Answer:
When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K
Explanation:
Step 1: Data given
A sample of sulfur hexafluoride gas occupies a volume of 5.10 L
Temperature = 198 °C = 471 K
The volume will be reduced to 2.50 L
Step 2 Calculate the new temperature via Charles' law
V1/T2 = V2/T2
⇒with V1 = the initial volume of sulfur hexafluoride gas = 5.10 L
⇒with T1 = the initial temperature of sulfur hexafluoride gas = 471 K
⇒with V2 = the reduced volume of the gas = 2.50 L
⇒with T2 = the new temperature = TO BE DETERMINED
5.10 L / 471 K = 2.50 L / T2
T2 = 2.50 L / (5.10 L / 471 K)
T2 = 230.9 K = -42.1
When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K
Using the Michaelis-Menten equation competitive inhibition, the Inhibition constant, Ki of the inhibitor is 53.4 μM.
<h3>What is the Ki for the inhibitor?</h3>
The Ki of an inhibitor is known as the inhibition constant.
The inhibition is a competitive inhibition as the Vmax is unchanged but Km changes.
Using the Michaelis-Menten equation for inhibition:
Making Ki subject of the formula:
where:
- Kma is the apparent Km due to inhibitor
- Km is the Km of the enzyme-catalyzed reaction
- [I] is the concentration of the inhibitor
Solving for Ki:
where
[I] = 26.7 μM
Km = 1.0
Kma = (150% × 1 ) + 1 = 2.5
Ki = 26.7 μM/{(2.5/1) - 1)
Ki = 53.4 μM
Therefore, the Inhibition constant, Ki of the inhibitor is 53.4 μM.
Learn more about enzyme inhibition at: brainly.com/question/13618533
Answer:
P_2 =0.51 atm
Explanation:
Given that:
Volume (V1) = 2.50 L
Temperature (T1) = 298 K
Volume (V2) = 4.50 L
at standard temperature and pressure;
Pressure (P1) = 1 atm
Temperature (T2) = 273 K
Pressure P2 = ??
Using combined gas law:
Answer:
An organic compound (such as acetylene or butane) containing only carbon and hydrogen and often occurring in petroleum, natural gas, coal, and bitumens.
Explanation:
Iooked it up
