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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Answer:
Explanation:
From the net ionic equation
Ba2+(aq) + SO42-(aq) ==> BaSO4(s) we see that 1 mole Ba2+ reacts with 1 mole SO42- to -> 1 mol BaSO4
Find moles of Ba2+ used: 0.250 moles/L x 0.0323 L = 0.008075 moles Ba2+
Find moles SO42- present: 0.008075 moles Ba2+ x 1 mol SO42-/1 mol Ba2+ = 0.008075 mol SO42-
Find mass of Na2SO4 present: 0.008075 mol SO42- x 1 mol Na2SO4/1 mol SO42- x 142.04 Na2SO4/mole = 1.14698 g = 1.15 g Na2SO4 (to 3 significant figures)
Answer:
The formal charge on nitrogen in 
is +1.
Explanation:
The structure of is as follows.
(In attachment)
From the structure, Nitrogen has no non bonding electrons. Nitrogen has four bonds and each bond corresponds to 2 electrons. Hence, nitrogen have eight bonding electrons and five valence electrons.
![Formal\,charge\,on\,nitrogen = 5-[0+ \frac{8}{2}]= +1](https://tex.z-dn.net/?f=Formal%5C%2Ccharge%5C%2Con%5C%2Cnitrogen%20%3D%205-%5B0%2B%20%5Cfrac%7B8%7D%7B2%7D%5D%3D%20%2B1)
Therefore, The formal charge on nitrogen in is +1.
The mass percentage is 15.1465%.
Answer:
586 kpa(kilopascal/1000 pascals)
Explanation:
given 1.24 atm(standard atmosphere), and 66.7 psi(pound force per square inch).
To find the total pressure we should use dalton's law of partial pressures which is the sum of the pressures of each individual gas.
then we convert them to pascals and divide by 1000 to get the measurement in kilopascal.
knowing that 1 atmosphere is proportional to around 14.696 psi. We can multiply our given measure of atm by that and sum it by psi like so. 1.24×14.6959 = 18.22298.
Then,
18.22298+ 66.7 = 84.92298
psi.
Since 1 psi is proportional to around 6894.76 pascals. 1 psi will be 68.9476 kilopascal. 84.92298 * 6.89476 = 585.523336 ≈ 586
