1.B
6.D
D because t<span>he mass of one mole (molar mass) of helium gas is </span>4.002602 g/mol. 4.002602 * 5=20.01309. Rounded equals 20. So, the answer is D.20 g.
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
Atomic radii increase when going down a group and decreases when going towards the anion periods. So A and D.
Answer:
1)Reactants
2)Light
3)An item that can increase reaction rates
4)Reactants must collide with each other
Less molecules lower the chance for collisions
The more collisions there are the higher the reaction rate
