Answer:
This reaction can be expressed as a word equation: copper + silver nitrate → silver + copper nitrate You can replace each part of the word equation by a chemical formula: Cu + AgNO3 → Ag + Cu(NO3)2 Looking at both sides of this equation, you can see that it is not balanced.
Explanation:
a) blue litmus paper turns red when u add acid in it .
b) red litmus paper , when an alkali dissolves in water it produce hydroxide ions
which cause it to become a alkaline
Answer:
A. ![r=k[NO]^2[H_2]](https://tex.z-dn.net/?f=r%3Dk%5BNO%5D%5E2%5BH_2%5D)
B. 
C. Second-order with respect to NO and first-order with respect to H₂
Explanation:
Hello,
In this case, for the reaction:
The rate law is determined by writing the following hypothetical rate laws:
![3.822x10^{-3}=k[0.3]^m[0.35]^n\\\\1.529x10^{-2}=k[0.6]^m[0.35]^n\\\\3.058x10^{-2}=k[0.6]^m[0.7]^n](https://tex.z-dn.net/?f=3.822x10%5E%7B-3%7D%3Dk%5B0.3%5D%5Em%5B0.35%5D%5En%5C%5C%5C%5C1.529x10%5E%7B-2%7D%3Dk%5B0.6%5D%5Em%5B0.35%5D%5En%5C%5C%5C%5C3.058x10%5E%7B-2%7D%3Dk%5B0.6%5D%5Em%5B0.7%5D%5En)
Whereas we can compute m as follows:
![\frac{3.822x10^{-3}}{1.529x10^{-2}} =\frac{[0.3]^m[0.35]^n}{[0.6]^m[0.35]^n} \\\\0.25=(0.5)^m\\\\m=\frac{log(0.25)}{log(0.5)} \\\\m=2](https://tex.z-dn.net/?f=%5Cfrac%7B3.822x10%5E%7B-3%7D%7D%7B1.529x10%5E%7B-2%7D%7D%20%3D%5Cfrac%7B%5B0.3%5D%5Em%5B0.35%5D%5En%7D%7B%5B0.6%5D%5Em%5B0.35%5D%5En%7D%20%5C%5C%5C%5C0.25%3D%280.5%29%5Em%5C%5C%5C%5Cm%3D%5Cfrac%7Blog%280.25%29%7D%7Blog%280.5%29%7D%20%5C%5C%5C%5Cm%3D2)
Therefore, the reaction is second-order with respect to NO. Thus, for hydrogen, we find n:
![\frac{1.529x10^{-2}}{3.058x10^{-2}} =\frac{[0.6]^2[0.35]^n}{[0.6]^2[0.7]^n} \\\\0.5=(0.5)^n\\\\n=\frac{log(0.5)}{log(0.5)}\\ \\n=1](https://tex.z-dn.net/?f=%5Cfrac%7B1.529x10%5E%7B-2%7D%7D%7B3.058x10%5E%7B-2%7D%7D%20%3D%5Cfrac%7B%5B0.6%5D%5E2%5B0.35%5D%5En%7D%7B%5B0.6%5D%5E2%5B0.7%5D%5En%7D%20%5C%5C%5C%5C0.5%3D%280.5%29%5En%5C%5C%5C%5Cn%3D%5Cfrac%7Blog%280.5%29%7D%7Blog%280.5%29%7D%5C%5C%20%5C%5Cn%3D1)
A) Therefore, the reaction is first-order with respect to H₂. In such a way, we conclude that that the rate law is:
B) Rate constant is computed from one kinetic data:
C. As mentioned before, reaction is second-order with respect to NO and first-order with respect to H₂.
Best regards.
Answer:
True becouse any graphs can show info
Answer:
0.200 M NaOH: 13.3 mL of the concentrated solution and complete to 20.0mL
0.150 M NaOH: 10.0 mL of the concentrated solution and complete to 20.0mL
0.100 M NaOH: 6.67 mL of the concentrated solution and complete to 20.0mL
0.050 M NaOH: 3.33 mL of the concentrated solution and complete to 20.0mL
0.025 M NaOH: 1.67 mL of the concentrated solution and complete to 20.0mL
Explanation:
It is possible to prepare a solution from a more concentrated one. In the problem, the concentrated solution is 0.300M NaOH. Thus, to prepare 20.0mL of each of the solutions you will need:
0.200 M NaOH: 20.0mL × (0.200M / 0.300M) = <em>13.3 mL of the concentrated solution and complete to 20.0mL</em>
<em>The ratio between the concentrated solution and the solution you want to prepare is called "dilution factor"</em>
<em />
0.150 M NaOH: 20.0mL × (0.150M / 0.300M) = <em>10.0 mL of the concentrated solution and complete to 20.0mL</em>
<em />
0.100 M NaOH: 20.0mL × (0.100M / 0.300M) = <em>6.67 mL of the concentrated solution and complete to 20.0mL</em>
<em />
0.050 M NaOH: 20.0mL × (0.050M / 0.300M) = <em>3.33 mL of the concentrated solution and complete to 20.0mL</em>
<em />
0.025 M NaOH: 20.0mL × (0.025M / 0.300M) = <em>1.67 mL of the concentrated solution and complete to 20.0mL</em>
