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bearhunter [10]
3 years ago
8

If you inactivate an enzyme that catalyses an intracellular chemical reaction, what would be the end result?

Chemistry
1 answer:
skelet666 [1.2K]3 years ago
8 0

Answer:

The intracellular chemical reaction will proceed at a slower rate.

Explanation:

Enzymes are proteinous substances referred to as "biological catalyst". Enzymes catalyze biochemical reactions by increasing the rate at which the reaction occurs. They do this by lowering the activation energy needed for the reaction to start.

Since enzymes speed up the rate of chemical reactions, an absence of such enzyme or its non-functionality will result in such reaction proceeding at a slower rate.

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Lauryl alcohol is a nonelectrolyte obtained from coconut oil and is used to make detergents. A solution of 8.80 g of lauryl alco
kipiarov [429]

Answer:

The molar mass of lauryl alcohol is approximately 180 g/mol

Explanation:

Step 1: Data given

Mass of lauryl alcohol = 8.80 grams

Mass of benzene = 0.100 kg = 100 grams

Freezing point of Benzene is 5.5 °C

Kf value for benzene = 5.12 °C/molal

Step 2: Calculate the freezing point depression

ΔTf = 5.5 - 3.0 °C = 2.5 °C

Step 3: Calculate molality

⇒ with ΔTf = the freezing point depression = 2.5 °C

⇒ with i = the van't Hoff factor = 1

⇒ with kf = the free point depression constat pf benzene = 5.12°C/m

⇒ with m =the molality = TO BE DETERMINED

molality = ΔTf / kf

molality = 2.5 °C / 5.12 °C /m

molality =  0.488 molal

Step 4: Calculate moles lauryl alcohol

Molality = moles lauryl alcohol / mass benzene

Moles lauryl alcohol = molality * mass benzene

Moles lauryl alcohol = 0.488 m * 0.100 kg

Moles lauryl alcohol = 0.0488 moles

Step 5: Calculate molar mass of lauryl alcohol

Molar mass lauryl alcohol = mass lauryl alcohol/ moles lauryl alcohol

Molar mass lauryl alcohol = 8.80 grams / 0.0488 moles

Molar mass lauryl alcohol = 180 g/mol

The molar mass of lauryl alcohol is approximately 180 g/mol

3 0
3 years ago
How does molecular shape affect polarity?
Andrej [43]

Answer:

Letter D

Explanation:

6 0
3 years ago
Read 2 more answers
A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential (s)(aq)(aq)(l)
miv72 [106K]

Answer:

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)  

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

E°cell = 1.10 V

Explanation:

<em>The half-reactions are missing, but I will propose some to show you the general procedure and then you can apply it to your equations.</em>

<em>Suppose we have the following half-reactions.</em>

<em>Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V</em>

<em>Zn²⁺(⁺aq) + 2 e⁻ → Zn(s)    E°red = -0.76 V</em>

<em />

To identify how to make a spontaneous cell, we need to consider the standard reduction potentials (E°red). The half-reaction with the higher E°red will occur as a reduction (in the cathode), whereas the one with the lower E°red will occur as an oxidation (in the anode).

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        E°red = -0.76 V

To get the overall equation we add both half-reactions.

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

E°cell = 0.34 V - (-0.76 V) = 1.10 V

Since E°cell > 0, the reaction is spontaneous.

5 0
3 years ago
What is the half-life in minutes of a compound if 75.0 percent of a given sample decomposes in 40.0 minutes? Assume first-order
yaroslaw [1]

Answer:

See below

Explanation:

.75 = 1/2^(40/h)      

log .75 / ( log 1/2) = 40 / h

<u>h = half life =   96.37683 min</u>

8 0
2 years ago
The worst feeling is when you are hurting so badly inside but not a single tear comes out.
guapka [62]

Answer:

chemistry wow

Explanation:

8 0
3 years ago
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