We may apply a mass balance here. The total moles of NaCl will remain constant before and after the mixing of the solutions. We have three solutions:
1) 25 ml of 5 M NaCl
2) 100 ml of 0 M NaCl
3) A mixture of these solutions with volume 125 ml and concentration C
The moles of a substance are equal to the product of the concentration and volume. Thus:
25 * 5 + 100 * 0 = 125 * C
C = 1 M
The concentration of the final solution is 1 M.
Answer:
<h3>The answer is 235.29 mL</h3>
Explanation:
The volume of a substance when given the density and mass can be found by using the formula

From the question.
density = 0.017 g/mL
mass = 4 g
We have

We have the final answer as
<h3>235.29 mL</h3>
Hope this helps you
Answer:
The answer to your question is: 6.55 x 10 ²³ atoms of Br
Explanation:
CH2Br2 = 37.9 g
MW CH2Br2 = (12 x 1) + (2 x 1) + (80 x 2) = 174 g
174 g of CH2Br2 ------------------ 160 g of Br2
37.9 g of CH2Br2 --------------- x
x = 37.9 x 160/174 = 34.85 g of Br
1 mol of Br ----------------- 160 g Br2
x ---------------- 174 g Be2
x = 174 x 1 /160 = 1.088 mol of Br2
1 mol of Br ----------------- 6.023 x 10 ²³ atoms
1.088 mol of Br ------------- x
x = 1.088 x 6.023 x 10 ²³ / 1 = 6.55 x 10 ²³ atoms
Answer:
2,2,4-Trimethyl-pentane
Explanation:
Structural characteristics of the compound is as follows:
- Has five methyl group
- Has one quaternary carbon
- No. double bond
- Gives four monochloro substitution products
Compound must have straight chain of 5 carbons.
Three methyl substituents are attached to 2 and 4 carbons.
Therefore, IUPAC name of the compound will be 2,2,4-Trimethyl-pentane.
1) As can be seen from any 1H NMR chemical shift ppm tables, hydrogens which have δ values from 2ppm to 2.3ppm are hydrogens from carbon which is bonded to a carbonyl group. From this, we can conclude that our hydrogens belong to the type, but from 2 different alkyl groups because of 2 different signals.
2) So, one alkyl group is CH3 and second one can be CH or CH2.
3) If we know that ratio between two types of hydrogens is 3:2, it can be concluded that second alkyl group is CH2.
4) Finally, we don't have any other signals and it indicates that part of the compound which continues on CH2 is exactly the same as the first part.
The ratio remains the same, 3:2 ie 6:4