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.
Explanation:
mass H2O2 = 55 mL(1.407 g/mL) = 80.85 g
molar mass H2O2 = 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol
moles H2O2 = 80.85 g/34.02 g/mol = 2.377 moles H2O2
For each mole of H2O2 you obtain 0.5 mole of O2 (see the equation).
moles O2 = 2.377 moles H2O2 (1 mole O2)/(2 moles H2O2) = 1.188 moles O2
Now, you need the temperature. If you are at STP (273 K, and 1.00 atm) then 1 mole of an ideal gas at STP has a volume of 22.4 L. Without temperature you are not really able to continue. I will assume you are at STP.
Volume O2 = 1.188 moles O2(22.4 L/mole) = 0.0530 L of O2.
which is 53 mL.
Answer:
5200 ppm
Explanation:
As per the definition, parts per million of a contaminant is a measure of the amount of mass of contaminant present per million amount of the solution. It is denoted by ppm.
Given in the question,
Water = 250 ml = 250 g
Lead = 1.30 g
So,
ppm of Lead = 
= 
= 5200 ppm
So, as calculated above, there is 5200 ppm of lead present in 250 ml of water.
Answer:
boron (B), germanium (Ge), and tellurium (Te)
