Answer:
Option c → Tert-butanol
Explanation:
To solve this, you have to apply the concept of colligative property. In this case, freezing point depression.
The formula is:
ΔT = Kf . m . i
When we add particles of a certain solute, temperature of freezing of a solution will be lower thant the pure solvent.
i = Van't Hoff factor (ions particles that are dissolved in the solution)
At this case, the solute is nonvolatile, so i values 1.
ΔT = Difference between fussion T° of pure solvent - fussion T° of solution.
T° fussion paradichlorobenzene = 56 °C
T° fussion water = 0°
T° fussion tert-butanol = 25°
Water has the lowest fussion temperature and the paradichlorobenzene has the highest Kf. But the the terbutanol, has the highest Kf so this solvent will have the largest change in freezing point, when all the molalities are the same.
<u>Answer:</u> The expression for 
is written below.
<u>Explanation:</u>
Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of the products and the reactants each raised to the power their stoichiometric ratios. It is expressed as
For a general chemical reaction:
The expression for is written as:
The partial pressure for solids and liquids are taken as 1.
For the given chemical equation:
The expression for for the following equation is:
The partial pressure of 
will be 1 because it is solid.
So, the expression for now becomes:
Hence, the expression for is written above.
Answer:
C The water had adequate nitrogen and phosphorus, so it is likely iron limited.
Explanation:
Phytoplankton are single- cell organisms that live in oceans.
They require nitrogen, phosphorus and trace amount of iron to survive.
From the scientist's results after testing the water for nitrogen and phosphorus,there are reasonable amount of these elements.
Therefore insufficient iron in the water is the reason why he could find plenty phytoplankton in the ocean.
<span>
• </span>Volume of the marshmallow:
V = 2.75 in^3 (but, 1 in^3 = 16.39 cm^3)
V = 2.75 × 16.39 cm^3
V = 2.75 × 16.39 cm^3
V = 45.0725 cm^3
• Density:
d = 0.242 g/cm^3
<span>• </span>Mass:
m = d × V
m = (0.242 g/cm^3) × (45.0725 cm^3)
m = (0.242 g/cm^3) × (45.0725 cm^3)
m = 10.907545 g
m ≈ 10.9 g <——<span>— this is the answer.
I hope this helps. =)
</span>
Hey there!:
From the given data ;
Reaction volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )
Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:
[E]t in molar = g/L * mol/g
[E]t = 0.01 g/L * 1 / 45,000
[E]t = 2.22*10⁻⁷
Vmax = 0.758 umole/min/ per mL
= 758 mmole/L/min
=758000 mole/L/min => 758000 M
Therefore :
Kcat = Vmax/ [E]t
Kcat = 758000 / 2.2*10⁻⁷ M
Kcat = 3.41441 *10¹² / min
Kcat = 3.41441*10¹² / 60 per sec
Kcat = 5.7*10¹⁰ s⁻¹
Hence kcat of xyzase is 5.7*10¹⁰ s⁻¹
Hope that helps!
