Answer:
density = 4.763pounds per liter
Explanation:
1g/cm³ = 2.205pounds per liter
2.16g/cm³ = (unknown)pounds per liter
(unknown)pounds per liter = 2.205 x 2.16 = 4.763
density = 4.763pounds per liter
Answer: The given statement is true.
Explanation:
When we increase the amount of solvent which is water in this case then it means there will occur an increase in the molecules. Hence, there will be more number of collisions to take place with increase in number of molecules.
Therefore, more is the amount of interaction taking place between the molecules of a solution more will be its rate of hydrolysis.
Thus, we can conclude that the statement increasing the amount of water in which the sugar is dissolved will increase the frequency of collisions between the sucrose molecules and the water molecules resulting in an increase in the rate of hydrolysis, is true.
Answer: 8.28g Na
Explanation: use ideal gas law
PV= nRT
Solve for moles of Cl2
n= PV/ RT
Substitute:
= 1 atm x 4.0 L / 0.08205 L.atm/ mol. K x 273 K
= 0.18 moles Cl2
Do stoichiometry to solve for m of Na
2 Na + Cl2 => 2 NaCl2
=0.18 moles Cl2 x 2 mol Na/ 1 mol Cl2 x 23g Na / 1 mol Na
= 8.28 g Na.
Answer: -
24 grams per kilogram.
Explanation: -
We know that
The mixing ratio = actual (measured) mass of water vapor (in parcel) in grams / mass of dry (non water vapor) air (in parcel) in kilogram
The saturation mixing ratio = mass of water vapor required for saturation (in parcel) in grams/ mass of dry (non water vapor) air (in parcel) in kilograms
Relative humidity = actual (measured) water vapor content/ maximum possible water vapor amount (saturation)
Thus saturation mixing ratio = Mixing ratio / relative humidity
= 6 / (25/100)
= 24
Answer:
15.4 g of sucrose
Explanation:
Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m
ΔT = Freezing T° of pure solvent - Freezing T° of solution
Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m
0.56°C / 1.86 m/°C = m → 0.301 mol/kg
m → molality (moles of solute in 1kg of solvent)
Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg
0.301 mol/kg . 0.150kg = 0.045 moles.
We determine the mass of sucrose, by the molar mass:
0.045 mol . 342 g/1mol = 15.4 g
