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pashok25 [27]
2 years ago
9

Find the "natural" equilibrium concentration of NO in air in units of molecules/cm3 .

Chemistry
1 answer:
Vedmedyk [2.9K]2 years ago
6 0

Answer:

6400 molecules / cm^3

Explanation:

10.6*10^-16 mol/L * 6.022*10^23 molecules/mol * 1 L / 1000 cm^3 * 1 / 100 L = 6400 molecules / cm^3

You might be interested in
Calculate the pH of a 0.30 M NaF solution. The Ka value for HF is 7.2*10^-4
Licemer1 [7]
We are given with the equilibrium constant of acid, HF and is asked to calculate the pH of 0.30 M NaF solution. The formula to be followed is 
Ka = [H+][F-]/[HF]Ka = 7.2 x 10 -4 = x^2/[0.3-x]x = [H+]= pH = -log (H+) = 1.84
5 0
3 years ago
Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O
Bond [772]

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ  

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= <em>XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)</em>

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

<em>ΔH = -793,6 kJ</em>

I hope it helps!

5 0
3 years ago
Why is there DNA in our food?
sukhopar [10]

Answer:

probably because most of our food comes from animals-

Explanation:

3 0
3 years ago
Read 2 more answers
Marissa blows up balloons for a party. She decides how big or small to make each balloon. Why does the air she blows into each b
Oxana [17]
Air is it gas so it fills its container, the balloon, completely.
5 0
3 years ago
47.9 ml hrdrogen is collected at 26° Celsius and 718 torr. Find the volume occupied at STP
Temka [501]

Answer:

41.45 mL

Explanation:

Applying the general gas equation,

PV/T = P'V'/T'............... Equation 1

Where P = Initial pressure of hydrogen, V = Initial volume of hydrogen, T= Initial Temperature of hydrogen, P' = Final pressure of hydrogen, V' = Final Volume of Hydrogen, T' = Final Temperature.

make V' the subject of the equation

V' = PVT'/TP'................ Equation 2

Given: P = 718 torr = (718×133.322) N/m² = 95725.196 N/m², V = 47.9 mL = 0.0479 dm³, T = 26 °C = (26+273) = 299 K, T' = 273 K, P' = 101000 N/m²

Substitute these values into equation 2

V' = ( 95725.196×0.0479×273)/(299×101000)

V' = 0.04145 dm³

V' = 41.45 mL

4 0
3 years ago
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