Answer:
1.6 L is the volume of NaOH that has reacted
Explanation:
The balanced reaction is:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
This is a neutralization reaction between a strong acid and a strong base. The products are the correspond salt and water.
We propose this rule of three:
1 mol of sulfuric acid needs 2 mol of NaOH to react to react
Then, 2.4 moles of H₂SO₄ will react with (2.4 . 2) / 1 = 4.8 moles of NaOH
As molarity is 3M, we can determine the volume of our solution
Molarity (M) = mol / volume(L) → Volume(L) = mol / Molarity
Volume(L) = 4.8 mol / 3 M = 1.6 L
Answer:
Depends, but in most cases, 2.
It's best to use as many digits as possible to keep it accurate.
Explanation:
This varies between teachers, as most schools go with 2 decimal places.
This is something that depends in your situation.
You technically want as many decimals as possible to keep it as accurate, but most people stick with 2.
I personally do 3, and commonly do 5 sometimes.
Answer:
250000 μL
Explanation:
If 1 L = 1000 mL
Then X L = 250 mL
X = (1 × 250) / 1000 = 0.25 L
Now we can calculate the number of microliters (μL) in 0.25 L:
if 1 μL = 10⁻⁶ L
then X μL = 0.25 L
X = (1 × 0.25) / 10⁻⁶ =250000 μL
Ok first, we have to create a balanced equation for the dissolution of nitrous acid.
HNO2 <-> H(+) + NO2(-)
Next, create an ICE table
HNO2 <--> H+ NO2-
[]i 0.230M 0M 0M
Δ[] -x +x +x
[]f 0.230-x x x
Then, using the concentration equation, you get
4.5x10^-4 = [H+][NO2-]/[HNO2]
4.5x10^-4 = x*x / .230 - x
However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable,
assume 0.230-x ≈ 0.230
4.5x10^-4 = x^2/0.230
Then, we solve for x by first multiplying both sides by 0.230 and then taking the square root of both sides.
We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.01M.
Then to find percent dissociation, you do final concentration/initial concentration.
0.01M/0.230M = .0434 or
≈4.34% dissociation.
Answer:
Recall that, at the boiling point, we observe that both liquid and gas are at equilibrium with one another. This is true at every combination of applied pressure and boiling point temperature. ... The applied pressure will be greater than the vapor pressure, and all of the gas will condense into the liquid
