Answer:
Option D. AlCl₃, MgC₂
Explanation:
We need to dissociate all the salts, to determine the i. (Van't Hoff factor).
The salt who has the highest value, will be the better conductor of electricity
CsCl → Cs⁺ + Cl⁻ i = 2
CaCl → Ca²⁺ + Cl⁻ i = 2
CaS → Ca²⁺ + S⁻² i = 2
Li₂S → 2Li⁺ + S⁻² i = 3
KBr → K⁺ + Br⁻ i = 2
AlCl₃ → Al³⁺ + 3Cl⁻ i = 4
MgC₂ → Mg²⁺ + 2C⁻ i = 3
KI → K⁺ + I⁻ i = 2
K₂S → 2K⁺ + S⁻² i = 3
The biggest i, is in pair D.
Answer:
C6H14O3F
Explanation:
The first step is to divide each compound by its molecular weight
Carbon
= 39.10/12
= 3.258
Hydrogen
= 7.67/1
= 7.67
Oxygen
= 26.11/16
= 1.63
Phosphorous
= 16.82/31
= 0.542
Flourine
= 10.30/19
= 0.542
The next step is to divide by the lowes value
3.258/0.542
= 6 mol of C
7.67/0.542
= 14 mol of H
1.63/0.542
= 3 mol of O
0.542/0.542
= 1 mol of P
0.542/0.542
= 1 mol of F
Hence the molecular formula is C6H14O3F
Answer:
The mass of water is 36 g.
Explanation:
Mass of hydrogen = 4 g
Mass of water = ?
Solution:
First of all we will write the balance chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of hydrogen = mass / molar mass
Number of moles of hydrogen = 4 g/ 2 g/mol
Number of moles of hydrogen = 2 mol
Now we compare the moles of water with hydrogen from balance chemical equation.
H₂ : H₂O
2 : 2
Mass of water = moles × molar mass
Mass of water = 2 mol × 18 g/mol
Mass of water = 36 g
If the water oxygen is in excess than mass of water would be 36 g.
Answer: C REDUCTION
Explanation:
Guessed after knowing oxidation isn't the answer. Got right
Answer:
2.893 x 10⁻³ mol NaOH
[HCOOH] = 0.5786 mol/L
Explanation:
The balanced reaction equation is:
HCOOH + NaOH ⇒ NaHCOO + H₂O
At the endpoint in the titration, the amount of base added is just enough to react with all the formic acid present. So first we will calculate the moles of base added and use the molar ratio from the reaction equation to find the moles of formic acid that must have been present. Then we can find the concentration of formic acid.
The moles of base added is calculated as follows:
n = CV = (0.1088 mol/L)(26.59 mL) = 2.892992 mmol NaOH
Extra significant figures are kept to avoid round-off errors.
Now we relate the amount of NaOH to the amount of HCOOH through the molar ratio of 1:1.
(2.892992 mmol NaOH)(1 HCOOH/1 NaOH) = 2.892992 mmol HCOOH
The concentration of HCOOH to the correct number of significant figures is then calculated as follows:
C = n/V = (2.892992 mmol) / (5.00 mL) = 0.5786 mol/L
The question also asks to calculate the moles of base, so we convert millimoles to moles:
(2.892992 mmol NaOH)(1 mol/1000 mmol) = 2.893 x 10⁻³ mol NaOH
