Answer:
Q1: 1.67 L.
Q2: Saturated solution.
Explanation:
<u><em>Q1:</em></u>
- We have the role that the no. of millimoles before dilution is equal to the no. of millimoles after dilution.
<em>(MV) before dilution = (MV) after dilution</em>
M before dilution = 10.0 M, V before dilution = 0.5 L.
M after dilution = 3.0 M, V after dilution = ??? L.
<em>∴ V after dilution = (MV) before dilution / M after dilution</em> = (10.0 M)(0.5 L) / (3.0 M) =<em> 1.67 L.</em>
<em></em>
<em></em>
<u><em>Q2:</em></u>
From the given curve, it is clear that the solubility of sodium nitrate at 35.0°C is 100 g per 100 g of water.
<em>So, the mentioned solution is a saturated solution at this T.</em>
<em></em>
<em>A saturated solution is a chemical solution containing the maximum concentration of a solute dissolved in the solvent.</em>
Answer:
96.09 g/mol
Explanation:
You just need to first get the atomic weights of the elements involved. You can easily get these from your periodic table.
If you are going to do this properly, please use the weight with at least two decimal places for accuracy (e.g. 15.99 g/mol).
Also, please take note that I will be using the unit g/mol for all the weights. Thus,
Step 1
N = 14.01 g/mol
H = 1.008 g/mol
O = 16.00 g/mol
C = 12.01 g/mol
Since your compound is
(
N
H
4
)
2
C
O
3
, you need to multiply the atomic weights by their subscripts. Therefore,
Step 2
N = 14.01 g/mol × 2 =
28.02 g/mol
H = 1.008 g/mol × (4×2) =
8.064 g/mol
O = 16.00 g/mol × 3 =
48.00 g/mol
C = 12.01 g/mol × 1 =
12.00 g/mol
To get the mass of the substance, we need to add all the weights from Step 2.
Step 3
molar mass of
(
NH
4
)
2
CO
3
=
(28.02 + 8.064 + 48.00 + 12.01) g/mol
=
96.09 g/mol
this is a google search and a example i hope is helps to solve
(i) 1 mole of Ar =6.022×10^23
atoms of Ar
52 mol of Ar=52×6.022×10 ^23
atoms of Ar =3.131×10 ^25 atoms of Ar
(ii) Atomic mass of He = 4amu
Or
4amu is the mass of He atoms = 1
Therefore 52 amu is the mass of He atoms=0.25×52=13 atoms of He
(iii) Gram atomic mass of He = 4g
Or
4g of He contains =6.022×10 ^23
atoms
Therefore 52 g of He contains=
6.022×10 ^23/4×52 =7.83×10 ^24 atoms
