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lesantik [10]
3 years ago
10

How many moles of ethylene (C2H4) can react with 12.9 liters of oxygen gas at 1.2 atmospheres and 297 Kelvin? C2H4(g) + 3O2(g) y

ields 2CO2(g) + 2H2O(g) 0.21 moles 0.32 moles 0.63 moles 0.84 moles
Chemistry
1 answer:
7nadin3 [17]3 years ago
5 0
1) Convert 12.9 liters of Oxygen to mol at the given conditions:

PV = nRT ⇒ n = PV/RT

n = [1.2atm*12.9 l] / [0.082 atm l /K mol * 297K]

n = 0.636 mol of O2

2) use the stoichiometry derived from the balanced chemical equation

 1mol C2H4 / 3  mol O2 =  x mol C2H4 / 0.636 mol O2

x = 0.636 / 3 mol O2 = 0.212 mol O2.

Answer: 0.212 mol O2
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8 0
3 years ago
How many grams of carbon are present in 45.0 g of CCl4?
lisabon 2012 [21]
To determine the amount of a certain element in a compound, we use the ratio of the elements from the compound. We calculate is follows:

45.0 g CCl4 ( 1 mol CCl4 / 153.82 g CCl4 ) ( 1 mol C / 1 mol CCl4 ) ( 12.01 g C / 1 mol C ) = 3.5135 g carbon present

Hope this answers the question. Have a nice day.
5 0
3 years ago
2. Calculate the pl of the following amino acids(use their Pka values) a. Arginine b. Glutamic acid of water an c. Asparagine d.
Savatey [412]
<h2>♨ANSWER♥</h2>

pl (25*C)

Arginine -----> 10.76

Glutamic -----> 3.08

Asparagine -----> 5.43

Tyrosine -----> 5.63

<u>☆</u><u>.</u><u>.</u><u>.</u><u>hope this helps</u><u>.</u><u>.</u><u>.</u><u>☆</u>

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5 0
2 years ago
Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is t
Rasek [7]

Answer:

4.6*10^{-14} M

Explanation:

Concentration of Cu^{2+} = [Cu(NO_3)_2] = 0.020 M

Constructing an ICE table;we have:

                                 Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}

Initial (M)             0.020          0.40                        0

Change (M)         -  x                - 4 x                       x

Equilibrium (M)  0.020 -x        0.40 - 4 x              x

Given that: K_f =1.7*10^{13}

K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}

1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}

Since x is so small; 0.40 -4x = 0.40

Then:

1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}

1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}

1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x

8.704*10^9-4.352*10^{11}x =x

8.704*10^9 = 4.352*10^{11}x

x = \frac{8.704*10^9}{4.352*10^{11}}

x = 0.0199999999999540

Cu^{2+}= 0.020 - 0.019999999999954

Cu^{2+} = 4.6*10^{-14} M

8 0
3 years ago
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