Heat = 1.74 kJ
<h3>Further explanation</h3>
Given
melts at 328 ℃ + 273 = 601 K
mass = 23 g = 0.023 kg
initial temperature = 297 K
Final tmperature = 702 K
Required
Heat
Solution
1. raise the temperature(297 to 601 K)
c of lead = 0.130 kJ/kg K
Q = 0.023 x 0.13 x (601-297)
Q = 0.909 kJ
2. phase change(solid to liquid)
Q = m.Lf (melting/freezing)
Q = 0.023 x 23 kj/kg = 0.529 kJ
3. raise the temperature(601 to 702 K)
Q = 0.023 x 0.13 x (702-601)
Q = 0.302 kJ
Total heat = 1.74 kJ
Answer:
Explanation:
Hello there!
In this case, since this imaginary gas can be modelled as an ideal gas, we can write:
Which can be written in terms of density and molar mass as shown below:
Thus, by computing the pressure in atmospheres, the resulting density would be:
Best regards!
Answer: Heat of reaction ∆H = -13.43kJ
Explanation:
The number of moles of NaOH = the number of moles of HCL = N
N = concentration × volume= CV = 0.5M × 500mL/1000ml/L
N= 0.5 × 0.5= 0.25mol
Since the Molar enthalpy is given by Hm = -53.72kJ/mol
Heat of reaction ∆H = N×Hm
∆H= 0.25mol × -53.72kJ/mol = -13.43kJ
Heat of reaction ∆H = -13.43kJ
All of the above is the answer
Answer:
well it is because of mainly what we call reaction the petrol are highly reactive with the flames
