So the acceleration has actually slowed down the ball because it was going in the direction opposite the velocity. Now see what happens as the ball falls back down to Earth. The ball has zero velocity, but the acceleration due to gravity accelerates the ball downward at a rate of –9.8 m/s2.
hope it helps
Explanation:
hope it helps you understand moles
Answer:
The molarity of the solution increases.
Explanation:
Molarity is the measure of the concentration of the solute in the solution. In this case, the solvent is the sugar solution and the solute is the sugar.
If sugar is ADDED to the already sugary solution, then there would be more sugar. Therefore, the sugar (solute) would increase in number.
This means that the answer is the third choice: The molarity of the solution increases.
The answer would not be the first or second choice because there isn't anything in the question that implies water. It just says sugar solution.
The answer is not the last choice because the sugar concentration does not decrease after you have added more sugar to it. It increases.
Answer:
The molality of isoborneol in camphor is 0.53 mol/kg.
Explanation:
Melting point of pure camphor= T =179°C
Melting point of sample =
= 165°C
Depression in freezing point = 

Depression in freezing point is also given by formula:

= The freezing point depression constant
m = molality of the sample
i = van't Hoff factor
We have:
= 40°C kg/mol
i = 1 ( organic compounds)



The molality of isoborneol in camphor is 0.53 mol/kg.
Answer:
Concentration of sodium carbonate in the solution before the addition of HCl is 0.004881 mol/L.
Explanation:

Molarity of HCl solution = 0.1174 M
Volume of HCl solution = 83.15 mL = 0.08315 L
Moles of HCl = n



According to reaction , 2 moles of HCl reacts with 1 mole of sodium carbonate.
Then 0.009762 mol of HCl will recat with:

Moles of Sodium carbonate = 0.004881 mol
Volume of the sodium carbonate containing solution taken = 1L
Concentration of sodium carbonate in the solution before the addition of HCl:
![[Na_2CO_3]=\frac{0.004881 mol}{1 L}=0.004881 mol/L](https://tex.z-dn.net/?f=%5BNa_2CO_3%5D%3D%5Cfrac%7B0.004881%20mol%7D%7B1%20L%7D%3D0.004881%20mol%2FL)