2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O
v(H₂SO₄)=0.25 L
c(H₂SO₄)=2.00 mol/L
v(NaOH)=2.00 L
n(H₂SO₄)=c(H₂SO₄)v(H₂SO₄)
n(NaOH)=2n(H₂SO₄)=2c(H₂SO₄)v(H₂SO₄)
c(NaOH)=n(NaOH)/v(NaOH)=2c(H₂SO₄)v(H₂SO₄)/v(NaOH)
c(NaOH)=2*2.00*0.25/2.00=0.5 mol/L
the concentration of the NaOH solution is 0.5 mol/L
The resulting temperature is 46.5°C.
Details:
According to Charles's law, the volume of gas, while maintaining constant pressure, correlates directly with temperature in Kelvin.
The formula representing Charles's law is expressed as follows:
We need to determine T2, thus:
V1 = 736 ml = 0.736 L
T1 = 15 ° C
V2 = 2.28 L
Substituting the values gives us:
T2 = 
= 46.5°C
It is evident that as the volume increases, the temperature also rises.
First convert grams of C4H10 to moles using its molar mass of 58.1 g/mol:
3.50 g C4H10 × (1 mol C4H10 / 58.1 g C4H10) = 0.06024 mol C4H10
Next convert moles to molecules using Avogadro’s number:
0.06024 mol C4H10 × (6.022×10^23 molecules C4H10 / 1 mol C4H10) = 3.627×10^22 molecules C4H10
Each butane molecule contains 4 carbon atoms, so:
3.627×10^22 molecules C4H10 × (4 atoms C / 1 molecule C4H10) = 1.45×10^23 carbon atoms present.
I hope you can decipher it and that it's also accurate. Wishing you all the best.
