Answer: 16.32 g of 
as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of 
require = 1 mole of
Thus 0.34 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
Moles of left = (0.68-0.17) mol = 0.51 mol
Mass of 
Thus 16.32 g of as excess reagent are left.
None of the alpha particles fired at the foil are being repelled back, like they were in the Rutherford atom simulation.I hope this correct.
Answer: The steepness of a ramp affects it by making it easier or harder.
Explanation: It's a bit situational. If you were going up a steep ramp with a heavy load, it will increase the work necessary, whereas if you were going down a ramp, it would decrease the work necessary. If you need this simply put, think about biking up and down a hill. It would be easier going down than up.
Molarity is defined as the number of moles of solute in 1 L of solution
the mass of Ca(NO₃)₂ present - 8.50 g
therefore number of moles of Ca(NO₃)₂ - 8.50 g / 164 g/mol = 0.0518 mol
the volume of solution prepared is 755 mL
therefore if there are 0.0518 mol in 755 mL
then in 1000 mL the number of moles - 0.0518 mol / 0.755 L
molarity is therefore - 0.0686 M
Answer:
Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Explanation:
Let's apply the formula for freezing point depression:
ΔT = Kf . m
ΔT = 74.2°C - 73.4°C → 0.8°C
Difference between the freezing T° of pure solvent and freezing T° of solution
Kf = Cryoscopic constant → 5.5°C/m
So, if we replace in the formula
ΔT = Kf . m → ΔT / Kf = m
0.8°C / 5.5 m/°C = m → 0.0516 mol/kg
These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg
0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:
Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
