The chemical equation is:
<span>HCO3^ -> H^+ + CO3^2- </span>
We know that the formula
for Ka is:
Ka = [H^+][CO3^2-]/[HCO3^-]
log Ka = log[H^+] + log[CO3^2-]/[HCO3^-]
pKa = pH - log[CO3^2-]/[HCO3^-]
log[CO3^2-]/[HCO3^-] = pH - pKa = 10.75 - 10.329 = 0.421
<span>[CO3^2-]/[HCO3^-] = Antilog (0.421) = 2.636 </span>
Answer:
<span>2.636</span>
Answer:
[CH₃COOH] = 1.70 M
Explanation:
When we talk about concentration we can determine Molarity
Molarity determines the moles of solute that are contained in 1L of solution.
In this case our solute is the acetic acid.
M = mol/L
M = 0.99 mol /0.58L → 1.70 M
We can also make a rule of three
In 0.58 L of solution we have 0.99 moles of solute
In 1 L of solution we may have (1 . 0.99) / 0.58 = 1.70 moles
Acetic acid is a weak acid, partially dissociated in water.
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
Production of materials and transportation are the examples of three carbon emission.
Extraction and production of purchased materials and transportation of purchased fuels are the examples of three carbon emission. Scope 3 emissions refers to all indirect emissions that occur in the chain of the reporting company that is included in both upstream and downstream emissions.
Big machineries are used for the production and extraction of materials as well as the transportation requires fossil fuels for working which releases carbondioxide gas in the atmosphere so we can conclude that production of materials and transportation are the examples of three carbon emission.
Answer:
Answer : The number of atoms present in 159 g of calcium are
Explanation :
First we have to calculate the moles of calcium.
Molar mass of calcium = 40 g/mole
Now we have to calculate the number of atoms of calcium.
As, 1 mole of calcium contains number of atoms of calcium
So, 3.975 mole of calcium contains number of atoms of calcium
Therefore, the number of atoms present in 159 g of calcium are
Explanation:
Answer:
The answer to your question is V2 = 825.5 ml
Explanation:
Data
Volume 1 = 750 ml
Temperature 1 = 25°C
Volume 2= ?
Temperature 2 = 55°C
Process
Use the Charles' law to solve this problem
V1/T1 = V2/T2
-Solve for V2
V2 = V1T2 / T1
-Convert temperature to °K
T1 = 25 + 273 = 298°K
T2 = 55 + 273 = 328°K
-Substitution
V2 = (750 x 328) / 298
-Simplification
V2 = 246000 / 298
-Result
V2 = 825.5 ml
