Answer:
The volume of the gas is 89.60
Explanation:
When converting from moles to volume, you would multiply by 22.4 because 1 mole equals 22.4 liters at STP.
So...
4.0 moles * 22.4 liters =
89.60 liters
Answer: 4.18925 kJ heat is needed to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.
Explanation:
Temperature of Solid 
Melting temperature of Solid 
Temperature of liquid 
Specific heats of solid ethanol = 0.97 J/gK
Specific heats of liquid ethanol = 2.3 J/gK
Heat required to melt the the 25 g solid 
at 159 K
= 159 K - 138 K = 21 K
Heat required to melt and raise the temperature of upto 223 K
= 223 K - 159 K = 64 K
Total heat to convert solid ethanol to liquid ethanol at given temperature :
(1kJ=1000J)
Hence, 4.18925 kJ of heat will be required to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.
Combustion reaction for menthol is as follows;
CxHyOz + O₂ ---> xCO₂ + H₂O
Mass of CO₂ formed - 28.16 mg
Therefore number of moles formed - 28.16/ 44 g/mol = 0.64 mmol
Mass of water formed - 11.53 mg
number of water moles formed - 11.53 mg/18 g/mol = 0.64 mmol
From CO₂,
1 mol of CO₂ - 1 mol of C and 2 mol of O
therefore number of C moles - 0.64 mmol
O moles - 1.28 mmol
from H₂O
1 mol of H₂O - 2 mol of H and 1 mol of O
number of H moles - 1.28 mmol
O moles - 0.64 mmol
Mass of menthol initially - 10 mg
in reactions, the masses of products are equal to the masses of reactants. The excess mass to the products formed is due to O₂ in air
Original mass of menthol - 10 mg
mass of water and CO₂ - 11.53 mg + 28.16 mg = 39.69
Difference in mass - 39.69 - 10 = 29.69 mg
This difference comes from O moles in air - 29.69 mg/ 16 g/mol = 1.8556 mmol
then O moles coming from menthol - (1.28 + 0.64) - 1.8556 = 0.064 mmol
In menthol
C moles - 0.64 mmol
H moles - 1.28 mmol
O moles - 0.064 mmol
ratios of C:H:O
C H O
0.64 1.28 0.064
x1000 x1000 x1000 to get whole numbers
640 1280 64
10 20 1
Simplest ratio of C:H:O is 10:20:1
therefore empirical formula of menthol is C₁₀H₂₀O
Noble gases
also known as Inert gases
Mass is the property of a physical body and the resistance to acceleration when a net force is applied on the body.
The atomic mass of sodium (Na) is = 22.98
The atomic mass of nitrate (N) is = 14.00
The atomic mass of oxygen (O) is = 15.99
The sodium nitrate (NaNO3) consists of the atomic masses of Na+N+(O)3 = 85 grams
Therefore, the mass of 6.5 mol of sodium nitrate is = 6.5 * 1 mol of NaNO3
= 6.5 * (85)
= 552.50 grams
