Answer:
The answer to the question is
The specific heat capacity of the alloy = 1.77 J/(g·°C)
Explanation:
To solve this, we list out the given variables thus
Mass of alloy = 45 g
Initial temperature of the alloy = 25 °C
Final temperature of the alloy = 37 °C
Heat absorbed by the alloy = 956 J
Thus we have
ΔH = m·c·(T₂ - T₁) where ΔH = heat absorbed by the alloy = 956 J, c = specific heat capacity of the alloy and T₁ = Initial temperature of the alloy = 25 °C , T₂ = Final temperature of the alloy = 37 °C and m = mass of the alloy = 45 g
∴ 956 J = 45 × C × (37 - 25) = 540 g·°C×c or
c = 956 J/(540 g·°C) = 1.77 J/(g·°C)
The specific heat capacity of the alloy is 1.77 J/(g·°C)
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol
15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol 2 mol
0.100 mol 0.200 mol
We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M
Concentration of NO3⁻ is 0.667 M.
Answer : The fuel value and the fuel density of pentane is, 49.09 kJ/g and 
respectively.
Explanation :
Fuel value : It is defined as the amount of energy released from the combustion of hydrocarbon fuels. The fuel value always in positive and in kilojoule per gram (kJ/g).
As we are given that:
Fuel value = 
Molar mass of pentane = 72 g/mol
Fuel value = 
Fuel value = 49.09 kJ/g
Now we have to calculate the fuel density of pentane.
Fuel density = Fuel value × Density
Fuel density = (49.09 kJ/g) × (0.626g/mL)
Fuel density = 30.73 kJ/mL =
Thus, the fuel density of pentane is
