<span>There
are a number of ways to express concentration of a solution. This includes
molarity and molality. Molarity is expressed as the number of moles of solute per volume of
the solution. MOlality is expressed as moles per kg solution.
5.25 mol H2SO4 / kg solution ( 1 kg / 1000 g ) ( 1.266 g / mL ) ( 1000 mL / 1L ) = 6.6 M H2SO4</span>
Answer:
Molarity: 0.522M
Percentage by mass: 2.36 (w/w) %
Explanation:
Formic acid, HCOOH reacts with NaOH as follows:
HCOOH + NaOH → NaCOOH + H₂O
To solve this question we must find the moles of NaOH added = Moles formic acid. Taken into account the dilution that was made we can find the moles -And molarity of formic acid and its percentage by mass as follows:
<em>Moles NaOH = Moles HCOOH:</em>
0.01580L * (0.1322mol / L) =0.002089 moles HCOOH
<em>Moles in the original solution:</em>
0.002089 moles HCOOH * (25mL / 10mL) = 0.005222 moles HCOOH
<em>Molarity of the solution:</em>
0.005222 moles HCOOH / 0.01000L =
<h3>0.522M</h3>
<em>Mass HCOOH in 1L -Molar mass: 46.03g/mol-</em>
0.522moles * (46.03g / mol) = 24.04g HCOOH
<em>Mass solution:</em>
1L = 1000mL * (1.02g / mL) = 1020g solution
<em>Mass percent:</em>
24.04g HCOOH / 1020g solution * 100
2.36 (w/w) %
Well ask yourself why don't we count it in moles and you should get your answer.
Answer:
Question 2: Na3PO4, KOH; Question 3: Na3PO4, KOH
Explanation:
Question 2
The reactants in a chemical equation are the species on the left side of the reaction arrow.
Thus the reactants are Na3PO4, KOH (sodium phosphate and potassium hydroxide).
Question 3.
The products in a chemical equation are the species on the right side of the reaction arrow.
Thus the products are NaOH, K3PO4 (sodium hydroxide and potassium phosphate).
