Answer:
138 mg
Explanation:
A company is testing drinking water and wants to ensure that Ca content is below 155 ppm (= 155 mg/kg), that is, <em>155 milligrams of calcium per kilogram of drinking water</em>. We need to find the maximum amount of calcium in 890 g of drinking water.
Step 1: Convert the mass of drinking water to kilograms.
We will use the relation 1 kg = 1000 g.
Step 2: Calculate the maximum amount of calcium in 0.890 kg of drinking water
Hydrogen bonding and dipole-dipole forces.
Answer:
ΔG°rxn = +50.8 kJ/mol
Explanation:
It is possible to obtain ΔG°rxn of a reaction at certain temperature from ΔH°rxn and S°rxn, thus:
<em>ΔG°rxn = ΔH°rxn - T×S°rxn (1)</em>
In the reaction:
2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l)
ΔH°rxn = 3×ΔHfNO2 + ΔHfH2O - (2×ΔHfHNO3 + ΔHfNO)
ΔH°rxn = 3×33.2kJ/mol + (-285.8kJ/mol) - (2×-207.0kJ/mol + 91.3kJ/mol)}
ΔH°rxn = 136.5kJ/mol
And S°:
S°rxn = 3×S°NO2 + S°H2O - (2×S°HNO3 + S°NO)
ΔH°rxn = 3×0.2401kJ/molK + (0.0700kJ/molK) - (2×0.146kJ/molK + 0.2108kJ/molK)
ΔH°rxn = 0.2875kJ/molK
And replacing in (1) at 298K:
ΔG°rxn = 136.5kJ/mol - 298K×0.2875kJ/molK
<em>ΔG°rxn = +50.8 kJ/mol</em>
<em />
Answer:
The mole fraction of ethanol is 0.6. A 10 mL volumetric pipette must be used for to measure the 10 mL of ethanol. The vessel should be clean and purged.
Explanation:
For calculating mole fraction of ethanol, the amount of moles ethanol must be calculated. Using ethanol density (0.778 g/mL), 10 mL of ethanol equals to 7.89 g of ethanol and in turn 0.17 moles of ethanol. The same way for calculate the amount of water moles (ethanol density=0.997 g/mL). 2 mL of water correspond to 0.11. The total moles are: 0.17+0.11=0.28. Mole fraction alcohol is: 0.17/0.28=0.6
