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poizon [28]
3 years ago
6

What is the percent composition of phosphoric acid H3PO4

Chemistry
1 answer:
Agata [3.3K]3 years ago
5 0
Hey there!:

Molar mass H3PO4 = <span>97.9952 g/mol

Atomic Masses :

H = </span><span>1.00794 a.m.u

</span>P = <span>30.973762 a.m.u

</span>O =  15.9994 a.m.u<span>

H % =  [ ( 1.00794 * 3 ) / </span> 97.9952  ]  * 100

H% = <span>3.0857 %

P % = [ ( </span>30.973762 * 1 ) / 97.9952 ] * 100

P% = <span>31.6074 %

O % = [ ( </span>15.9994 * 4 ) / 97.9952 ] * 100

O% = <span>65.3069 %


Hope this helps!</span>
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2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.
azamat

Answer:

K=1.12x10^9

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

K=\frac{[NO_2]^2}{[NO]^2[O_2]}

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

4 0
3 years ago
How is acids and bases connected to Physical Science?
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8 0
3 years ago
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maks197457 [2]

Answer:

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Explanation:

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3 0
2 years ago
A gas mixture of Ne and Ar has a total pressure of 4.00 atm and contains 16.0 mol of gas. If the partial pressure of Ne is 2.75
Stolb23 [73]

Answer:

5 moles of Argon is present in the mixture.

Explanation:

Total pressure of the gaseous mixture = 4 atm

Total number of moles = 16

Partial pressure of Ne = 2.75 atm

By Dalton's law of partial pressure, the total pressure of gaseous mixture is the sum of partial pressures of individual gases which are non-reactive.

Hence:

P_{total}=P_{Ar}+P_{Ne}\\4=P_{Ar}+2.75\\P_{Ar}=1.25\ atm

Also :

Partial pressure = mole fraction*total pressure

P_{Ar}=X_{Ar}P_{total}

X_{Ar}=\frac{1.25}{4}=0.3125

\frac{n_{Ar}}{n_{total}}=0.3125\\n_{Ar}=5

∴Number of moles of Argon = 5

4 0
3 years ago
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