Answer:
Explanation: Hydrocarbons are the compounds which contain only carbon and hydrogen as their constituent elements. These are considered to be as the non polar species as because of the less electronegativity difference between the two components.
Now as it is a non polar species thus it will dissolve only a non polar solute following the principle of 'Like dissolves Like'. Thus the most non polar species given in the options is pentane. Cyclohexane will definitely dissolve this pentane as both are non polar in nature.
, HI and NaBr are polar specie which cannot be dissolved in non polar species.
Answer:
The temperature difference of the body after 3 hours = 5.16 K
Explanation:
we know that the number of moles of O₂ inhaled are 0.02 mole/min⁻¹
or, 1.2 mole.h⁻¹
The average heat evolved by the oxidation of foodstuffs is then:
⇒ Q avg = = 7.2 kj.h⁻¹.Kg⁻¹
the heat produced after 3 h would be:
= 7.2 kj. h⁻¹.Kg⁻¹ x 3 h
= 21.6 kj. kg⁻¹
= 21.6 x 10³ j kg⁻¹
We know Qp = Cp x ΔT
Assume the heat capacity of the body is 4.18 J g⁻¹K⁻¹
⇒ ΔT =
⇒ ΔT =
⇒ ΔT = 5.16 K
Explanation:
As it is known that there are two types of properties. These are extensive and intensive.
Extensive properties : Properties that depend on the size or amount of system. For example, mass, volume etc.
Intensive properties : Properties that do not depend on the size or amount of system. For example, density, melting point, specific heat capacity etc.
On the basis of these properties water and ethanol are distinguished as follows.
Thus, we can conclude that properties like density, melting point, specific heat capacity can help a chemist distinguish between ethanol and water.
Answer:
0.03g/mL
Explanation:
Given parameters include:
Five μL of a 10-to-1 dilution of a sample; This implies the Volume of dilute sample is given as 5 μL
Dilution factor = 10-to-1
The absorbance at 595 nm was 0.78
Mass of the diluted sample = 0.015 mg
We need to first determine the concentration of the diluted sample which is required in calculating the protein concentration of the original solution.
So, to determine the concentration of the diluted sample, we have:
concentration of diluted sample =
= (where ∝ was use in place of μ in the expressed fraction)
= 0.003 mg/μL
The dilution of the sample is from 10-to-1 indicating that the original concentration is ten times higher; as such the protein concentration of the original solution can be calculated as:
protein concentration of the original solution = 10 × concentration of the diluted sample.
= 10 × 0.003 mg/μL
= 0.03 mg/μL
= 0.03g/mL
Hence, the protein concentration of the original solution is known to be 0.03g/mL