This should help :)
Example 1: A 36.0 g sample of water is initially at 10.0 °C.
How much energy is required to turn it into steam at 200.0 °C? (This
example starts with a temperature change, then a phase change followed
by another temperature change.)
Solution:
<span>q = (36.0 g) (90.0 °C) (4.184 J g¯1 °C¯1) = 13,556 J = 13.556 kJ
q = (40.7 kJ/mol) (36.0 g / 18.0 g/mol) = 81.4 kJ
q = (36.0 g) (100.0 °C) (2.02 J g¯1 °C¯1) = 7272 J = 7.272 kJ
q = 102 kJ (rounded to the appropriate number of significant figures)
</span>
Answer:
Q = 28.9 kJ
Explanation:
Given that,
Mass of Aluminium, m = 460 g
Initial temperature, 
Final temperature, 
We know that the specific heat of Aluminium is 0.9 J/g°C. The heat required to raise the temperature is given by :
So, 28.9 kJ of heat is required to raise the temperature.
Answer:
D. All of the above
Explanation:
Because all of these have something to do with how glaciers are affecting the planet.
Answer:
50 mL
Explanation:
In case of titration , the following formula is used -
M₁V₁ = M₂V₂
where ,
M₁ = concentration of acid ,
V₁ = volume of acid ,
M₂ = concentration of base,
V₂ = volume of base .
from , the question ,
M₁ = 0.50M
V₁ = 100 mL
M₂ = 1.0M
V₂ = ?
Using the above formula , the volume of base , can be calculated as ,
M₁V₁ = M₂V₂
substituting the respective values ,
0.50M * 100 mL = 1.0M * V₂
V₂ = 50 mL
