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Ludmilka [50]
4 years ago
10

What is the empirical formula of a compound that is 66.6% c, 11.2% h, and 22.2% o by mass?

Chemistry
1 answer:
viktelen [127]4 years ago
7 0
The empirical formula of a compound is the simplest ratio of components making up the compound.
In 100 g of compound,there's <span>66.6 g of C, 11.2 g of H and 22.2 g of O
lets calculate for 100 g of compound 
                              C                                 H                             O
mass </span>                    66.6 g                         11.2 g                      22.2 g
number of moles   66.6/12 g/mol           11.2/1 g/mol              22.2/ 16 g/mol
                               = 5.55 mol                =11.2 mol                 =1.3875 mol
 divide by the least number of moles 
                               5.55/1.3875              11.2/1.3875             1.3875/1.3875
                               = 4                             = 8.08                     = 1
round them off to the nearest whole number 
C - 4
H - 8
O - 1
Therefore empirical formula of compound is C₄H₈O
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If helium effuses through a porous barrier in 1.34 min, how much time (in min) would it take the same amount of ammonia to effus
timofeeve [1]

Answer:

The time taken the same amount of ammonia to effuse through the same barrier under the same conditions is 2.76 minutes.

Explanation:

Let the volume of the helium gas be = V

Time taken by the helium gas = t = 1.34 min

Effusion rate of helium gas = R=\frac{V}{1.34 min}

If V volume of ammonia effuse through same porous barrier the effusion rate of ammonia gas will be given as:

R'=\frac{V}{t'}

Using Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

Molar mass of helium gas = M = 4 g/mol

Molar mass of ammonia gas = M' = 17 g/mol

\frac{R}{R'}=\sqrt{\frac{M'}{M}}

\frac{\frac{V}{1.34 min}}{\frac{V}{t'}}=\sqrt{\frac{17 g/mol}{4 g/mol}}

t'=1.34 min\times \sqrt{\frac{17 g/mol}{4 g/mol}}=2.76 min

The time taken the same amount of ammonia to effuse through the same barrier under the same conditions is 2.76 minutes.

8 0
3 years ago
Which of the following molecules would be the best hydrogen bond donor?
leva [86]

Answer: Option (d) is the correct answer.

Explanation:

When a hydrogen atom comes in contact with an electronegative atom then it results in the formation of a chemical bond.

More is the electronegativity of combining atom, more stronger will be the bond with hydrogen atom. As a result, the compound formed will not easily give up hydrogen atom upon dissociation.

Whereas less is the electronegativity of atom combining with hydrogen atom, easily it will donate the hydrogen atom upon dissociation.

Since, out of the given option sulfur (S) atom has low electronegativity as compared to oxygen and nitrogen atom.

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Answer:

Number 1 is false

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3.- Gasoline has partial positive and negative charges which allows it to dissolve in water. This is true, gasoline is soluble in water.

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7 0
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NARA [144]

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brainly.com/question/11755303

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dezoksy [38]

Answer:

See detailed explanation.

Explanation:

Hello!

In this case, for the reaction:

I_2+H_2\rightleftharpoons 2HI

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b) Yes, it stops at the point in which the following expression:

\frac{[HI]^2}{[I_2][H_2]}

Equals the equilibrium constant.

Best regards!

6 0
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