You can solve this problem through dimensional analysis.
First, find the molar mass of NaHCO3.
Na = 22.99 g
H = 1.008 g
C = 12.01 g
O (3) = 16 (3) g
Now, add them all together, you end with with the molar mass of NaHCO3.
22.99 + 1.008 + 12.01 + 16(3) = 84.008 g NaHCO3. This number means that for every mole of NaHCO3, there is 84.008 g NaHCO3. In simpler terms, 1 mole NaHCO3 = 84.008 g NaHCO3.
After finding the molar mass of sodium bicarbonate, now you can use dimensional analysis to solve for the number of moles present in 200. g of sodium bicarbonate.
Cross out the repeating units which are g NaHCO3, and the remaining unit is mole NaHCO3
200. * 1 = 200
200/ 84.008 = 2.38
Notice how there are only 3 sig figs in the answer. This is because the given problem only gave three sig figs.
Your final answer is 2.38 mol NaHCO3.
Answer:
They allow particles to stay close together.
The attractive forces (bonds) in a liquid are strong enough to keep the particles close together, but weak enough to let them move around each other. For example, Liquids are useful in car brake systems because they flow and cannot be compressed.
Explanation:
Hope this helps :)
Answer:
The weight/weight % or percent by mass of the solute is 1.04g.mL
Explanation:
We assume, (i) that the solvent is water, and (ii) that dissolution occurs without change in volume.
So given the standard density of water, the mass of the solution is simply,
(4.87+86.4)g
For solution,
=
= 1.04 g. mL
Because nitrate salts are soluble stuff, you could get significantly higher solution densities if you dissolved more salt (volume change would be fairly negligible).