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mash [69]
3 years ago
9

A hydrocarbon is a compound that contains mostly carbon and hydrogen. Calculate the percent composition (by mass) of the followi

ng hydrocarbon: C5H12. Enter the percentages of carbon and hydrogen numerically to four significant figures, separated by commas. View Available Hint(s)
Chemistry
1 answer:
Shalnov [3]3 years ago
5 0

Answer: The percentages of carbon and hydrogen numerically to four significant figures is 83.33\% and 16.67\% respectively.

Explanation:

Molar mass of the compound is the mass of 1 mole of compound which is the sum of masses of each element.

Mass of 1 mole of compound= mass of carbon + mass of hydrogen+ = 5(12) + 12(1)= 72 g.

a) Percentage by mass of carbon =\frac{\text {mass of carbon}}{\text {Total mass}}\times 100\ %

Percentage by mass of carbon =\frac{60}{72}\times 100\%=83.33\%

b) Percentage by mass of hydrogen =\frac{\text {mass of hydrogen}}{\text {Total mass}}\times 100\ %

Percentage by mass of hydrogen =\frac{12}{72}\times 100\%=16.67\%

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If 8.500 g CH is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final
mojhsa [17]

The final temperature = 36 °C

<h3>Further explanation</h3>

The balanced combustion reaction for C₆H₆

2C₆H₆(l)+15O₂(g)⇒ 12CO₂(g)+6H₂O(l)  +6542 kJ

MW C₆H₆ : 78.11 g/mol

mol C₆H₆ :

\tt \dfrac{8.5}{78.11}=0.109

Heat released for 2 mol C₆H₆ =6542 kJ, so for 1 mol

\tt \dfrac{0.109}{2}\times 6542=356.539~kJ/mol

Heat transferred to water :

Q=m.c.ΔT

\tt 356.539=5.691~kg\times 4.18~kj/kg^oC\times (t_2-21)\\\\t_2-21=15\rightarrow t_2=36^oC

3 0
2 years ago
I need y’all’s help please
nikitadnepr [17]

Answer:

6 mass

Explanation:

because the si unit of gram is mass

7 0
2 years ago
Complete the charge balance equation for an aqueous solution of h2co3 that ionizes to hco−3 and co2−3.
Zielflug [23.3K]

The charge balance equation for an aqueous solution of H₂CO₃ that ionizes to HCO₃⁻ and CO₃⁻² is [HCO₃⁻] =  2[CO₃⁻²] + [H⁺] + [OH⁻]

<h3>What is Balanced Chemical Equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

The equation for aqueous solution of H₂CO₃ is

H₂CO₃ → H₂O + CO₂

The charge balance equation is

[HCO₃⁻] =  2[CO₃⁻²] + [H⁺] + [OH⁻]

Thus from the above conclusion we can say that The charge balance equation for an aqueous solution of H₂CO₃ that ionizes to HCO₃⁻ and CO₃⁻² is [HCO₃⁻] =  2[CO₃⁻²] + [H⁺] + [OH⁻]

Learn more about the Balanced Chemical equation here: brainly.com/question/26694427
#SPJ4

8 0
2 years ago
Charles’ Law can be expressed mathematically as V=kT. If the constant k for 6.0 L of gas is 0.020 L/K (liters per kelvin), what
kozerog [31]

Answer: 300 K

Explanation:

Charles' Law: This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

V\propto T    (At constant pressure and number of moles)

V=kT

Given : V= 6.0 L

k= 0.020 L/K

T=?

6.0=0.020LK^{-1}\times T

T=300 K

Thus temperature of the gas is 300 K.

3 0
3 years ago
The essential oil found in cloves, eugenol, can be isolated by steam distillation because it is insoluble in water and has a mea
Damm [24]
We know that:
Molar Mass H2O: 18 g/mol 
<span>Molar Mass of Eugenol: 164 g/mol </span>
<span>Boiling point of H2O: 100 degrees C </span>
<span>Boiling point of Eugenol: 254 degrees C </span>
<span>Density of water: 1.0 g/mL </span>
<span>Density of Eugenol: 1.05 g/mL </span>

<span>Using formula:
V= [mole fraction x molar mass] / density </span>

<span>mH20: 0.9947 * 18
          = 17.9046 / 1 g/mL
          = 17.9046 </span>
<span>morg: 0.0053 * 164  
        = 0.8692/ 1.05 g/mL
        = 0.8278 </span>

<span>V% = Vorg/(Vorg + VH2O) * 100 </span>
<span>(0.8278/18.7324) * 100 = 4.419% </span>

Yotal volume = 30 mL; therefore, 
<span>0.0442 = (volume eugenol/30) </span>

<span>(m eug/mH2O) = (peug*164/pH2O*18) </span>
<span>(m eug/30) = (4*164/760*18) </span>
<span>m eug = about 1.44g and </span>
<span>
volume = mass/density
            = 1.44/1.05
            = about 1.37 mL </span>
6 0
3 years ago
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