Answer: Within each element square, information on the element's symbol, atomic number, atomic mass, electronegativity, electron configuration, and valence numbers can be found. At the bottom of the periodic table is a two row block of elements that contain the lanthanoids and actinides.
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:
D
Explanation:
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Answer:
mm = 1043.33 g/mol
Explanation:
osmotic pressure (π):
∴ π = 17.8 torr = 0.0234 atm
∴ Cb: solute concentration
∴ T = 25°C = 298 K
∴ R = 0.082 atm.L/K.mol
⇒ Cb = π/RT
⇒ Cb = (0.0234 atm)/((0.082 atm.L/K.mol)(298 K))
⇒ Cb = 9.585 E-4 mol/L
molar mass (mm):
⇒ mm = (1.00 g/L)(L/9.585 E-4 mol)
⇒ mm = 1043.33 g/mol
Answer:
The net ionic equation for the given reaction :
Explanation:
...[1]
..[2]
...[3]
Replacing 
, NaI and 
in [1] by usig [2] [3] and [4]
Removing the common ions present ion both the sides, we get the net ionic equation for the given reaction [1]:
