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3 years ago

8

when one plots enzyme rate vs. substrate concentration, the resulting curve is called a. an inhibition plot b. a saturation curv

e c. a straight line d. none of the above

1 answer:

marusya05 [52]3 years ago

4 0

Answer:

a. an inhibition

Explanation:

When one plots enzyme rate vs. substrate concentration, the resulting curve is called an inhibition.

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Iron (III) oxide and hydrogen react to form iron and water, like this: Fe 03(s)+3H9)2Fe(s)+3HO) At a certain temperature, a chem

belka [17]

The question is incomplete, here is the complete question:

Iron (III) oxide and hydrogen react to form iron and water, like this:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

At a certain temperature, a chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound Amount

Fe₂O₃ 3.95 g

H₂ 4.77 g

Fe 4.38 g

H₂O 2.00 g

Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

<u>Answer:</u> The value of equilibrium constant for given equation is $1.0\times 10^{-4}$

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

<u>For hydrogen gas:</u>

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

$\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M$

<u>For water:</u>

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

$\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M$

For the given chemical equation:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

The expression of equilibrium constant for above equation follows:

$K_{eq}=\frac{[H_2O]^3}{[H_2]^3}$

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

$K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}$

Hence, the value of equilibrium constant for given equation is $1.0\times 10^{-4}$

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The elements from which group would be most likely to react

timofeeve [1]

Answer:

Group 15

That is answer c

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flammable liquid is being pumped out of a drum into a bucket using a hand pump. describe an appropriate grounding and bonding pr

lutik1710 [3]

When transferring flammable liquids from storage drums to smaller electrically conductive containers, bonding and grounding are required. In any workplace, you should do the same whenever you move these liquids between conductive containers.

<h3>What exactly occurs in this case?</h3>

Ground dispensing drums in the area for storing and dispensing flammable liquids. Connecting the container to an already grounded, electrically conductive object grounds it. This could be a grounded metal construction framework, a buried metal plate, a metallic subsurface gas piping system, or metal water pipes. Sparks are prevented from discharging by bonding the two containers together and grounding one of them. All connections for bonding and grounding must be made from bare metal to bare metal.

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When an atom loses or gains a proton the atomic number is changed making it change what element it is.

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lidiya [134]

Answer:

A

Explanation:

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