Answer:
The answer to your question is Volume = 11.4 L
Explanation:
Data
Volume 1 = V1 = 6 L
Pressure 1 = P1 = 1 atm
Temperature 1 = T1 = 22°C
Volume 2 = V2 = ?
Pressure 2 = 0.45 atm
Temperature 2 = -21°C
Process
1.- Convert temperature (°C) to °K
T1 = 273 + 22 = 295°K
T2 = 273 + (-21) = 252°K
2.- Use the combined gas law to solve this problem
P1V1 / T1 = P2V2 / T2
-Solve for V2
V2 = P1V1T2 / T1P2
-Substitution
V2 = (6)(1)(252) / (295)(0.45)
- Simplification
V2 = 1512 / 132.75
- Result
V2 = 11.38 L
Answer:
you can use the idea of molecular masses to calculate easily
The balanced chemical reaction is:
<span>Ca + Cl2 = CaCl2
</span>
We are given the amount of calcium metal to be used for this reaction. This will be the starting point for the calculations.
56 g Ca ( 1 mol Ca / 40.08 g Ca) (1 mol Cl2 / 1 mol Ca) ( 22.414 L Cl2 / 1 mol Cl2 ) = 31.32 L Cl2 gas produced from the reaction
If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
<h3>
What is base dissociation constant?
</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 2.8× 10^(-9)
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) /{2.8×10^(-9) }
= 3.5× 10^(-6)
Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
DISCLAIMER: The above question have mistake. The correct question is given as
Question:
Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?
learn more about base dissociation constant:
brainly.com/question/9234362
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