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Lena [83]
3 years ago
5

Write the full ground-state electron configuration for each element. (a) s:

Chemistry
1 answer:
DIA [1.3K]3 years ago
7 0
[Ne] 3s² 3p⁴ is the electronic configuration of sulfur in ground state
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What is the partial pressure of water vapor in an air sample when the total pressure is 1.00 atm, the partial pressure of nitrog
Inga [223]

Answer:

0.0156 atm

Explanation:

Let the partial pressure of water vapor be P_w.

Given:

Total pressure of air sample is, P_T=1.00\ atm

Partial pressure of nitrogen is, P_n=0.79\ atm

Partial pressure of oxygen is, P_o=0.19\ atm

Partial pressure of all other gases is, P_{other}=0.0044\ atm

From Dalton's law of partial pressure, we know that, the total pressure of a mixture of gases is equal to sum of partial pressure of each individual gas.

So, total pressure of air sample is equal to the sum of partial pressure of nitrogen, oxygen, other gases and water vapor.

Therefore, framing in equation form, we have:

P_n+P_o+P_{other}+P_w=P_T

Plug in the given values and solve for P_w. This gives,

0.79+0.19+0.0044+P_w=1.00\\\\0.9844+P_w=1.00\\\\P_w=1.00-0.9844\\\\P_w=0.0156\ atm

Therefore, the partial pressure of water vapor in an air sample is 0.0156 atm.

7 0
3 years ago
Which factor does not influence the rate of a chemical reaction
KonstantinChe [14]
<span>Factors that influence the reaction rates of chemical reactions include the concentration of reactants, temperature, the physical state of reactants and their dispersion, the solvent, and the presence of a catalyst.</span>
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3 years ago
Research is being carried out on cellulose as a source of chemicals for the production of fibers, coatings, and plastics. Cellul
Luden [163]

Answer:

+ 636 KJ

Explanation:

We want to arrive to the equation

C6H12O6(s) ---------> 6 H2CO(g) ΔH ° rxn = ?

by manipulating algebraically the first four  given equations.

We notice the first one has our product H2CO(g) as a reactant. This indicates we must take the inverse of that equation. Also we need 6 mol of H2CO(g), thus it also needs to be multiplied by 6

6 CO2(g) + 6 H2O(g)  ------->6H2CO(g) + 6O2(g)  ΔH °comb = + 572.9 KJ/x 6

Now we want C6H12O6(s) as a reactant and it  is a product in the second one, therefore lets reverse it

C6H12O6(s)  -------> 6 C(s) + 6 H2(g) + 3 O2(g)   ΔH ° f = + 1274.4 KJ/mol

Now if take the third equation and multiply it by six we will cancel the C(s) with the above equation

6 C(s) + 6O2(g) ---------> 6 CO2(g) ΔH ° f = - 393.5 KJ/mol x 6

Finally by multiplying the last equation by 6 and adding all the equations we will arrive at our desired one

6 H2(g) + 3 O2(g) -----------> 6H2O(g) ΔH ° f = - 285.8 KJ/mol x 6

then lets add them to get ΔH ° rxn:

  6 CO2(g) + 6 H2O(g)  ------->6H2CO(g) + 6O2(g)  ΔH °comb = + 3437.4 KJ

+ C6H12O6(s)  -------> 6 C(s) + 6 H2(g) + 3 O2(g)      ΔH ° f = + 1274.4 KJ

+ 6C(s) + 6O2(g) ---------> 6 CO2(g)                            ΔH ° f = - 2361.0 KJ

+6 H2(g) + 3 O2(g) -----------> 6H2O(g)                      ΔHº f  = - 1714.8 KJ

<u>                                                                                                                            </u>

C6H12O6(s) ---------> 6 H2CO(g)  

ΔH ° rxn =  3437.4 + 1274.4 - 2361.0 - 1714.8 =  636 KJ

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6 0
3 years ago
Read 2 more answers
Why do we need to calibrate equipment​
Vlad1618 [11]

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