I divided the volume by a 1000 because it must be in litres, but it wont make any difference because they will cancel out, but you must always convert it to L or dm3 unless it says in the question something else.
Answer:
The thermal energy (heat) needed, to raise the temperature of oil of mass 'm' kilogram and specific heat capacity 'c' from 20°C to 180°C is 160·m·c joules
Explanation:
The heat capacity, 'C', of a substance is the heat change, ΔQ, required by a given mass, 'm', of the substance to produce a unit temperature change, ΔT
∴ C = ΔQ/ΔT
ΔQ = C × ΔT
C = m × c
Where;
c = The specific heat capacity
ΔT = The temperature change = T₂ - T₁
∴ ΔQ = m × c × ΔT
Therefore, the thermal energy (heat) needed, ΔQ, to raise the temperature of oil of mass 'm' kilogram and specific heat capacity, 'c' from 20°C to 180°C is given as follows;
ΔQ = m × c × (180° - 20°) = 160° × m·c
ΔQ = 160·m·c joules
Answer:
0.33moles
Explanation:
RAM of;
Na = 23, H=1, C=12, O=16
molar mass of NaHCO= 23+1+12+16= 52g/mol
mole= mass / molar mass
mole = 17.31/52
mole= 0.33moles
Answer:
The answer is 139.1 g of sodium azide.
Explanation:
To solve this question we need to find first the stoichometry of the reaction to find the moles involved.
1. The balanced formula:
2NaN3 ⇒ 3N2 + 2Na
(important: the nitrogen is a gas, therefore, is always diatomic N2)
2. The moles:
We need to convert the 90 g of N2 into moles to know the relations of amounts between the other compounds in the chemical equation
90 g of N2 x (1 mol of N2/ 28 g of N2) = 3.21 mol of N2
3. Stoichometry:
The relations of the amounts of moles of N2 and NaN3 using the balanced formula
3.21 mol of N2 x (2 mol of NaN3/3 mol of N2) = 2.14 mol of NaN3
4. Moles into grams
then to answer the question we convert the moles value of NaN3 into grams of NaN3
2.14 mol of NaN3 x ( 65 g of NaN3/ 1 mol of NaN3) = 139.1 g of NaN3.
and this is the answer of the question.
