P be the initial pressure.
From the question given above, the following data were obtained:
Initial pressure (P1) = P
Initial volume (V1) = 1.04 L
Final pressure (P2) = double the initial pressure = 2P
Final volume (V2) =?
The new volume (V2) of the gas can be obtained by using the the Boyle's law equation as shown below:
P1V1 = P2V2
P × 1.04 = 2P × V2
1.04P = 2P × V2
Divide both side by 2P
V2 = 1.04P /2P
V2 = 0.52 L
Thus, the new volume of the gas is 0.52 L.
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]](https://tex.z-dn.net/?f=rate%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
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]](https://tex.z-dn.net/?f=H_2%20and%20%5Btex%5DI_2%2C%20the%20initial%20rate%20of%20reaction%20is%200.120%20M%2Fs.%20What%20would%20the%20initial%20rate%20of%20the%20reaction%20be%20if%20the%20concentration%20of%20%5Btex%5DH_2%20were%20halved.%3C%2Fp%3E%3Cp%3E%3Cstrong%3EAnswer%20%3A%20The%20initial%20rate%20of%20the%20reaction%20will%20be%2C%200.03%20M%2Fs%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3EExplanation%20%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3ERate%20law%20expression%20for%20the%20reaction%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%5Btex%5Drate%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
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]](https://tex.z-dn.net/?f=0.120%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
....(1)
Expression for rate law for second observation:
![R=k(\frac{[H_2]}{2})^2[NH_3]](https://tex.z-dn.net/?f=R%3Dk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BNH_3%5D)
....(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]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7B0.120%7D%3D%5Cfrac%7Bk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BNH_3%5D%7D%7Bk%5BH_2%5D%5E2%5BNH_3%5D%7D)
Therefore, the initial rate of the reaction will be, 0.03 M/s
