Answer:
127.3° C, (This is not a choice)
Explanation:
This is about the colligative property of boiling point.
ΔT = Kb . m . i
Where:
ΔT = T° boling of solution - T° boiling of pure solvent
Kb = Boiling constant
m = molal (mol/kg)
i = Van't Hoff factor (number of particles dissolved in solution)
Water is not a ionic compound, but we assume that i = 2
H₂O → H⁺ + OH⁻
T° boling of solution - 118.1°C = 0.52°C . m . 2
Mass of solvent = Solvent volume / Solvent density
Mass of solvent = 500 mL / 1.049g/mL → 476.6 g
Mol of water are mass / molar mass
76 g / 18g/m = 4.22 moles
These moles are in 476.6 g
Mol / kg = molal → 4.22 m / 0.4766 kg = 8.85 m
T° boling of solution = 0.52°C . 8.85 m . 2 + 118.1°C = 127.3°C
Answer:
visible light falls between which set of waves within the electromagnetic spectrum?
D.) infrared and Ultraviolet
Moles of helium is required to blow up a balloon to 87.1 liters at 74 C and 3.5 atm is 021.65 mole
Mole is the unit of amount of substances of specified elementary entities
According to the ideal gas law he number of moles of a gas n can be calculated knowing the partial pressure of a gas p in a container with a volume V at an absolute temperature T from the equation
n =pV/RT
Here given data is volume = 87.1 liters
Temperature = 74 °C means 347.15 k
Pressure = 3.5 atm
R = 0.0821
Putting this value in ideal gas equation then
n =pV/RT
n = 3.5 atm×87.1 liters / 0.0821 ×347.15 k
n = 021.65 mole
Moles of helium is required to blow up a balloon to 87.1 liters at 74 C and 3.5 atm is 021.65 mole
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Answer: 208g/mol
Explanation:
First of all we have to write the balance equation for the reaction.
BaCl2 + Fe2(SO4)3____> BaSO4 + FeCl3
After balancing we have.
3BaCl2 +Fe2(SO4)3_____> 3BaSO4 +2FeCl3
Looking at the equation, we find out that 3 moles of barium chloride reacts with 1 mole of iron iii sulfate
Therefore we have
3moles of BaCl2 _____> 400g/mole of iron iii sulfate
Xmole of BaCl2 _____> 200g/mole of iron iii sulfate
X = 2 * 200g/mol divide by 400g/mol
X = 1mole
1 mole of BaCl2 will be need to react with 200g/mol of iron iii sulfate.
This 1 mole of BaCl2 is equivalent to 208g/mol of BaCl2.
Therefore the gram of barium chloride that must be present is = 208g/mol//
