First we have to find moles of C:
Molar mass of CO2:
12*1+16*2 = 44g/mol
(18.8 g CO2) / (44.00964 g CO2/mol) x (1 mol C/ 1 mol CO2) =0.427 mol C
Molar mass of H2O:
2*1+16 = 18g/mol
As there is 2 moles of H in H2O,
So,
<span>(6.75 g H2O) / (18.01532 g H2O/mol) x (2 mol H / 1 mol H2O) = 0.74mol H </span>
<span>Divide both number of moles by the smaller number of moles: </span>
<span>As Smaaler no moles is 0.427:
So,
Dividing both number os moles by 0.427 :
(0.427 mol C) / 0.427 = 1.000 </span>
<span>(0.74 mol H) / 0.427 = 1.733 </span>
<span>To achieve integer coefficients, multiply by 2, then round to the nearest whole numbers to find the empirical formula:
C = 1 * 2 = 2
H = 1.733 * 2 =3.466
So , the empirical formula is C2H3</span>
For a p type of semiconductor we need a dopant which is from 13th group in periodic table
Al , B, Ga, In Tl
So the correct element will be In : Indium
The other elements belongs to 15th group and hence will give n type semiconductor
Answer: <u>It is, as all stars are, a hot ball of gas made up mostly of Hydrogen. The Sun is so hot that most of the gas is actually plasma, the fourth state of matter. ... As we heat up liquid, the liquid turns to gas. Gas is the third state of matter</u>
Explanation:
Answer:
[H⁺] = 6.083x10⁻⁴ M, [C₆H₅OO⁻] = 6.083x10⁻⁴ M, [C₆H₅OOH] = 3.98x10⁻³M, pH = 3.22
Explanation:
Data: we have 0.56 gr of benzoic acid, disolved in 1Lt of water. Kₐ = 6.4x10⁻⁵
M (molar mass) of BA (Benzoic Acid) = 122 g/mol
Then, the inicial concentration is 0.56/122 = 4.59x10⁻³ M
We should consider the equation once it reaches the equilibrium:
C₆H₅COOH ⇄ C₆H₅COO⁻ + H⁺
C - x x x
And, for the Kₐ:
Kₐ = [H⁺][C₆H₅COO⁻]/[C₆H₅COOH] = x²/(C-x) , where C = 4.59x10⁻³
Then: x² + Kₐx - KₐC = 0
x² + 6.4x10⁻⁵ - 2.9x10⁻⁷ = 0
Resolving this cuadratic equation (remember to use Baskara equation), we obtain:
x = 6.083x10⁻⁴ M
Then: [H⁺] = [C₆H₅COO⁻] = 6.083x10⁻⁴ M
[C₆H₅COOH] = C - x = 3.98x10⁻³ M
pH = -Log [H⁺] = 3.22
