Answer:
2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)
Explanation:
K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)
The complete ionic equation for the above equation can be written as follow:
In solution, K2CO3 and CuF will dissociate as follow:
K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)
CuF(aq) —› Ca^2+(aq) + 2F¯(aq)
Thus, we can write the complete ionic equation for the reaction as shown below:
K2CO3(aq) + 2CuF(aq) —›
2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)
Answer:
9.1
Explanation:
Step 1: Calculate the basic dissociation constant of propionate ion (Kb)
Sodium propionate is a strong electrolyte that dissociates according to the following equation.
NaC₃H₅O₂ ⇒ Na⁺ + C₃H₅O₂⁻
Propionate is the conjugate base of propionic acid according to the following equation.
C₃H₅O₂⁻ + H₂O ⇄ HC₃H₅O₂ + OH⁻
We can calculate Kb for propionate using the following expression.
Ka × Kb = Kw
Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.3 × 10⁻⁵ = 7.7 × 10⁻¹⁰
Step 2: Calculate the concentration of OH⁻
The concentration of the base (Cb) is 0.24 M. We can calculate [OH⁻] using the following expression.
[OH⁻] = √(Kb × Cb) = √(7.7 × 10⁻¹⁰ × 0.24) = 1.4 × 10⁻⁵ M
Step 3: Calculate the concentration of H⁺
We will use the following expression.
Kw = [H⁺] × [OH⁻]
[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴/1.4 × 10⁻⁵ = 7.1 × 10⁻¹⁰ M
Step 4: Calculate the pH of the solution
We will use the definition of pH.
pH = -log [H⁺] = -log 7.1 × 10⁻¹⁰ = 9.1
Answer:
the average molar mass of this air sample can be calculated as
addition of the product of the average molar weights of the component gases and their percentage compositions
1. Average Molar mass of Air = 0.7803 x 28 + 0.2099 x 32 + 0.00033 x 44 = 28.58g/mol
2. The partial pressures of N2, O2, and CO2 in atm.
From Ideal gas law, at stp, Volume of air V=22.4L/mol
PV =nRT
Since, at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas
Total Pressure P=1atm
Partial Pressure p = mol fraction x P
Volume of N2 = 0.7803 x 22.4L = 17.47L, Partial Pressure = 0.7803atm
Volume of O2 = 0.2099 x 22.4L = 4.68L Partial Pressure = 0.209atm
Volume of CO2 = 0.00033 x 22.4L = 0.00739L, Partial Pressure = 0.033atm
C) change to water at the same temperature
Explanation:
Adding 334Joules of heat to one gram of ice at STP will cause ice to change to water at the same temperature.
- The heat of fusion is the amount of energy needed to melt a given mass of a solid
- It is also conversely the amount of energy removed from a substance to freeze it.
- The addition of this energy does not cause a decrease or increase in temperature.
- Only a phase change occurs.
Learn more:
Heat of fusion brainly.com/question/4050938
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Answer:
1. The product has a higher Rf value on a silica gel TLC plate because it is more polar than the starting methyl benzoate.
2. False
3. True
Explanation:
In chromatography, there is a stationary phase and a mobile phase. The ratio of the distance moved by a component and the distance moved by the solvent gives the retention factor (Rf).
Since silica gel is a polar solvent, it will retain the more polar product methyl m-nitrobenzoate compared to the methyl benzoate starting material.
In comparing the electrophillic aromatic substitution of m-nitrobenzoate and methyl benzoate, we must remember that the presence of electron withdrawing groups (such as -NO2 and -CHO) on the aromatic compound deactivates the compound towards electrophillic aromatic substitution hence, methyl m-nitrobenzoate is less reactive than methyl benzoate in Electrophilic Aromatic Substition and Methyl benzoate is less reactive than benzene in Electrophilic Aromatic Substition
