Answer:
for given question is 2.79 and 
is 0.52
{i- vant hoff’s constant ; Kb- constant ; m molarity }
M = no. of moles of the solute present in one kg of solution
Let the weight of amount of solute be “w” and its molecular mass be “M”
Let the mass of the solvent in the given question be “x”
Remember that:
number of moles = mass/molar mass
First, we get the molar mass of the nitrogen gas molecule:
It is known the the nitrogen gas is composed of two nitrogen atoms, each with molar mass 14 gm (from the periodic table)
Therefore, molar mass of nitrogen gas = 14 x 2 = 28 gm
Second we calculate the mass of the precipitate:
we have number of moles = 0.03 moles (given)
and molar mass = 28 gm (calculated)
Using the equation mentioned before,
mass = number of moles x molar mass = 0.03 x 28 = 0.84 gm
Answer:
.0924 moles of NaCl
Explanation:
So you know you have 5.4 g of NaCl and you need to know how many moles there are in this amount of NaCl
- You'll need to find the atomic mass of the compound NaCl to help you solve for moles
- Sodium (Na) on the periodic table has a mass of 22.99
- Chlorine (Cl) on the periodic table has a mass of 35.45
Add these two together----> 22.99 + 35.45 = 58.44
Now you can calculate for moles
<u>Written-out method:</u>
<u>5.4 grams of NaCl | 1 mole of NaCl </u>
| 58.44 grams NaCl = .0924 moles of NaCl
<u>Plug into calculator method:</u>
(5.4 g of NaCl/ 58.44g NaCl= .0925 moles)
Answer: A quantitative observation
Explanation: Quantitative observations deal with numbers or amounts. As opposed to <em>qualitative</em> observations, which are observations made with the 5 senses.
