Answer:
Mass stays the same but weight decreases due to a lack of gravity
In metals, some of the electrons (often one per atom) are not stuck to individual atoms but flow freely among the atoms. Of course, that's why metals are such good conductors of electricity. Now if one end of a bar is hot, and the other is cold, the electrons on the hot end have a little more thermal energy- random jiggling- than the ones on the cold end. So as the electrons wander around, they carry energy from the hot end to the cold end, which is another way of saying they conduct heat.
Here, sodium is a metal which possesses an extra (valence) electron carries the heat around its body as it is a free electron, which enables sodium to conduct thermal energy.
Hope this help :)
Answer:
A. c. Keq=[H2]^2[S2]/[H2S]^2
B. b. Keq=[COCl2]/[CO][Cl2]
Explanation:
Hello,
In this case, considering the law of mass action which states that the equilibrium expression is written in terms of the concentration of products divided by the concentration of reactants considering the stoichiometric coefficients as powers we obtain:
A. For the reaction:
The equilibrium expression is:
![Keq=\frac{[H_2]^2[S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BH_2%5D%5E2%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
Therefore, answer is c. Keq=[H2]^2[S2]/[H2S]^2.
B. For the reaction:
The equilibrium expression is:
![Keq=\frac{[COCl_2]}{[CO][Cl_2]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BCOCl_2%5D%7D%7B%5BCO%5D%5BCl_2%5D%7D)
Therefore, answer is b. Keq=[COCl2]/[CO][Cl2].
Regards.
Answer:
Volume of ammonia produced = 398.7 dm³
Explanation:
Given data:
Volume of N₂ = 200 dm³
Pressure and temperature = standard
Volume of ammonia produced = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Number of moles of N₂:
PV = nRT
1 atm× 200 L = n× 0.0821 atm.L/mol.K × 273 K
n = 200 atm.L /22.41 atm.L/mol
n = 8.9 mol
Now we will compare the moles of ammonia and nitrogen.
N₂ : NH₃
1 : 2
8.9 : 2/1×8.9 = 17.8 mol
Volume of ammonia:
1 mole of any gas occupy 22.4 dm³ volume
17.8 mol ×22.4 dm³/1 mol = 398.7 dm³
