Answer:
a. 2 HgO(s) ⇒ 2 Hg(l) + O₂(g)
b. 0.957 g
Explanation:
Step 1: Write the balanced equation
2 HgO(s) ⇒ 2 Hg(l) + O₂(g)
Step 2: Convert 130.0 °C to Kelvin
We will use the following expression.
K = °C + 273.15
K = 130.0°C + 273.15
K = 403.2 K
Step 3: Calculate the moles of O₂
We will use the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 1 atm × 0.0730 L/0.0821 atm.L/mol.K × 403.2 K
n = 2.21 × 10⁻³ mol
Step 4: Calculate the moles of HgO that produced 2.21 × 10⁻³ moles of O₂
The molar ratio of HgO to O₂ is 2:1. The moles of HgO required are 2/1 × 2.21 × 10⁻³ mol = 4.42 × 10⁻³ mol.
Step 5: Calculate the mass corresponding to 4.42 × 10⁻³ moles of HgO
The molar mass of HgO is 216.59 g/mol.
4.42 × 10⁻³ mol × 216.59 g/mol = 0.957 g
Answer:
This description needs a unit describing the system of measurement.
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!
Can we talk here?
I don’t have what you want to talk on
Here are some examples of chemical properties:
Reactivity with other chemicals.
Toxicity.
Coordination number.
Flammability.
Enthalpy of formation.
Heat of combustion.
Oxidation states.
Chemical stability. HOPE THIS HELPS!
