Answer:
24.9%
Explanation:
According to this question, mole fraction of NaCl in an aqueous solution is 0.0927. This means that the mole percent of NaCl in the solution is:
0.0927 × 100 = 9.27%
Let's assume that the solution contains water (solvent) + NaCl (solute), hence, the mole fraction of water will be;
100% - 9.27% = 90.73%
THEREFORE, it can be said that, NaCl contains 0.0927moles while H2O contains 9.073moles
N.B: mole = mass/molar mass
Given the Molar Mass
NaCl: 58.44 g/mol
H2O: 18.016 g/mol
For NaCl;
0.0927 = mass/58.44
mass = 0.0927 × 58.44
5.42g
For H2O;
9.073 = mass/18.016
mass = 9.073 × 18.016
= 16.35g
Total mass of solution = 16.35g + 5.42g = 21.77g
Mass percent of NaCl = mass of NaCl/total mass × 100
% mass of NaCl = 5.42g/21.77g × 100
= 0.249 × 100
= 24.9%
Answer:
See explanation.
Explanation:
I highly suggest you watch OChem Tutor's videos on IUPAC nomenclature because the actual naming would take a lot of time to teach in text-based format. But here is how to name them:
1) I think there are two seperate pictures for number 1. The molecule on the left is 1-pentene and the one on the right is 4-methyl-1-pentene. If the whole thing is one molecule but there is just a bond missing where the red marker numbers are, that molecule would be 9-methyl-1,6-decadiene.
2) 4-methyl-2-pentene
3) 2,4-octadiene
4) 1,5-nonadiene
5) 2,5-dimethyl-3-hexene
6) 3,6-dimethyl-2,4-heptadiene
7) 2,5,5-trimethyl-2-hexene
Answer:
initial temperature=
Explanation:
Assuming that the given follows the ideal gas nature;
mole of gass will remain same at any emperature:
putting all the value we get:
initial temperature=
Answer:
12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.
Explanation:
For first solution of sulfuric acid :
C₁ = 40% , V₁ = ?
For second solution of sulfuric acid :
C₂ = 10% , V₂ = ?
For the resultant solution of sulfuric acid:
C₃ = 28% , V₃ = 20L
Also,
<u>V₁ + V₂ = V₃ = 20L</u> ......................................(1)
Using
<u>C₁V₁ + C₂V₂ = C₃V₃</u>
<u>40×V₁ + 10×V₂ = 28×20</u>
So,
40V₁ + 10V₂ = 560........................................(2)
Solving 1 and 2 as:
V₂ = 20 - V₁
Applying in 2
40V₁ + 10(20 - V₁) = 560
40V₁ + 200 - 10V₁ = 560
30V₁ = 360
<u>V₁ = 12 L</u>
So,
<u>V₂ = 20 - V₁ = 8L</u>
<u><em>12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.</em></u>
Nylon 6,6 is a common example of a polyamide.
<em>Polyamides</em> are polymers that contain <em>repeating amide (-CO-NH-) linkages</em>.
The structure of Nylon 6,6 is
[-NH-(CH_2)_6-<u>NH-CO</u>-(CH_2)_4-CO-]_<em>n</em>
where <em>n</em> is a large number.
The numbers in the name showow that there are six carbon atoms on either side of an amide linkage.
