Answer:
Following are the responses to the given choices:
Explanation:
- The RBC crenation is implied through NaCl by 2,67 percent(m/v) because that solution becomes hypertonic to RBC because of the water within the RBC that passes externally towards the outskirts. RBC thus shrinks.
- 1.13% (m/v), because the low concentration or osmotic that all this solution shows is hypotonic regarding RBC because of the water which has reached the resulting swelling in RBC.
- Distilled H2 implies hemolytic distillation.
- Glucose is indicated by crenation at 8.69 percent (m/v).
- 5.0% (m/v) glucose and 0.9% (m/v) (Crenation is indicated by NaCl.v)
Mass = 5 g
volume = 20 cm³
density = mass / volume
therefore:
D = m / V
D = 5 / 20
D = 0.25 g/cm³
<span>A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60. g/mol.
</span>Q1)
Empirical formula is the simplest ratio of whole numbers of components making up a compound.
the percentages have been given, therefore we can calculate for 100 g of the compound.
C H O
Mass in 100 g 40.0 g 6.7 g 53.5 g
Molar mass 12 g/mol 1 g/mol 16 g/mol
Number of moles 40.0/12= 3.33 6.7/1 = 6.7 53.5/16 = 3.34
Divide by the least number of moles
3.33/3.33 = 1 6.7/3.33 = 2.01 3.34/3.33 = 1.00
after rounding off
C - 1
H - 2
O - 1
Empirical formula - CH₂O
Q2)
Molecular formula is the actual number of components making up the compound.
To find the number of empirical units we have to find the mass of one empirical unit.
Mass of one empirical unit = CH₂O - 12 + (1x2) + 16 = 30 g
Mass of one mole of compound = 60 g
Number of empirical units = 60 g / 30 g = 2
Therefore molecular formula - 2(CH₂O)
Molecular formula - C₂H₄O₂
Answer: A. Cilla Is Correct.