C. Temperature of the chamber. As temperature increases, pressure increases. As temp decreases, pressure decreases.
Answer:
⁹⁶₄₂Mo + ²₁H ⇒ ⁹⁸₄₃Tc
Explanation:
During the bombardment of a molybdenum target with deuterium ions, an element not found in nature was produced.The name of the element formed in the process is technetium. It is an element with the symbol Tc, mass number of 98, and atomic number 43. The nuclear equation is shown below:
⁹⁶₄₂Mo + ²₁H ⇒ ⁹⁸₄₃Tc
Technetium is the lightest element whose isotopes are all radioactive in nature except the fully ionized state of isotope ⁹⁷Tc. It is silvery gray in color and can be found between manganese and rhenium in group seven of the periodic table.
Answer:
mole fraction hexane 0.64
Explanation:
We know from Raoult's law of partial pressures that for a binary mixture, the vapor pressure above solution is the sum of the partial pressures of each gas:
P total = P(A) + P(B)
Also the partial pressure of a component is the product of its mole fraction , X, times its pure vapor pressure:
Ptotal = X (A)Pº(A) + X (B)Pº(B)
Since for solution of two components the mole fractions add to one, we now have all the information required to compute the answer.
250 torr = X(hexane)151 torr + ( 1 - X(hexane) )425 torr
250 = 151X(hexane) + 425 - 425 X(hexane)
274 X(hexane) = 175
X(hexane) = 1 -0.6 = 0.64
The answer is 13.6 moles c
Answer : The rate law for formation of NOBr based on this mechanism is, ![\frac{k_1\times k_2}{k_1^-}[Br_2][NO]^2](https://tex.z-dn.net/?f=%5Cfrac%7Bk_1%5Ctimes%20k_2%7D%7Bk_1%5E-%7D%5BBr_2%5D%5BNO%5D%5E2)
Explanation :
The overall reaction is:

Rate law = ![k[Br_2][NO]^2](https://tex.z-dn.net/?f=k%5BBr_2%5D%5BNO%5D%5E2)
The first step of the overall reaction is:


Rate law 1 = ![k_1[Br_2][NO]](https://tex.z-dn.net/?f=k_1%5BBr_2%5D%5BNO%5D)
Rate law 2 = ![k_1^-[NOBr_2]](https://tex.z-dn.net/?f=k_1%5E-%5BNOBr_2%5D)
The second step of the overall reaction is:

Rate law 3 = ![k_2[NOBr_2][NO]](https://tex.z-dn.net/?f=k_2%5BNOBr_2%5D%5BNO%5D)
Now rate law of overall reaction can be obtained as follows.
We are multiplying rate law 1 and rate law 3 and dividing by rate law 2, we get:
Rate law = ![\frac{[k_1[Br_2][NO]]\times [k_2[NOBr_2][NO]]}{[k_1^-[NOBr_2]]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bk_1%5BBr_2%5D%5BNO%5D%5D%5Ctimes%20%5Bk_2%5BNOBr_2%5D%5BNO%5D%5D%7D%7B%5Bk_1%5E-%5BNOBr_2%5D%5D%7D)
Rate law = ![\frac{k_1\times k_2}{k_1^-}[Br_2][NO]^2](https://tex.z-dn.net/?f=%5Cfrac%7Bk_1%5Ctimes%20k_2%7D%7Bk_1%5E-%7D%5BBr_2%5D%5BNO%5D%5E2)
Rate law = ![k[Br_2][NO]^2](https://tex.z-dn.net/?f=k%5BBr_2%5D%5BNO%5D%5E2)
The rate law for formation of NOBr based on this mechanism is, ![\frac{k_1\times k_2}{k_1^-}[Br_2][NO]^2](https://tex.z-dn.net/?f=%5Cfrac%7Bk_1%5Ctimes%20k_2%7D%7Bk_1%5E-%7D%5BBr_2%5D%5BNO%5D%5E2)