Answer is: 39,083kJ.
m(coal) = 2,00g.
m(water) = 500g.
ΔT = 43,7°C - 25°C = 18,7°C, <span>difference at temperatures.</span>
c(water) = 4,18 J/g·°C, <span>specific heat of water
</span>Q = m(water)·ΔT·c(water), heat of reaction.
Q = 500g·18,7°C·4,18J/g·°C.
Q = 39083J = 39,083kJ.
Answer:
The correct answer is - 29.45 / 100 x 25.6 = 7.5392 grams
Explanation:
It is given in the question that in 100 gms of CaSO4 there are 29.45 grams of Ca present and there is 25.6 gram of total CaSO4 sample present, So, to calculate the exact value of calcium in this given sample is:
mass of Ca = total amount of sample*percentage of calcium in sample /100
M of Ca =25.6*29.45/100
M of Ca = 7.5392 grams
Thus, the correct procedure is given by 29.45 / 100 x 25.6 = 7.5392 grams
Wait is that suppose to be a question??!!
Answer:
52.5 mol O2
Explanation:
4 FeCl3 + 6 O2 -> 2 Fe2O3 +6 Cl2
4 mol FeCl3 -> 6 mol O2
35.0 mol FeCl3 -> x
x= (35.0 mol FeCl3 * 6 mol O2)/4 mol FeCl3
x=52.5 mol O2
