Answer:
The concentration of chloride ions in the final solution is 3 M.
Explanation:
The number of moles present in a solution can be calculated as follows:
number of moles = concentration in molarity * volume
In 100 ml of a 2 M KCl solution, there will be (0.1 l * 2mol/l) 0.2 mol Cl⁻
For every mol of CaCl₂, there are 2 moles of Cl⁻, then, the number of moles of Cl⁻ in 50 l of a 1.5 M solution will be:
number of moles of Cl⁻ = 2 * number of moles of CaCl₂
number of moles of Cl⁻ = 2 ( 50 l * 1.5 mol / l ) = 150 mol Cl⁻
The total number of moles of Cl⁻ present in the solution will be (150 mol + 0.2 mol ) 150.2 mol.
Assuming ideal behavior, the volume of the final solution will be ( 50 l + 0.1 l) 50.1 l. The molar concentration of chloride ions will be:
Concentration = number of moles of Cl⁻ / volume
Concentration = 150.2 mol / 50.1 l = 3.0 M
Answer:
A. 6N
B. 4H, 2O
C. 4H, 4N, 12O
D. 2Ca, 4O, 4H
E. 3Ba, 6Cl, 18O
F. 5Fe, 10N, 30O
G. 12Mg, 8P, 32O
H. 4N, 16H, 2S, 8O
I. 12Al, 18Se, 72O
J. 12C, 32H
I am 90% sure this is correct
Answer:
(II) only correctly rank the bonds in terms of increasing polarity.
Explanation:
Bond polarity is proportional to difference in electronegativity between bonded atoms.
Atoms Electronegativity Bond Electronegativity difference
Cl 3.0 Cl-F 1.0
Br 2.8 Br-Cl 0.2
F 4.0 Cl-Cl 0
H 2.1 H-C 0.4
C 2.5 H-N 0.9
N 3.0 H-O 1.4
O 3.5 Br-F 1.2
I 2.7 I-F 1.3
Si 1.9 Cl-F 1.0
P 2.2 Si-Cl 1.1
Si-P 0.3
Si-C 0.6
Si-F 2.1
So, clearly, order of increasing polarity : O-H > N-H > C-H
So, (II) only correctly rank the bonds in terms of increasing polarity
<span>The atomic weight of 13C should be pretty close to 13.0. (If you have the exact mass, use it in the problem.) So,
9.00 g / 13.0 g/mol = 0.692 moles
Therefore, the answer should be 0.692 moles are in 9.00 g of 13C.</span>
Well it breaks down into small parts
