Answer:
Electricity
Explanation:
Nonmetals are defined as elements that generally cannot conduct electricity
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.
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The average rate of reaction over a given interval can be calculated by taking the difference of concentration on a particular given reactant, and dividing it by the total time. In this case, (1.00 M - 0.655 M)/30 s = 0.0115 M/s, or 0.0115 mol/L-s, and this is the final rate of reaction.
Answer: decrease the temperature of the reaction
Explanation:
CH3COOH + CH3CH2OH ⇌ CH3COOCH2CH3 + H2O
The formation of an ester is an exothermic reaction. Therefore, a decrease in the temperature of the reaction will favour the production of ester.
Answer:
356.484 g.
Explanation:
- Molarity (M) is defined as the no. of moles of solute (NaCl) dissolved in a 1.0 L of the solvent.
<em>M = (no. of moles of NaCl)/(volume of the solution (L))</em>
M = 6.10 M, volume of the solution = 1.0 L.
∴ No. of moles of NaCl = (M)(volume of the solution (L)) = (6.10 M)(1.0 L) = 6.10 mol.
<em>∵ no. of moles of NaCl = (mass of NaCl)/(molar mass of NaCl)</em>
∴ Mass of NaCl = (no. of moles of NaCl)(molar mass of NaCl) = (6.10 mol)(58.44 g/mol) = 356.484 g.
