Answer: V= 3.13 L
Explanation: solution attached:
Use combine gas law equation:
P1 V1 / T1 = P2 V2/ T2
Derive to find V2
V2 = P1 V1 T2 / T1 P2
Convert temperatures in K
T1= 13.0°C + 273 = 286 K
T2= 22.5°C + 273 = 295.5 K
Substitute the values.
0.6137 g of KHP contains 1.086 × 10^21 acidic protons.
Number of moles of KHP = mass of KHP/molar mass of KHP
Molar mass of KHP = 204.22 g/mol
Mass of KHP = 0.6137 g
Number of moles of KHP = 0.6137 g/204.22 g/mol = 0.003 moles of KHP
Now, 1 each molecule of KHP contains 1 acidic proton.
For 0.003 moles of KHP there are; 0.003 × 1 × NA
Where NA is Avogadro's number.
So; 0.003 moles of KHP contains 0.003 × 1 × 6.02 × 10^23
= 1.086 × 10^21 acidic protons.
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A. The Average Ecological footprint will need to decrease . 6-11-18
The ore contains 55.4% calcium phosphate (related to the mineral apatite) so the amount of Ca3(PO4)2 is 55.4%x=1000g so x=1000/0.554= 1.805kg. Now for the % of P in this amount of calcium phosphate, use all the masses of the elements in Ca3PO4= Ca=40.078 x 3= 120.23 and (PO4)2= (30.974+64)2=189.95 (NB oxygen is 16 mass x 4 =64) so the total mass is 310.2 and we have 61.95 of P (Pmass x 2) so 61.95/3102.= 0.19 or 19% P. So of the 1.805 x 0.19= 0.34kg of phosphorus.
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!