Answer:
it would be OA that is negatively charged
Answer:
1.60 is the van't Hoff factor for ammonium chloride in X.
Explanation:
...(1)
where,
=Elevation in boiling point =
i = van't Hoff factor
= Freezing point constant
m = molality
1) When 70.4 g of benzamide are dissolved in 850. g of a certain mystery liquid X.
Mass of benzamide = 70.4 g
Molar mass of benzamide = 121 g/mol
i = 1 (organic molecule)
Mass of liquid X = 850 g = 0.850 kg
= Freezing point constant of liquid X= ?
Putting all value in a (1):
2) When 70.4 g of ammonium chloride are dissolved in 850. g of a certain mystery liquid X.
Mass of ammonium chloride= 70.4 g
Molar mass of ammonium chloride = 53.5 g/mol
i = ? (ionic molecule)
Mass of liquid X = 850 g = 0.850 kg
Putting all value in a (1):
i = 1.6011 ≈ 1.60
1.60 is the van't Hoff factor for ammonium chloride in X.
The van't Hoff factor, which measures ionization, for ammonium chloride in the mysterious liquid X can be calculated to be approximately 1.01. This is calculated by first determining the cryoscopic constant from the observed depression of the freezing point by benzamide (which does not ionize), and then utilizing this value to calculate the theoretical freezing point depression for ammonium chloride (pretending it does not ionize either). Since the observed depression was 9.9℃ and the calculated was 9.8℃, the van't Hoff factor is their quotient, or approximately 1.01.
To solve this problem, we need to understand that the van't Hoff factor (i) is a measure of the extent of ionization in solution. It can be calculated using the formula i = ΔTf observed / ΔTf calculated, where ΔTf observed is the observed freezing point depression and ΔTf calculated is the theoretical freezing point depression if no ionization occurs.
First, we calculate the theoretical freezing point depression for ammonium chloride. We know that this is given by the benzamide that reduces the freezing point of the same amount of liquid X by 2.7℃. Therefore we assume the van't Hoff factor of benzamide is 1 (since it does not ionize) and we get the cryoscopic constant (Kf) of X from ΔTf = Kf * m * i. Substituting into the formula and rearranging gives Kf = ΔTf / (m * i) = 2.7 ℃/(70.4 g/850 g) = 2.7 ℃/0.082824 = 32.6 ℃ kg/mol.
We then use this Kf to calculate the ΔTf calculated for ammonium chloride: ΔTf calculated = Kf * m * i (where we again assume i=1) = 32.6 ℃ kg/mol * (70.4 g/850 g) = 9.8 ℃. Finally we can calculate the van't Hoff factor for ammonium chloride using the original formula: i = ΔTf observed / ΔTf calculated = 9.9 ℃ / 9.8 ℃ = 1.01.
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B) 40.00
C) 160.0
D) 6,400
Will mark brainliest to correct answers
1. What happens in a chain reaction?
A) Products that start a new reaction are released.
B) Reactants that have two parts split
C) Products that are radioactive are lost
D) Radioactive reactants are deposited on control rods
2. Which type of coolant(s) usually is (are) used to remove heat from nuclear reactor core?
A) water only
B) liquid sodium only
C) liquid sodium or water
D) CFCs
3. A reaction that results in the combining of smaller atomic nuclei is ___.
A) chemical
B) fission
C) fussion
D) ionization
4. What does neutron absorption accomplish in a nuclear reactor?
A) It slows down the reaction.
B) It speeds up the reaction
C) It increases the rate of heat absorption
D) It recycles the fuel
Answer:
C B A D
Explanation:
Answer: The ionic characteristic of oxygen is that it tends to gain 2 electrons and form anions with a charge of -2 in order to achieve a stable electron configuration.
Explanation: