Partial pressure is the amount of pressure or force that is exerted by the atoms into the outer environment. it is dependent on the temperature and pressure of the present surroundings. in this case, we are asked in this problem to determine the partial pressure of oxygen at 16oC and 1 atm. We have to look into a solubility data table commonly found in handbooks and determined via experiments and correlations. According to literature, the value of the partial pressure is equal to 0.617 mM.This is under the assumption that the salinity of the water in which oxygen is dissolved is equal to zero.
Answer:
0.192 mol.
Explanation:
- To calculate the no. of moles of a substance (n), we use the relation:
<em>n = mass / molar mass.</em>
mass of AsH₃ = 15.0 g.
molar mass of AsH₃ = 77.95 g/mol.
∴ The number of moles in 15.0 g AsH₃ = mass / molar mass = (15.0 g) / (77.95 g/mol) = 0.192 mol.
Answer:
8kJ/mol
Explanation:
since the forward reaction is -8kJ/mol, the backward reaction has the same enthalphy but reversed
The simplified solubility of glucose at 30°C is 1.25 g/g of water. Considering that the density of water at 30°C is 1 g/mL, the equivalent mass of 400 mL of water is also 400g.
The concentration of the solution in water is,
550 g/400g of water = 1.375 g glucose / g of water
Since the concentration is higher compared to the solubility of glucose at the specified temperature, it can be said that the solution is SATURATED.
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!
