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.
Answer:
21 g of N₂ are produced by the decomposition
Explanation:
The reaction is: 2 NaN3 → 2 Na + 3 N2
2 moles of sodium nitride decompose in order to produce 2 moles of Na and 3 moles of nitrogen gas.
According to stoichiometry, ratio is 2:3. Therefore we say,
2 moles of nitride can produce 3 moles of N₂
Then, 0.5 moles of NaN₃ will produce (0.5 . 3) / 2 = 0.75 moles of N₂
We convert the moles to mass, to find the answer
0.75 mol . 28 g / 1 mol = 21 g
Answer:
Yes, it is possible. Let us consider an example of two solutions, that is, solution A having 20 percent mass RbCl (rubidium chloride) and solution B is having 15 percent by mass NaCl or sodium chloride.
It is found that solution A is having more concentration in comparison to solution B in terms of mass percent. The formula for mass percent is,
% by mass = mass of solute/mass of solution * 100
Now the formula for molality is,
Molality = weight of solute/molecular weight of solute * 1000/ weight of solvent in grams
Now molality of solution A is,
m = 20/121 * 1000/80 (molecular weight of RbCl is 121 grams per mole)
m = 2.07
Now the molality of solution B is,
m = 15/58.5 * 1000/85
m = 3.02
Therefore, in terms of molality, the solution B is having greater concentration (3.02) in comparison to solution A (2.07).
An anion has a negative charge, while a cation has a positive charge.
(sample g/1) X (1 mole/40.078(MW of Ca)) = moles of sample (moles of sample)(6.022 x 10^23( no of atoms)/ 1 mole) = # of atoms in a 120 g sample of calcium Avogadro's number=6.022x 10^23 atoms in 1 mole
