In the petroleum industry,a process is carrie | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The chemical equation is: $C_7H_{14}(\ell) ightarrow C_7H_8(\ell) + 3H_2(g)$.Calculate the molar masses:$M(C_7H_{14}) = (7 	imes 12) + (14 	im

Vedclass · Accepted Answer

$6.1$

Vedclass · Answer

(C) The chemical equation is: $C_7H_{14}(\ell) ightarrow C_7H_8(\ell) + 3H_2(g)$.Calculate the molar masses:$M(C_7H_{14}) = (7 	imes 12) + (14 	imes 1) = 84 + 14 = 98 \ g/mol$.$M(C_7H_8) = (7 	imes 12) + (8 	imes 1) = 84 + 8 = 92 \ g/mol$.Since $C_7H_{14}$ and $C_7H_8$ are liquids,the initial weight of the liquid is $98 \ g$ and the final weight of the liquid is $92 \ g$.The decrease in the weight of the liquid is $98 \ g - 92 \ g = 6 \ g$.The percentage decrease in the weight of the liquid is calculated as:$	ext{Percentage decrease} = \frac{	ext{Decrease in weight}}{	ext{Initial weight}} 	imes 100$$	ext{Percentage decrease} = \frac{6}{98} 	imes 100 \approx 6.12\%$.

Similar Questions

The weight of a residue obtained by heating $2.76 \, g$ of silver carbonate is......$g$

The molarity of pure water is ...... .

The volume of air needed for complete conversion of $100 \ cm^3$ of $NO$ to $NO_2$ will be (assuming $20 \%$ $O_2$ by volume in air).
$2 NO_{(g)} + O_{2(g)} \rightarrow 2 NO_{2(g)}$ (in $cm^3$)

The formula of a metallic oxide is $MO$. What will be the formula of its phosphate?

How many grams of concentrated nitric acid solution should be used to prepare $250 \ mL$ of $2.0 \ M \ HNO_3$? The concentrated acid is $70\% \ HNO_3$ by mass.

Vedclass Test Series

Exam Paper Generator

Online Exam Module