Answer:
50.127 J
Explanation:
The following data were obtained from the question:
Initial temperature (T₁) = –5 °C
Final temperature (T₂) = 37 °C
Mass (M) = 3.10 g
Specific heat capacity (C) = 0.385 J/gºC
Heat (Q) absorbed =?
Next, we shall determine the change in temperature. This can be obtained as follow:
Initial temperature (T₁) = –5 °C
Final temperature (T₂) = 37 °C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 37 – (–5)
ΔT = 37 + 5
ΔT = 42 °C
Finally, we shall determine the heat absorbed. This can be obtained as follow:
Mass (M) = 3.10 g
Specific heat capacity (C) = 0.385 J/gºC
Change in temperature (ΔT) = 42 °C
Heat (Q) absorbed =?
Q = MCΔT
Q = 3.10 × 0.385 × 42
Q = 50.127 J
Thus, the heat absorbed is 50.127 J.
Answer:
Tc
Explanation:
You just have to follow the rows with the exponents. Just remember that when we get to d, the number in the front is a period lower. Hope this helps!
5= Elemental hydrogen, nitrogen, oxygen, fluorine and chlorine are all gases at room temperature,
<span> the </span>vapor pressure<span> of the liquid at a temperature T</span>2<span> ... Now, </span>it's<span> important to realize that the </span>normal boiling point<span> of a substance is measured at an atmoshperic ... ΔHvap=−ln(</span>134mmHg760mmHg<span> )⋅8.314J mol−1K−1 (1(273.15+</span>0)−1(273.15+40))K−1 ... Give equations that can be used tocalculate<span> the .
Now try it yourself :)</span>
