<span>_____Balanced Chemical eqn___
C6H5OH + 7O2 --> 6C02 + 3H20
Firstly, heat of rxn will be absorbed by bobm calorimaeter, so find the qcal
qcal = CΔT
= (11.66 KJ/°C)(26.37°C- 21.36°C)
= 58.42 KJ
since the heat of rxn absorbed by calorimeter, assume:
qrnx = -qcal
= - 58.42 KJ
_____q GRAM OF C6H5OH______
Find the number of moles you have by dividing the given grams of C6H5OH by its molar mass
(1.8 g C6H5OH)(94 g/mole C6H5OH) = 0.0191 mol C6H5OH
then divide qrnx by the number of moles.
(-58.42 KJ)/(0.0191 mol C6H5OH) = -3058.64 KJ/mol C6H50H</span>
Answer:
0.31M is the new concentration of the solution
Explanation:
The solution is diluted from 125mL to 1.0L = 1000mL. The dilution is of:
1000mL / 125mL = 8 times.
That means the concentration of the diluted solution is 8 times lower than the original solution. That is:
2.5M / 8 times =
<h3>0.31M is the new concentration of the solution</h3>
Salts dissociate when dissolved in water:
PbSO4 -> Pb2+ + SO4 2-
So there are 2 different particles. 1 mol
of PbSO4 produces 1 mol of Pb2+, and 1 mol of SO4 2- (one of each).
Kps = [Pb2+] [SO4 2-]
We will call “s” the molarity of PbSO4
So the molarity of Pb2+ is also “s” (same
number)
And the molarity of SO4 2- is also “s”
(same number)
Kps = s*s = s2
1.82
× 10-8 = s2
So,
in order to find s, we have to make the square root of 1.82 × 10-8, which is: 0,00013491 M
89.1% Au
The molar mass of Au2O3 is 2(196.97) + 3(16.00) = 441.94 g/mol
The mass percent of Au is therefore 2(196.97) / 441.94, which is 393.94/441.94 = 0.891 —> 89.1%
To determine the number of moles of the gas, we need to relate moles to pressure, temperature and volume. To simplify things, we assume the gas is ideal so we use the ideal gas equation PV=nRT.
PV = nRT
n = PV / RT
n = 3.5 (9.0) / 0.08205 (74+273.15)
n = 1.1059 mol
