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tia_tia [17]
2 years ago
7

Upon combustion, a 0.8009 g sample of a compound containing only carbon, hydrogen, and oxygen produces 1.6004 g co2 and 0.6551 g

h2o. determine the empirical formula of the compound.
Chemistry
1 answer:
dlinn [17]2 years ago
3 0
The compound contains Carbon, Hydroxide and Oxide
1 mole of carbon iv oxide contains 44 g, out of which 12 g are carbon.
Therefore, 1.6004 g of CO2 will contain;
    1.6004 ×12/44 = 0.4365 g of carbon
1 mole of water contains 18 g of which 2 g is hydrogen,
Therefore, 0.6551 g of H2O will hace ;
         0.6551 × 2/18 = 0.0728 g of hydrogen.
The total mass of the compound is 0.8009 g,
Thus the mass of oxygen = 0.8009 -(0.4365 +0.0728)
                                         = 0.2916 g
To get the empirical formula we first get the number of moles of each element;\
 Carbon = 0.4365/12= 0.036375 moles
Hydrogen = 0.0728/1 = 0.0728 moles
Oxygen = 0.2916/16 = 0.018225 moles
Then, to get the smallest ratio we divide each with the smallest value;
Carbon :                        Hydrogen :                    Oxygen
= (0.036375/0.018225) : (0.0728/0.018225) : ( 0.018225/0.018225)
= 1.996                         :   3.995                    : 1
≈ 2 : 4 : 1
Therefore, the empirical formula is C2H4O


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Answer:

41.3 minutes

Explanation:

Since the reaction is a first order reaction, therefore, half life is independent of the initial concentration, or in this case, pressure.

t_{1/2}= \frac{0.693}{K}

So, fraction of original pressure = \frac{1}{2}^2

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⇒ n= 3

it took 124 minutes to drop pressure to 1/8 of original value, half life = 124/3= 41.3 minutes.

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2 years ago
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COCl2(g) decomposes according to the equation above. When pure COCl2(g) is injected into a rigid, previously evacuated flask at
mestny [16]

<u>Answer:</u> The value of K_p for the reaction at 690 K is 0.05

<u>Explanation:</u>

We are given:

Initial pressure of COCl_2 = 1.0 atm

Total pressure at equilibrium = 1.2 atm

The chemical equation for the decomposition of phosgene follows:

                  COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)

Initial:            1                    -         -

At eqllm:       1-x                 x        x

We are given:

Total pressure at equilibrium = [(1 - x) + x+ x]

So, the equation becomes:

[(1 - x) + x+ x]=1.2\\\\x=0.2atm

The expression for K_p for above equation follows:

K_p=\frac{p_{CO}\times p_{Cl_2}}{p_{COCl_2}}

p_{CO}=0.2atm\\p_{Cl_2}=0.2atm\\p_{COCl_2}=(1-0.2)=0.8atm

Putting values in above equation, we get:

K_p=\frac{0.2\times 0.2}{0.8}\\\\K_p=0.05

Hence, the value of K_p for the reaction at 690 K is 0.05

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2 years ago
How many grams are there in 7.5 X10^23 molecules of H2SO4?
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<span>Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles.

 </span><span>7.5 X10^23 molecules of H2SO4 ( 1 mol / 6.022x10^23 molecules ) (98.08 g / 1 mol ) = 122.15 grams H2SO4</span>
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Ancient Egyptians used a variety of lead compounds as white pigments in their cosmetics, including PbS, PbCO3, PbCl(OH), and Pb2
ser-zykov [4K]

Answer : The compound contains the highest percentage of lead (by mass) is, PbS.

Explanation :

To calculate the percentage of lead in sample, we use the equation:

\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100

<u>For PbS :</u>

Mass of PbS = 239.3 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{239.3g}\times 100=86.58\%

The percentage of lead in the PbS is 86.58 %.

<u>For PbCO_3 :</u>

Mass of PbCO_3 = 267.2 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{267.2g}\times 100=77.55\%

The percentage of lead in the PbCO_3 is 77.55 %.

<u>For PbCl(OH) :</u>

Mass of PbCl(OH) = 259.7 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{259.7g}\times 100=79.78\%

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<u>For Pb_2Cl_2CO_3 :</u>

Mass of Pb_2Cl_2CO3_ = 545.3 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{545.3g}\times 100=37.99\%

The percentage of lead in the Pb_2Cl_2CO3_ is 37.99 %.

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2 years ago
A 0.652-g sample of a pure strontium halide reacts with excess sulfuric acid. the solid strontium sulfate formed is separated, d
seraphim [82]

Answer:

The formula of the original halide is SrCl₂.

Explanation:

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SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X is the halide.

  • From the equation stichiometry, 1.0 mole of strontium halide will result in 1.0 mole of SrSO₄.
  • The number of moles of SrSO₄ <em>(n = mass/molar mass) </em>= (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.
  • The number of moles of SrX are  4.11 x 10⁻³ moles from the stichiometry of the balanced equation.
  • n = mass / molar mass, n =  4.11 x 10⁻³ moles and mass = 0.652 g.
  • The molar mass of SrX₂ = mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.
  • The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).
  • The atomic mass of halide X = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2  g/mole = 35.5 g/mole.
  • This is the atomic mass of Cl.
  • <em>So, the formula of the original halide is SrCl₂</em>.
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