Considering ideal gas behavior, the volume of 1 mol of gas at STP is 22.4 L; then the volume occupied by 1.9 moles is 1.9mol*22.4L/mol
The answer is 43 L if I am correct.
H3O4 is your answer but u should use your chart
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
<span>
this is because nonmetals. the elements that form covalent bonds want to share the electrons to creat an ocet when this happens the positively charged nucleus has and attraction for the valence, or outer most, electrons of its own atoms as well as for the electrons in the atom that it is trying to share electrons with
</span>
Answer:
(4) Water, stirring, and filtering
Explanation:
The added water will dissolve the sugar but not the sand. When you filter the mixture, the sand will be trapped in the filter paper and the dissolved sugar will pass through the pores of the paper.
(1) and (2) are wrong. You don't separate a mixture of sugar and sand by adding more sand. Furthermore, neither substance will boil at 100 °C.
(3) is wrong. You can dissolve the sugar in water but, if you boil the water away, the sugar and sand will still be together.