Answer:
Adding more substrate would overcome the effect of the compound
Explanation:
- Enzymes are biochemical catalysts that speed up chemical reactions. They act on specific substrate to convert them to products.
- Compounds known as inhibitors slow down the rate of enzyme activity.
- Inhibitors are classified as competitive and non-competitive inhibitors.
- Competitive inhibitors will compete with the substrate to bind the active sites on the enzyme. The effect of competitive inhibitors may be reduced by increasing the concentration of the substrate.
- The compound added by the biologist was a competitive inhibitor and therefore adding more substrate would overcome its effect on enzyme catalysis
- Non-competitive inhibitors binds the active site of the enzyme permanently and prevents the substrate from accessing the active sites.
It is B. Thank you later please and do good on the test!
Answer:
Let's say we were Subtracting 3-2=? To Find the Answer we would Subtract 2 from 3 which is 1 Simple our answer is 1 But let's say the Question is 3 - 1= ? to find this answer we would subtract 1 from 3 which is 2 Let's say you were subtracting 3-3=? to do this we take 3 away from 3 now 3 is 0 so our answer is 0 so there are 3 different problems we can make with 3 we could make more but I'm just telling the basics Hope I Helped Bye :)
Explanation:
Hi there!
One of the main causes of air pollution is smoking.
Smoking affects not only humans, but pets and animals as well. Animals can get cancer too!
Another big cause of pollution is cars. Gasoline of course comes out into the air after it is used. It also affects animals.
A lot of people know that we should change this, but hardly anyone is really doing anything to try and stop it.
I know this is a short answer, but I hope it helps!
When PH + POH = 14
∴ POH = 14 -7 = 7
when POH = -㏒[OH-]
7 = -㏒ [OH-]
∴[OH-] = 10^-7
by using ICE table:
Mn(OH)2(s) ⇄ Mn2+ (aq) + 2OH-(aq)
initial 0 10^-7
change +X +2X
Equ X (10^-7 + 2X)
when Ksp = [Mn2+][OH-]^2
when Ksp of Mn(OH)2 = 4.6 x 10^-14
by substitution:
4.6 x 10^-14 = X*(10^-7+2X)^2 by solving this equation for X
∴ X =2.3 x 10-5 M
∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol
= 0.002 g/ L
