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3 years ago

What is the molecular formula of the hydrocarbon whose molar mass is 536 g/mol and contains 89.55 w/w% carbon?

Chemistry

1 answer:

leonid [27]3 years ago

7 0

Given the percentage composition of HC as C → 81.82 % and H → 18.18 %

So the ratio of number if atoms of C and H in its molecule can will be:

C : H = 81.82 12 : 18.18 1 C : H = 6.82 : 18.18 = 6.82 6.82 : 18.18 6.82 = 1 : 2.66 ≈ 3 : 8

So the Empirical Formula of hydrocarbon is:

C 3 H 8

As the mass of one litre of hydrocarbon is same as that of C O 2 The molar mass of the HC will be same as that of C O 2 i.e 44 g mol

Now let Molecular formula of the HC be ( C 3 H 8 ) n

Using molar mass of C and H the molar mass of the HC from its molecular formula is:

( 3 × 12 + 8 × 1 ) n = 44 n So 44 n = 44 ⇒ n = 1

Hence the molecular formula of HC is C 3 H 8

Does that help?

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the half life of I -137 is 8.07 days. if 24 grams are left after 40.35 days, how many grams were in the original sample?

Ray Of Light [21]

Answer:

768g

Explanation:

We can use to formula $N(A) = N_0(\frac{1}{2})^\frac{t}{t_{1/2}}$ . Here, N(A) is the final amount. N0 is the initial amount. t is the time elapsed, and $t_{1/2}$ is the half life. Plugging in, we get the answer above.

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3 years ago

A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %

kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

$H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)$

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and $y_{C-6}$ and $y_{N_2}$ are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

$H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol$

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

$Q=-40036+(-672.21)=-40708.21J$

Finally we convert this result to kJ:

$Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ$

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6 0

2 years ago

Which term names a metamorphic change in rocks over a wide area from tectonic forces in Earth's crust and mantle? A. local metam

Pavlova-9 [17]

D is the answer. A, B,C are absurd. Doing a little word right there.

8 0

3 years ago

Read 2 more answers

A scientist measures out equal volumes and concentrations of hydrochloric acid and sodium hydroxide solutions. When mixed, these

olasank [31]

Answer:

Neutral solution is formed.

Explanation:

When the hydrochloric acid and sodium hydroxide which is a strong base are combined together, it produces sodium chloride which is a salt and water. This solution is known as Neutral solution because the solution do not have the characteristics or properties of either an acid or a base. If the concentration of one of the reactant is higher as compared to another reactant so the product has the characteristics of that reactant.

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3 years ago

Read 2 more answers

PLEASE I NEED HELP ASAP!!!!

miv72 [106K]

The percent yield of the calcium hydroxide is 84.5%.

<h3>What is stoichiometry?</h3>

Stoichiometry enables us to obtain the mass of a substance form the equation of the reaction.

The equation of the reaction is;

CaCO3 + 2HCl -----> CaCl2 + CO2 + H2O

Number of moles of X = 40.0 grams/100 g/mol = 0.4 moles

Number of moles of HCl = 850/1000 * 1 M = 0.85 moles

If 1 mole of CaCO3 reacts with 2 moles of HCl

0.4 moles of CaCO3 reacts with 0.4 moles * 2 moles/1 mole

= 0.8 moles of HCl

Hence X is the limiting reactant.

The reaction is 1:1 then the amount of CO2 produced is 0.4 moles

Mass of CO2 = 0.4 CO2 * 44 g/mol = 17.6 g

2) The reaction equation is; 2NaOH + CaCO3 ---> Ca(OH)2 + Na2CO3

Number of moles of X = 25.0 grams/100 g/mol = 0.25 moles

Number of moles of NaOH= 40/1000 L * 2 M = 0.08 moles

If 1 mole of X reacts with 2 moles of NaOH

0.25 moles reacts with 0.25 moles * 2 moles /1 mole

= 0.5 moles

NaOH is the limiting reactant

2 moles of NaOH produces 1 mole of CO2

0.08 moles of NaOH produces 0.08 moles * 1 mole/2 moles

= 0.04 moles of CO2

Theoretical yield of CO2 = 0.04 moles of CO2 * 74 g/mol = 2.96 g

Percent yield = 2.5 g/ 2.96 g * 100

= 84.5%

Learn ore about percent yield:brainly.com/question/17042787

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5 0

2 years ago