B) <span>Radiation moves from a warmer object to a cooler object</span>
That’s how you draw it , I guess just copy it ?
Answer:
pH = 5.35
Explanation:
Given 1.60 grams sodium acetate (NaOAc(aq))*** added to 50ml of 0.10M acetic acid (HOAc(aq)) solution.
Applying common ion effect keeping in mind that the addition of NaOAc provides the common-ion (OAc⁻).
HOAc(aq) ⇄ H⁺(aq) + OAc⁻(aq)
I 0.10m 1.32 x 10⁻³M ≈ ∅M* (1.6g/82.03g/mol) / 0.050L = 0.39M
C -x +x 0.39M + x ≈ 0.39M**
E 0.10M - x x 0.39M
≈ 0.10M
Ka = [H⁺][OAC⁻]/[HOAC] => [H⁺] = Ka·[HOAc] / [OAc⁻]
[H⁺] = (1.75 X 10⁻⁵)(0.10) / (0.39) = 4.5 x 10⁻⁶M
∴ pH = -log[H⁺] = -log(4.5 x 10⁻⁶) = -(-5.35) = 5.35
_______________________________________________
* [H⁺] before adding NaOAc = SqrRt(Ka · [HOAc]) = SqrRt(1.75 x 10⁻⁵· 0.10) = 1.32 x 10⁻³M. Since this concentration value is so small, the initial [H⁺] is assumed to be zero molar (∅M).
** The added [H⁺] is negligible and dropped in the ICE table. That is, adding ~[H⁺] in the order of 10⁻³M does not change the H⁺ ion concentration sufficiently to affect problem outcome and is therefore dropped in the ICE table.
*** Acetic Acid and Sodium Acetate are frequently written HOAc and NaOAc where the OAc⁻ anion is the acetate ion (CH₃COO⁻) for brevity.
Answer:The below file contains the answer
Explanation:
The mass of solute per 100 mL of solution is abbreviated as (m/v). Mass is not technically the same thing as weight, but the abbreviation (w/v) is also common. 262 grams of sucrose are needed to make 655 mL of a 40.0% (w/v) sucrose solution
<h3>Define Solute</h3>
A solute is a material that dissolves in a solution. The amount of solvent present in fluid solutions is greater than the amount of solute. The two most common examples of solutions in daily life are salt and water. Salt is the solute because it dissolves in water.
<h3>forms of ratios for product concentration or yield:-</h3>
- w/v:- Weight by volume or weight per volume are the terms used. Any solid compound's concentration in a liquid can be calculated using it. It is measurable in gm/ml.
- Weight by weight ratio is referred to as w/w.It is employed to determine the final yield of the compound obtained from the starting compound. as in —mg/—gm.
It provides the real yield of the substance or item.
- Volume/volume. It is used to specify a liquid's composition or percent in a liquid compound.
using w/v we can calculate the weight of sucrose:-
40.0% means 40 g sucrose/ 100 g solution
40.0g sucrose x (655/100)=grams of sucrose
262 grams of sucrose are needed to make 655 mL of a 40.0% (w/v) sucrose solution.
Learn more about Solute here:-
brainly.com/question/14397121
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