Weak acid dissociates partially in water to form hydronium ion and conjugate base of the acid. Since the acid is weak the conjugate base is strong. Acetic acid is a weak acid which dissolves in water to form hydronium ion and acetate ions (conjugate base).
CH₃COOH + H₂O → H₃O⁺ + CH₃COO⁻
b. 3.0
c. 0.20
d. 20
1- 5 grams O2 / 1 gram C3H8
2- 5 moles O2 / 1 mole C3H8
3- 10 grams O2 / 11 grams C3H8
4- 10 moles O2 / 11 moles C3H8
Mole measure the number of elementary entities of a given substance that are present in a given sample. Therefore, 5 moles oxygen / 1 mole propane. The correct option is option C among all the given options.
The SI unit of amount of substance in chemistry is mole. The mole is used to measure the quantity or amount of substance. We know one mole of any element contains 6.022×10²³ atoms which is also called Avogadro number.
The balanced equation for the given chemical reaction can be given as
CH + 5O 3CO + 4HO
Stoichiometry represents the number of number of moles of reactant and products. The mole ratio between oxygen and propane is 5:1. the correct option isoption 2.
Therefore, 5 moles oxygen / 1 mole propane. The correct option is option C among all the given options.
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B. Amperes
C. Charge
D. Resistance
Answer:
The correct option is C
Explanation:
A battery is an electrochemical device that converts chemical energy into electrical energy. Electrochemical device contains two electrodes (anode and cathode). These two electrodes are placed in contact with an electrolyte. A spontaneous chemical reaction (redox reaction) causes electrical energy/current to be produced. This chemical reaction causes a potential difference (which produces electric current/charge). However, this charge/electric current causes a voltage difference.
In 6.3 moles of calcium, there are approximately 3.79 x 10²⁴ atoms. This is calculated using Avogadro's number (6.022 x 10²³), which provides the number of atoms in a mole.
To find the number of atoms in 6.3 moles of calcium, you need to use Avogadro's number, which is 6.022 x 10²³. This number, known as Avogadro's constant, is the number of atoms or molecules in one mole of any substance. Therefore, you would calculate the number of atoms as follows:
6.3 moles of calcium × (6.022 x 10²³ atoms/mole) = 3.79 x 10²⁴ calcium atoms.
So, 6.3 moles of calcium have approximately 3.79 x 10²⁴ calcium atoms.
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10.20 g C6H12O6 x (1 mole C6H12O6 / 180 g C6H12O6) = 0.0567 mole C6H12O6
Then convert grams of water to kilograms of water by dividing by 1000.
355 g / 1000 = .355 Kg of H2O
molality (m) = (0.0567 mole C6H12O6 / .355 Kg of H2O) = 0.160 m
Then plug it in to the formula Δtf = Kfm
Δtf = (-1.86 °C/m) x (0.160 m)
Δtf = -0.298 °C
Assume the CaCI2 dissolves completely in the water.. The density of water at 25.0°C is 0.997 g/L, and the KF of H20 is 1.858°C kg/mol.
Answer:
-4741 °C
Something is strange, because this is a weird number.
Explanation:
ΔT = Kf . m. i
That's the colligative property of freezing point depression.
Kf = Cyroscopic constant
m = molality (moles of solute in 1kg of solvent)
i = Van't Hoff factor (numbers of ions dissolved)
We assume 100% dissociation:
CaCl₂ → Ca²⁺ + 2Cl⁻ i = 3
ΔT = Freezing point of pure solvent - Freezing point of solution
Let's determine molality
Solute = CaCl₂
Moles of solute = 23.5 g . 1 mol/ 110.98 g = 0.212 moles
We determine the mass of solvent by density. Density's data is in g/L. We need to convert the volume from mL to L
250 mL . 1L / 1000 mL = 0.250 L
0.997 g/L = mass of water / volume of water → 0.997 g/L . volume of water = mass of water
0.997 g/L . 0.250L = 0.249 g
Now, we convert the mass of water from g to kg
0.249 g . 1 kg / 1000 g = 2.49×10⁻⁴ kg
Molality = mol/kg → 0.212 mol / 2.49×10⁻⁴ kg = 850.5 m
We replace data:
0°C - Freezing point of solution = 1.858 °C . kg /mol . 850.5 mol/kg . 3
Freezing point of solution = -4741 °C
Answer:
The freezing point of the solution is -4.74 °C
Explanation:
Step 1: data given
Mass of calcium chloride CaCl2 = 23.50 grams
Step 2: Calculate moles CaCl2
Moles CaCl2 = mass CaCl2 / molar mass CaCl2
Moles CaCl2 = 23.50 grams / 110.98 g/mol
Moles CaCl2 = 0.212 moles
Step 3: Calculate mass H2O
Density = mass / volume
Mass = density * volume
Mass H2O = 997 g/L * 0.250 L
MAss H2O = 249.25 grams
Step 3: calculate molality of the solution
Molality = moles CaCl2 / mass H2O
Molality = 0.212 moles / 0.24925 kg
Molality = 0.851 molal
ΔT = i*Kf*m
⇒with ΔT = the freezing point depression = TO BE DETERMNED
⇒ with i = the van't Hoff factor = 3
⇒with Kf = the freezing point depression constant of water = 1.858 °C /m
⇒with m = the molality = moles CaCl2 / mass water = 0.851 molal
ΔT = 3 * 1.858 * 0.851
ΔT = 4.74 °C
Step 4: Calculate the freezing point of the solution
ΔT = T (pure solvent) − T (solution)
ΔT = 0°C - 4.74 °C
The freezing point of the solution is -4.74 °C
NOTE: the density of water = 0.997 kg/L or 997 g/L