Let's begin with the basic values that will be used in the solution.
The formula of propane is C3H8. It is an alkane, a hydrocarbon with the general formula of CnH2n+2. Notice that hydrocarbons have only Carbon and Hydrogen atoms. Its molar mass (M) is 44 g.
Molar Mass Calculation is done as like that
C=12 g/mol, H=1 g/mol. 1 mole propane has 3 moles Carbon atoms and 8 mole Hydrogen atoms. M(C3H8)= 3*12+ 8*1= 44 g
Combustion reaction of hydrocarbons gives carbon dioxide and water by releasing energy. That energy is called as enthalpy of combustion (ΔHc°).
ΔHc° of propane equals -2202.0 kj/mol. Burning of 1 mole C3H8 releases 2202 kj energy. Minus sign only indicates that the energy is given out ( an exothermic reaction ).
Let's write the combustion reaction.
C3H8 + O2 ---> CO2 + H20 (unbalanced) ΔHc° = -2202 kj/mol
Now, we calculate mole of 20 kg propane. Convert kilogram into gram since we use molar mass is defined in grams.
mole=mass/molar mass ; n=m/M ; n= 20000 g /44 (g/mol)=454 mole
1 mole propane releases 2202 kj energy.
454 mole propane release 2202 kj *454= 1000909 kj
The answer is 1000909 kj.
46 gram of ethanol ≡ 1 mole of ethanol
1 gram of ethanol ≡ 1/46 mole of ethanol
10 gram of ethanol ≡ 1*10/46 mole of ethanol
=0.217 mole of ethanol
I believe it's B: series circuit
good luck
Answer:
C
Explanation:
Neon is a noble gas, which is monoatomic.
Answer:
v = 534.5mL
m = 597.15g
Density = 9.23g/mL
Density = 9.125g/mL
Explanation:
Density = mass/ volume
For the first question
Density = 1.59g/mL
Mass = 834.01g
Volume = ?
Using the above formula we have 1.59 = 834.01/v
v = 834.01/1.59
v = 534.5mL
For the second question
Density =0.9167g/mL
Volume = 651.41mL
Mass =?
Using the above formula we have
0.9167 =m/651.41
Cross multiply
m = 0.9167 x 651.41
m = 597.15g
For the third question
Mass =803.44g
Volume=87.03mL
Density =?
Density = 803.44/87.03
= 9.23g/mL
For the fourth
Density = 56.85/6.23
= 9.125g/mL
