Answer:
q = -6464.9 kJ
Explanation:
We are given that the heat of combustion is ∆H° = −394 kJ per mol of carbon.Therefore what we need to do is calculate how many moles of C are in the lump of coal by finding its mass since the density is given.
vol = 5.6 cm x 5.1 cm x 4.6 cm = 131.38 cm³
m = d x v = 1.5 g/cm³ x 131.38 cm³ = 197.06 g
mol C = m/MW = 197.06 g/ 12.01g/mol = 16.41 mol
q = −394 kJ /mol C x 16.41 mol C = -6464.9 kJ
Answer:
The right answer is "58.3%".
Explanation:
The given values are:
Mass of lithium carbonate,
= 700 mg
i.e.,
= 0.7 g
Mass of tablet,
= 1.20 g
Now,
The percentage of active ingredient will be:
= 
On substituting the given values, we get
= 
= 
=
%
Answer:
Explanation:
We shall find out volume of air at NTP or at 273 K and 10⁵ Pa ( 1 atm )
Let it be V₂


V₂ = 7.87 litres
22.4 litres of any gas is equivalent to 1 mole
7.87 litres of air will be equivalent to
7.87 / 22.4 moles
= .35 moles .