True or False

The combustion of ethane ( C 2 H 6 ) produces carbon dioxide and steam. 2 C 2 H 6 ( g ) + 7 O 2 ( g ) ⟶ 4 CO 2 ( g ) + 6 H 2 O (

grigory [225]

Answer:

10.6 moles of CO₂ are produced in this combustion

Explanation:

The combustion reaction is:

2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (g)

We assume the ethane as the limiting reactant because the excersise states that the O₂ is in excess.

We make a rule of three:

2 moles of ethane can produce 4 moles of CO₂

Therefore 5.30 moles of ethane will produce (5.3 . 4) /2 = 10.6 moles

7 0

3 years ago

What is meant by chemical combination? what are the laws of chemical combination?

balu736 [363]

Answer:

In simple terms, this law states that matter can neither be created nor destroyed. In other words, the total mass, that is, the sum of the mass of reacting mixture and the products formed remains constant.

Explanation:

please give me brain list and follow

7 0

3 years ago

Wich of the following is an example of a cottonwood tree maintaining homeostasis

aleksley [76]

The answer I believe would be A

8 0

3 years ago

An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?

umka2103 [35]

Answer: The boiling point of solution is 100.53

Explanation:

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

$K_b$ = molal boiling point elevation constant = 0.51°C/m

$m_{solute}$ = Given mass of solute (CsCl) = 8.00 g

$M_{solute}$ = Molar mass of solute (CsCl) = 168.4 g/mol

$W_{solvent}$ = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC$

Hence, the boiling point of solution is 100.53

6 0

3 years ago

What mass of propane (c3h8(g)) must be burned to supply 2775 kj of heat? the standard enthalpy of combustion of propane at 298 k

Shalnov [3]

Answer is: 55.125 grams of propane must be burned.
Balanced chemical reaction: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
Make proportion: 1 mol(C₃H₈) : 2220 kJ = n(C₃H₈) : 2775kJ.
n(C₃H₈) = 2775 kJ·mol ÷ 2220 kJ.
n(C₃H₈) = 1.25 mol.
m(C₃H₈) = n(C₃H₈) · M(C₃H₈).
m(C₃H₈) = 1.25 mol · 44.1 g/mol.
m(C₃H₈) = 55.125 g.
n - amount of substance.

5 0

4 years ago