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lilavasa [31]
3 years ago
12

If an unknown quantity of gas is held at a temperature of 789 K in a container with a volume of

Chemistry
1 answer:
Orlov [11]3 years ago
8 0

Answer:

n = 0.34 mol

Explanation:

Given data:

Given temperature = 789 K

Pressure = 765 torr ( 765/760 = 1.0 atm

Volume of container = 22.0 L

Number of moles of gas = ?

Solution:

Formula:

PV = nRT

n = PV/RT

n = 1.0 atm × 22.0 L / 0.0821 atm.L/mol.K ×789 K

n = 22.0 L.atm /64.8 atm.L/mol

n = 0.34 mol

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Nitrogen monoxide, NO, is formed in automobile exhaust by the reaction of nitrogen and oxygen in the air: N2 (g) + O2 (g) ↔ NO (
Solnce55 [7]

Answer:

The equilibrium concentration of NO is 0.001335 M

Explanation:

Step 1: Data given

The equilibrium constant Kc is 0.0025 at 2127 °C

An equilibrium mixture contains 0.023M N2 and 0.031 M O2,

Step 2: The balanced equation

N2(g) + O2(g) ↔ 2NO(g)

Step 3:  Concentration at the equilibrium

[N2] = 0.023 M

[O2] = 0.031 M

Kc = 0.0025 = [NO]² / [N2][O2]

Kc = 0.0025 = [NO]² / (0.023)(0.031)

[NO] = 0.001335 M

The equilibrium concentration of NO is 0.001335 M

6 0
3 years ago
Please help me, I really don't want to fail but I don't know how to do this
Brrunno [24]

Answer:

A)

<u>4, 7, 4, 6</u>

B)

<u>12 moles</u>

Explanation:

NH_{3}(g) + O_{2}(g) \: → NO_{2} + H_{2}O(g)

__↑______↑

8.00 mol | 14.00 mol

________________

NH_{3}(g) + O_{2}(g) \: → NO_{2} + H_{2}O(g)

You can turn this into a system of variables which are solvable.

To do this, create variables for the coefficients of each compound in the reaction respectively.

a(NH_{3}(g)) + b(O_{2}(g)) → \\c(NO_{2}) + d(H_{2}O(g))

Because to be balanced, the count of atoms in each element of the compound correspond to the coefficient of the variable in that compound so that the count of the left (reactant) side is set equal to the right (product) side.

a corresponds to the coefficient of the first compound, b corresponds to the coefficient of the second compound, c corresponds to the coefficient of the third compound, and d corresponds to the coefficient of the fourth compound.

(Reactant = Product)

Reactant: 1a [N] Product: 1c.

Reactant: 3a [H] Product: 2d.

Reactant: 2b [O] Product: 2c + 1d.

Thus the system is:

1a = 1c

3a = 2d

2b = 2c + 1d.

Then just use the substitution methods to solve.

3 0
2 years ago
λmax for the π → π* transition in ethylene is 170 nm. Is the HOMO-LUMO energy difference in ethylene greater than or less than t
-Dominant- [34]

Answer:

the HOMO-LUMO energy difference in ethylene is greater than that of cis,trans−1,3−cyclooctadiene

Explanation:

The λmax is the wavelength of maximum absorption. We could use it to calculate the HOMO-LUMO energy difference as follows:

For ethylene

E= hc/λ= 6.63×10^-34×3×10^8/170×10^-9= 1.17×10^-18J

For cis,trans−1,3−cyclooctadiene

E= hc/λ=6.63×10^-34×3×10^8/230×10^-9=8.6×10^-19J

Therefore, the HOMO-LUMO energy difference in ethylene is greater than that of cis,trans−1,3−cyclooctadiene

8 0
2 years ago
: How much energy is required to heat an iron nail with a mass of 25.5 grams from 65°C until it becomes red hot at 720°C?
Evgen [1.6K]

Q = mct  

-Q= energy in Joules  

-m = mass in grams  

-c= specific heat capacity in J/g degree C  

-t = delta temperature in degrees Celsius  

So,  

Q = m c t  

Q = (7 grams)(0.448J/g C)(750 C - 25 C)  

Q = 2273.6 J  

Your final answer = 2273.6 Joules

8 0
3 years ago
In a combustion reaction, pentane (molar mass (MM) = 72.17 g/mol) burns, in the presence of oxygen (MM = 32.00 g/mol), to produc
ladessa [460]
It is C bc I’m right anyits a free country
3 0
2 years ago
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