Answer:
57 grams of H3PO4
Explanation:
M= moles/ liters
convert mL to L
234 mL x 1L/1000mL = 0.234L
Rearrange the Molarity formula to solve for moles.
moles= MxL
moles= 2.5M x 0.234L
moles= 0.585 mol
Use the molar mass of H3PO4 to get to grams
0.585 mol x 97.994 grams/1 mol = 57.326 grams of H3PO4
round to two sig figs for 57 grams
Answer:
This is my own explanation:
Explanation:
It is important to know the mixtures you input together because you may determine the type of substance you produced from different elements and several combinations of matter. This assists in identifying your specific substance.
Answer:
if you want to have answers, don't take it wrong ...but put more details !
Explanation:
I think you divide something
a. AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)
b. Ksp AgBr = s²
c. 5 x 10⁻¹³ mol/L
<h3>Further explanation</h3>
Given
solubility AgBr = 7.07 x 10⁻⁷ mol/L
Required
The dissolution reaction
Ksp
The solubility product constant
Solution
a. dissolution reaction of AgBr
AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)
b. Ksp
Ksp AgBr = [Ag⁺] [Br⁻]
Ksp AgBr = (s) (s)
Ksp AgBr = s²
c. Ksp AgBr = (7.07 x 10⁻⁷)² = 5 x 10⁻¹³ mol/L
