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: <u><em>Shape and position</em></u>
Explanation:
<u><em>I just took the test and I got it correct</em></u>
Find the moles of CaO
divide mass (2.0g) by the RFM which is 56 (Ca is 40 add that to O which is 16 making 56) this gives 0.0356 moles.
Find the theoretical mass by multiplying the moles of CaO (which is 0.0356 as there are no balancing number making the ratio 1:1) by the RFM of Ca(OH)2 which is 74 (40+16+16+1+1)
74 (Ca(OH)2 RFM) x 0.0357 (CaO moles) = 2.6g which is the theoretical mass of Ca(OH)2
Find percentage yield by dividing the actual mass of Ca(OH)2 by the theoretical and then x100 this Should give you 82.3%