Answer:
The answer to the question is;
The concentration of the Solution #1 in terms of molarity is
0.16704X moles/litre.
Explanation:
Let the concentration of the stock solution be X moles/liter
Therefore, 83.52 ml of the stock solution contains
83.52×(X/1000) moles
Dilution of 83.52 ml of X to 500 ml gives solution 1 with a concentration of
500 ml of solution 1 contains 83.52×(X/1000) moles
Therefore 1000 ml or 1 litre contains 2×83.52×(X/1000) moles = 0.16704X moles/litre
The molarity of solution 1 is 0.16704X moles/litre.
If anything you would use a protractor but that’s not a answer.... so I would pick whatever relates to a protractor
Answer:
Mole fraction for solute = 0.1, or 10%
Molality = 6.24 mol/kg
Explanation:
22.3% by mass → In 100 g of solution, we have 22.3 g of HCOOH
Mass of solution = 100 g
Mass of solute = 22.3 g
Mass of solvent = 100 g - 22.3g = 77.7 g
Let's convert the mass to moles
22.3 g . 1mol/ 46 g = 0.485 moles
77.7 g. 1mol / 18 g = 4.32 moles
Total moles = 4.32 moles + 0.485 moles = 4.805 moles
Xm for solute = 0.485 / 4.805 = 0.100 → 10%
Molality → mol/ kg → we convert the mass of solvent to kg
77.7 g. 1 kg / 1000g = 0.0777 kg
0.485 mol / 0.0777 kg = 6.24 m
Answer:
The object placed in the water has a volume of 19 cm³
Explanation:
<u>Step 1: </u>Data given
volume of the cylinder before adding the object = 28 mL = 28 cm³
After adding an object with volume X the volume rises to 47 mL = 47 cm³
<u>Step 2:</u> Calculate the volume of the object
Volume of the object = Final volume - initial volume
Volume of the object = 47 cm³ (or 47 mL) - 28 cm³ ( or 28 mL) = 19 cm³ (or 19 mL)
The object placed in the water has a volume of 19 cm³
