The solution to the problem is as follows:

a)

C + O2 = CO2

Molar mass CO2 = 44 g/mol

3.67 g CO2 * 1 mol / 44 g =

=0.0834 mol CO2 = 0.0834 mol C

I hope my answer has come to your help. God bless and have a nice day ahead!

4 0

3 years ago

Calculate the density of a 20.015 g object that occupies 5.44 cm3 .

Scilla [17]

Answer:

$density = \frac{mass}{volume} \\ \\ density = \frac{20.015}{5.44} \\ \\ { \boxed{ \boxed{density = 3.679 \: g {cm}^{ - 3} }}}$

5 0

2 years ago

Read 2 more answers

Assuming an efficiency of 30.80%, calculate the actual yield of magnesium nitrate formed from 147.4g of magnesium and excess cop

Reil [10]

Answer:

The answer to your question is 280 g of Mg(NO₃)₂

Explanation:

Data

Efficiency = 30.80 %

Mg(NO₃)₂ = ?

Magnesium = 147.4 g

Copper (II) nitrate = excess

Balanced Reaction

Mg + Cu(NO₃)₂ ⇒ Mg(NO₃)₂ + Cu

Reactants Elements Products

1 Mg 1

1 Cu 1

2 N 2

6 O 6

Process

1.- Calculate the theoretical yield

Molecular weight Mg = 24

Molecular weight Mg(NO₃)₂ = 24 + (14 x 2) + (16 x 6)

= 24 + 28 + 96

= 148 g

24 g of Mg -------------------- 148 g of Mg(NO₃)₂

147.4 g of Mg ------------------- x

x = (147.4 x 148) / 24

x = 908.96 g of Mg(NO₃)₂

2.- Calculate the Actual yield

yield percent = $\frac{actual yield}{theoretical yield}$

Solve for actual yield

Actual yield = Yield percent x Theoretical yield

Substitution

Actual yield = $\frac{30.8}{100}$ x 908.96

Actual yield = 279.95 ≈ 280g

4 0

3 years ago