If 4.9 kg of CO2 are produced during a combustion reaction, how many molecules of CO2 would be produced?
1 answer:
Answer:
6.7 x 10²⁶molecules
Explanation:
Given parameters
Mass of CO₂ = 4.9kg = 4900g
Unknown:
Number of molecules = ?
Solution:
To find the number of molecules, we need to find the number of moles first.
Number of moles =
Molar mass of CO₂ = 12 + 2(16) = 44g/mol
Number of moles = = 111.36mole
A mole of substance is the quantity of substance that contains the avogadro's number of particles.
1 mole = 6.02 x 10²³molecules
111.36 moles = 111.36 x 6.02 x 10²³molecules = 6.7 x 10²⁶molecules
You might be interested in
Answer:
The heat needed to warm 25.3 g of copper from 22°C to 39°C is 165.59 Joules.
Explanation:
Where:
Q = heat absorbed or heat lost
c = specific heat of substance
m = Mass of the substance
ΔT = change in temperature of the substance
We have mass of copper = m = 25.3 g
Specific heat of copper = c = 0.385 J/g°C
ΔT = 39°C - 22°C = 17°C
Heat absorbed by the copper :
The heat needed to warm 25.3 g of copper from 22°C to 39°C is 165.59 Joules.
Answer:
the initial concentration of SCN- in the mixture is 0.00588 M
Explanation:
The computation of the initial concentration of the SCN^- in the mixture is as follows:
As we know that
As it is mentioned in the question that KSCN is present 10 mL of 0.05 M
So, the total milimoles of SCN^- is
= 10 × 0.05
= 0.5 m moles
The total volume in mixture is
= 45 + 10 + 30
= 85 mL
Now the initial concentration of the SCN^- is
= 0.5 ÷ 85
= 0.00588 M
hence, the initial concentration of SCN- in the mixture is 0.00588 M