Volumen = constant => P/T = constant
T1 = 21°C + 273.15 = 294.15 k
P1 = 0.82 atm
T2 = - 3.5°C + 273.15 = 269.65 k
P2 = ?
P2 / 269.65k = 0.82atm / 294.15k
P2 = [0.82atm / 294.15k] * 269.65kg = 0.75 atm
You would need 67 if you add 56 and 89 = 145 and how i got 67. i went and subtract 212 and 145 = 67 and hope that answer you want if not well least i try
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!
Answer:
3853 g
Step-by-step explanation:
M_r: 107.87
16Ag + S₈ ⟶ 8Ag₂S; ΔH°f = -31.8 kJ·mol⁻¹
1. Calculate the moles of Ag₂S
Moles of Ag₂S = 567.9 kJ × 1 mol Ag₂S/31.8kJ = 17.858 mol Ag₂S
2. Calculate the moles of Ag
Moles of Ag = 17.86 mol Ag₂S × (16 mol Ag/8 mol Ag₂S) = 35.717 mol Ag
3. Calculate the mass of Ag
Mass of g = 35.717 mol Ag × (107.87 g Ag/1 mol Ag) = 3853 g Ag
You must react 3853 g of Ag to produce 567.9 kJ of heat
Answer: 2 mol
Explanation:
- According to the ideal gas law, One mole of an ideal gas at STP (standard temperature and normal pressure) occupies 22.4 liters.
- Using cross multiplication,
1 mol of (O2) → 22.4 L
? → 43.9 L
Therefore, the number of moles of oxygen in 43.9 L = (43.9 × 1)/ 22.4 = 1.96 mol≈ 2 mol..