If Ka for HCN is 6. 2×10^−10 at 25 °C, then the value of Kb for cn− at 25 °C is 1.6 × 10^(-5).
<h3>What is base dissociation constant? </h3><h3 />
The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.
The dissociation reaction of hydrogen cyanide can be given as
HCN --- (H+) + (CN-)
Given,
The value of Ka for HCN is 6.2× 10^(-10)
The correlation between base dissociation constant and acid dissociation constant is
Kw = Ka × Kb
Kw = 10^(-14)
Substituting values of Ka and Kw,
Kb = 10^(-14) /{6.2×10^(-10) }
= 1.6× 10^(-5)
Thus, the value of base dissociation constant at 25°C is 1.6 × 10^(-5).
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Answer:
Explanation:
The number of moles of solute is equal to product of the molar concentration (molarity) and the volume (in liters) of solution.
Since the volumes and the molar concentrations of the<em> NaOH </em>and <em>HCl </em>solutions mixed are equal, each one of them contributes the same number of moles of solute.
Since every mol of NaOH produces one mol of OH⁻ ions and every mol of HCl produces one mol of H⁺ ion, the number of moles of OH ⁻ and H⁺ in solution are equal.
Thus, OH⁻ and H⁺ ions will be neutralized by the reaction:
- OH⁻ (aq) + H⁺ (aq) ⇄ H₂O (l)
Which is strongly shifted to the right and has <em>neutral pH</em>.
Hence, you conclude that the approximate <em>pH of the solution is neutral.</em>
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Answer:
1.88 × 10²² Molecules of CO
Explanation:
At STP for an ideal gas,
Volume = Mole × 22.4 L/mol
Or,
Mole = Volume / 22.4 L/mol
Mole = 0.7 L / 22.4 L/mol
Mole = 0.03125 moles
Now,
No. of Molecules = Moles × 6.022 × 10²³ Molecules/mol
No. of Molecules = 0.03125 × 6.022 × 10²³ Molecules/mol
No. of Molecules = 1.88 × 10²² Molecules of CO
Answer:
Explanation:
We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.
The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.
- ΔT= final temperature - initial temperature
The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.
- ΔT= 88.9 °C - 58.8 °C = 30.1 °C
Now we know three variables:
- Q= 4500.0 J
- c= 0.4494 J/g°C
- ΔT = 30.1 °C
Substitute these values into the formula.
Multiply on the right side of the equation. The units of degrees Celsius cancel.
We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g
The units of Joules cancel.
The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.
The mass of the sample of metal is approximately <u>333 grams.</u>
