Answer:
C₅ H₁₂ O
Explanation:
44 g of CO₂ contains 12 g of C
30.2 g of CO₂ will contain 12 x 30.2 / 44 = 8.236 g of C .
18 g of H₂O contains 2 g of hydrogen
14.8 g of H₂0 will contain 1.644 g of H .
total compound = 12.1 out of which 8.236 g is C and 1.644 g is H , rest will be O
gram of O = 2.22
moles of C, O, H in the given compound = 8.236 / 12 , 2.22 / 16 , 1.644 / 1
= .6863 , .13875 , 1.644
ratio of their moles = 4.946 : 1 : 11.84
rounding off to digits
ratio = 5 : 1 : 12
empirical formula = C₅ H₁₂ O
Answer:
Y = 92.5 %
Explanation:
Hello there!
In this case, since the reaction between lead (II) nitrate and potassium bromide is:
Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:
Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:
And the resulting percent yield:
Regards!
Answer:
Higher boiling point and lower freezing point
Explanation:
Boiling point is the temperature at which substances goes to gaseous phase form their liquid phase. Liquids boil when vapour pressure of the liquid becomes equal to atmospheric pressure. So boiling point of a liquid is affected by the change in its vapour pressure and vapour pressure is affected by adding a non-volatile solute in it.
Effect of adding non -volatile solute on vapour pressure:
When a non-volatile solute is added to a liquid, it alters the interaction between the liquid particles and prevents the liquid particles to going into gaseous phase which results in decrease in vapour pressure. More energy is needed for the vapour pressure of the liquid to becomes equal to atmospheric pressure. Therefore, boiling point increases.
CaCl2 is a non-volatile solute and hence increases the boiling point of the water.
Effect of adding non-volatile solute on freezing point:
When a non-volatile solute is added to a liquid, it alters the interaction between the liquid molecules. The new solute and solvent interaction prevents the liquid to soldify and requires more lowering in temperature. Therefore, freezing point decreases.
Answer:
129 units cubed
Explanation:
There are ten spaces between 120 and 130, so we can infer that each line represents a change in volume of 1 unit. The volume of the substance in the graduated cylinder looks like it's 1 under 130, so its volume is 129 units.
Hopefully this was helpful! :)
