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Brilliant_brown [7]
3 years ago
15

What is the percent of oxygen by mass in a pure sample of fecr2o4? 1. 52% 2. 13% 3. 37% 4. 7.1% 5. 29%?

Chemistry
1 answer:
Marta_Voda [28]3 years ago
6 0
Let us suppose that we have 1 mol of FeCr2O4. I'm going to use approximate masses because I have no idea what your periodic table will say. Just put in the exact masses from your periodic table.

Fe = 56
Cr*2 = 2*52 = 104
O4 = 4*16 = 64
===========
Total = 56 + 104 + 64 = 224
The % oxygen = (64 / 224) * 100 = 28.5 % but none of your answers match this. Perhaps you are talking about Fe2(Cr2O4)3 The brackets make all the difference in the world.

Without going through all the detail I did before, The molecular mass is
Fe * 2 = 112
Cr * 6 = 312
O * 12 = 192
The total molecular mass = 616

The % Oxygen = (192 / 616) * 100 = 31% roughly. That answer isn't there either. 

Let's wait and see who else answers.

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So if there is 16 protons and 16 neutrons what would be the charge right now?​
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Be sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00−L container at 727°C, 1.30 mol
Sonbull [250]

<u>Answer:</u> The value of K_c for 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g) reaction is 5.13\times 10^2

<u>Explanation:</u>

We are given:

Initial moles of nitrogen gas = 1.30 moles

Initial moles of hydrogen gas = 1.65 moles

Equilibrium moles of ammonia = 0.100 moles

Volume of the container = 1.00 L

For the given chemical equation:

                N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

<u>Initial:</u>            1.30       1.65

<u>At eqllm:</u>       1.30-x    1.65-3x             2x

Evaluating the value of 'x'

\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol

The expression of K_c for above equation follows:

K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}

Equilibrium moles of nitrogen gas = (1.30-x)=(1.30-0.05)=1.25mol

Equilibrium moles of hydrogen gas = (1.65-x)=(1.65-0.05)=1.60mol

Putting values in above expression, we get:

K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}

Calculating the K_c' for the given chemical equation:

2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2

Hence, the value of K_c for 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g) reaction is 5.13\times 10^2

8 0
3 years ago
Read 2 more answers
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