Explanation:
First, we need to calculate the number of moles of sodium carbonate we have in a 25 g sample. To calculate this, we will
find the molar mass of sodium carbonate (Na2CO3):
⇒ 2 × Molar mass of sodium + Molar mass of carbon + 3×molar mass of oxygen
⇒ 2 × 23 + 12 + 3 × 16
⇒ 46 + 12 + 48
⇒ 106g/mol
Thus, the molar mass of Na2CO3 is 106g/mol.
Therefore, number of moles = 25 ÷ 106
=> 0.2358 mol
Now, we know that every mole of Na2CO3 have 0.2358 moles of Na+ ions. Hence, total moles of Na2CO3 is 0.4716 moles
Number of ions present = 6.022 × 1023 × 0.4716 mol = 2.84 × 1023ions
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:
grams O₂ = 134 grams
Explanation:
PV = nRT => n = PV/RT
P = 8.15atm
V = 12.2 Liters
R = 0.08206L·atm/mol·K
T = 16.0°C + 273 = 289K
n = (8.15atm)(12.2L)/(0.08206L·atm/mol·K)(289K) = 4.2 moles O₂
grams O₂ = 4.2 moles O₂ x 32g/mol = 134 grams O₂
Notice the subscript went from 83 to 84 and the 214 stayed the same that means it was beta decay.
