Answer:
188.5g of dextrose are needed
Explanation:
In Weight per volume percentage - %(w/v) -, the concentration is defined as the mass of solute in grams -In this case, dextrose-, in 100mL of solution.
As you want to prepare 725mL of a 26.0% (w/v) solution. you need:
725mL * (26g / 100mL) = 188.5g of solute =
<h3>188.5g of dextrose are needed</h3>
Answer:
Explanation:
THE CORECT QUESTION
A 50.0 mL solution of 0.127 M KOH is titrated with 0.254 M HCl. Calculate the pH of the solution after the addition of each of the given amounts of HCl.
SOLUTION
Get the concentration of the HCl first using titration formula
CA X V A / CB VB = NA/ NB
Equation of reation; KOH + HCl => KCl + H2O
CA = 0.254 M
CB = 0.127
VA = 1/0.254 = 3.937
CA (after the addition) = 0.127 x 50 / 3.937
= 1.612 M
But pH = - Log[hydrogen ion]
= -log 1.612
=
Answer:
160 gm
Explanation:
Five times as much water means you can dissolve 5 times as much potassium nitrate 5 x 32 = 160 gm <u> <===== this seems unlikely though as I doubt 32 g of potassium nitrate will dissolve in only 1 cm^3 of water 1 cm^3 of water is only 1 gm of water </u>
Delta H = q / mass * delta temperature
Answer:
The volume of the system will change and maybe the shape of the system boundary
