Answer:
https://www.quora.com/The-following-chemical-reaction-shows-the-decomposition-of-water-to-form-hydrogen-gas-and-oxygen-gas-2H2O-I-produces-2H2-g-O2-g-if-10-0-grams-of-water-reacted-and-you-found-1-11-grams-of-H2-formed-how-many-of-O2
Explanation: i hope ths helps you
We can write the balanced equation for the synthesis reaction as
H2(g) + Cl2(g) → 2HCl(g)
We use the molar masses of hydrogen chloride gas HCl and hydrogen gas H2 to calculate for the mass of hydrogen gas H2 needed:
mass of H2 = 146.4 g HCl *(1 mol HCl / 36.46 g HCl) * (1 mol H2 / 2 mol HCl) *
(2.02 g H2 / 1 mol H2)
= 4.056 g H2
We also use the molar masses of hydrogen chloride gas HCl and chlorine gas CL2 to calculate for the mass of hydrogen gas H2:
mass of CL2 = 146.4 g HCl *(1 mol HCl / 36.46 g HCl) * (1 mol Cl2 / 2 mol HCl) *
(70.91 g Cl2 / 1 mol Cl2)
= 142.4 g Cl2
Therefore, we need 4.056 grams of hydrogen gas and 142.4 grams of chlorine gas to produce 146.4 grams of hydrogen chloride gas.
Answer:

Explanation:
Hello,
In this case, by knowing the given reference reactions, one could rearrange them as follows:


Subsequently, to obtain the main reaction, we add the aforementioned reference rearranged reactions as shown below (just as reference):

Consequently, the equilibrium constant is computed as:
![Kp=\frac{[N_2][O_2]}{[NO]^2} * \frac{[NO_2]^2}{[N_2][O_2]^2} =Kp_2*Kp_3=4.35x10^{18}*7.056x10^{-13}=3.07x10^6](https://tex.z-dn.net/?f=Kp%3D%5Cfrac%7B%5BN_2%5D%5BO_2%5D%7D%7B%5BNO%5D%5E2%7D%20%2A%20%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2%5D%5BO_2%5D%5E2%7D%20%3DKp_2%2AKp_3%3D4.35x10%5E%7B18%7D%2A7.056x10%5E%7B-13%7D%3D3.07x10%5E6)
Best regards.
Answer:
0.9715 Fraction of Pu-239 will be remain after 1000 years.
Explanation:


Where:
= decay constant
=concentration left after time t
= Half life of the sample
Half life of Pu-239 =
[
![\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7B0.693%7D%7B24%2C000%20y%7D%3D2.8875%5Ctimes%2010%5E%7B-5%7D%20y%5E%7B-1%5D)
Let us say amount present of Pu-239 today = 
A = ?
![A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}](https://tex.z-dn.net/?f=A%3Dx%5Ctimes%20e%5E%7B-2.8875%5Ctimes%2010%5E%7B-5%7D%20y%5E%7B-1%5D%5Ctimes%201000%20y%7D)


0.9715 Fraction of Pu-239 will be remain after 1000 years.
<span>thermal energy
hope this helped</span>