Answer:
C. water acts as a radiation shield to reduce the radiation level
Answer:
![[NO]=\frac{k_{-1}}{k_1} [N_2O_2]](https://tex.z-dn.net/?f=%5BNO%5D%3D%5Cfrac%7Bk_%7B-1%7D%7D%7Bk_1%7D%20%5BN_2O_2%5D)
Explanation:
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In this case, since the reaction may be assumed in chemical equilibrium, we can write up the rate law as shown below:
![r=-k_1[NO]+k_{-1}[N_2O_2]](https://tex.z-dn.net/?f=r%3D-k_1%5BNO%5D%2Bk_%7B-1%7D%5BN_2O_2%5D)
However, since the rate of reaction at equilibrium is zero, due to the fact that the concentrations remains the same, we can write:
![0=-k_1[NO]+k_{-1}[N_2O_2]](https://tex.z-dn.net/?f=0%3D-k_1%5BNO%5D%2Bk_%7B-1%7D%5BN_2O_2%5D)
Which can be also written as:
![k_1[NO]=k_{-1}[N_2O_2]](https://tex.z-dn.net/?f=k_1%5BNO%5D%3Dk_%7B-1%7D%5BN_2O_2%5D)
Then, we solve for the concentration of NO to obtain:
Best regards!
The property of water which is described in the lines above is <u>cohesion.
</u>It refers to the fact that water molecules stick to other water molecules, i. e. to themselves, as opposed to sticking to another material, which happens with adhesion. Water and capillaries don't have any connection.<u>
</u>
Answer:
1.63425 × 10^- 18 Joules.
Explanation:
We are able to solve this kind of problem, all thanks to Bohr's Model atom. With the model we can calculate the energy required to move the electron of the hydrogen atom from the 1s to the 2s orbital.
We will be using the formula in the equation (1) below;
Energy, E(n) = - Z^2 × R(H) × [1/n^2]. -------------------------------------------------(1).
Where R(H) is the Rydberg's constant having a value of 2.179 × 10^-18 Joules and Z is the atomic number= 1 for hydrogen.
Since the Electrons moved in the hydrogen atom from the 1s to the 2s orbital,then we have;
∆E= - R(H) × [1/nf^2 - 1/ni^2 ].
Where nf = 2 = final level= higher orbital, ni= initial level= lower orbital.
Therefore, ∆E= - 2.179 × 10^-18 Joules× [ 1/2^2 - 1/1^2].
= -2.179 × 10^-18 Joules × (0.25 - 1).
= - 2.179 × 10^-18 × (- 0.75).
= 1.63425 × 10^- 18 Joules.
As hot as my cooter lips
hope this helped :)
