PH + pOH = 14
pH + 0.253 = 14
pH = 14 - 0.253
pH = 13.747
[ H+] = 10 ^ -pH
[ H+ ] = 10 ^- 13.747
[ H+ ] = 1.790x10⁻¹⁴ M
hope this helps!
Hi,
To solve the question, first of all we will find out the no. of moles of H2SO4 in 19 g of sulfuric acid.
As we know:
No . of moles = Mass/ Molar mass
No. of moles= 19 g/98.08
g
No. of moles= 0.1937
Now we know the no of moles of H2SO4 that will react with 2LiOH. We also know the molar equivalence of H2SO4 , and 2LiOH that will react.
So, the water that will be produced will be 2H2O and 1 Li2SO4 when H2SO4 that will react with 2LiOH.
0.1937 x 2x 18.01
=6.977
=6.98
Therefore, approximately 6.98 grams of water will be produced from 19 g of sulfuric acid.
Hope it helps!
Answer:
it will float if the object is 1g/cm^3(water 's density ) because it is less dense
Answer:
32.8%
Explanation:
All of the Pb⁺² species precipitated as lead(II) cromate, PbCrO₄ (we know this as excess K₂CrO₄ was used).
First we convert 0.130 g of PbCrO₄ into moles, using its molar mass:
- 0.130 g ÷ 323 g/mol = 4.02x10⁻⁴ mol PbCrO₄
There's 1 Pb⁺² mol per PbCrO₄ mol, so in total 4.02x10⁻⁴ moles of Pb⁺² were in the ethanoate sample.
We <u>convert those 4.02x10⁻⁴ moles of Pb into grams</u>:
- 4.02x10⁻⁴ mol * 207 g/mol = 0.083 g Pb
Finally we calculate the percentage composition of Pb:
- 0.083 g Pb / 0.254 g salt * 100% = 32.8%
