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shtirl [24]
1 year ago
14

A low level of carbon dioxide labeled with 14C is accidentally released into the atmosphere surrounding industrial workers as th

ey resume work following the lunch hour. Unknowingly, they breathe the contaminated air for 1 hour. Which of the following compounds will be radioactively labeled?
A) About one half of the carbon atoms of newly synthesized fatty acids.
B) All of the carbon atoms of newly synthesized fatty acid.
C) The carboxyl atom of newly synthesized fatty acids.
D) One half of the carbon atoms of newly synthesized acetyl CoA.
E) About one third of the carbons of newly synthesized malonyl CoA.
Chemistry
1 answer:
ivann1987 [24]1 year ago
7 0

Answer:

D) One half of the carbon atoms of newly synthesized acetyl CoA.

Explanation:

It will be radioactively labeled because Malonyl CoA which contains 3 Carbon molecule is synthesized from Acetyl CoA which has 2 Carbon molecule.

This happens with the addition of ‘CO2’ with the help of the enzyme called acetyl CoA carboxylase.

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What would be the products for the reaction Br2 + KI →? (Just identify the correct products for the reaction. You do not need to
Savatey [412]

<u>Answer:</u> The products of the reaction will be I_2\text{ and }KBr

<u>Explanation:</u>

Single displacement reaction is defined as the reaction in which more reactive element displaces a less reactive element from its chemical reaction.

The general chemical equation for the single displacement reaction follows:

A+BC\rightarrow AC+B

The given chemical equation follows:

Br_2+2KI\rightarrow I_2+2KBr

Bromine element is more reactive than iodine element. Thus, can easily replace iodine from its chemical reaction.

Hence, the products of the reaction will be I_2\text{ and }KBr

5 0
2 years ago
What is the hydronium ion concentration of a solution whose pH is 7.30
Assoli18 [71]
[ H₃O⁺] = 10 ^ - pH

[ H₃O⁺ ] = 10 ^ - 7.30

[ H₃O⁺ ] = 5.011 x 10⁻⁸ M

hope this helps!
6 0
1 year ago
For the reaction 2N2O5(g) &lt;---&gt; 4NO2(g) + O2(g), the following data were colected:
KonstantinChe [14]

Answer:

a) The reaction is first order, that is, order 1. Option C is correct.

b) The half life of the reaction is 23 minutes. Option B is correct

c) The initial rate of production of NO2 for this reaction is approximately = (3.7 × 10⁻⁴) M/min. Option has been cut off.

Explanation:

First of, we try to obtain the order of the reaction from the data provided.

t (minutes) [N2O5] (mol/L)

0 1.24x10-2

10 0.92x10-2

20 0.68x10-2

30 0.50x10-2

40 0.37x10-2

50 0.28x10-2

70 0.15x10-2

Using a trial and error mode, we try to obtain the order of the reaction. But let's define some terms.

C₀ = Initial concentration of the reactant

C = concentration of the reactant at any time.

k = rate constant

t = time since the reaction started

T(1/2) = half life

We Start from the first guess of zero order.

For a zero order reaction, the general equation is

C₀ - C = kt

k = (C₀ - C)/t

If the reaction is indeed a zero order reaction, the value of k we will obtain will be the same all through the set of data provided.

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = (0.0124 - 0.0092)/10 = 0.00032 M/min

At t = 20 minutes, C = 0.0068 M

k = (0.0124 - 0.0068)/20 = 0.00028 M/min

At t = 30 minutes, C = 0.0050 M

k = (0.0124 - 0.005)/30 = 0.00024 M/min

It's evident the value of k isn't the same for the first 3 trials, hence, the reaction isn't a zero order reaction.

We try first order next, for first order reaction

In (C₀/C) = kt

k = [In (C₀/C)]/t

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = [In (0.0124/0.0092)]/10 = 0.0298 /min

At t = 20 minutes, C = 0.0068 M

k = 0.030 /min

At t = 30 minutes, C = 0.0050 M

k = 0.0303

At t = 40 minutes

k = 0.0302 /min

At t = 50 minutes,

k = 0.0298 /min

At t = 60 minutes,

k = 0.031 /min

This shows that the reaction is indeed first order because all the answers obtained hover around the same value.

The rate constant to be taken will be the average of them all.

Average k = 0.0302 /min.

b) The half life of a first order reaction is related to the rate constant through this relation

T(1/2) = (In 2)/k

T(1/2) = (In 2)/0.0302

T(1/2) = 22.95 minutes = 23 minutes.

c) The initial rate of production of the product at the start of the reaction

Rate = kC (first order)

At the start of the reaction C = C₀ = 0.0124M and k = 0.0302 /min

Rate = 0.0302 × 0.0124 = 0.000374 M/min = (3.74 × 10⁻⁴) M/min

3 0
1 year ago
The solubility of glucose at 30°C is 125 g/100 g water. Classify a solution made by adding 550 g of glucose to 400 mL of water a
seraphim [82]
The simplified solubility of glucose at 30°C is 1.25 g/g of water. Considering that the density of water at 30°C is 1 g/mL, the equivalent mass of 400 mL of water is also 400g. 

The concentration of the solution in water is,
                            550 g/400g of water = 1.375 g glucose / g of water

Since the concentration is higher compared to the solubility of glucose at the specified temperature, it can be said that the solution is SATURATED.
4 0
2 years ago
Read 2 more answers
The bond dissociation energy to break 4 bond(s) in 1 mole of CH₄ molecules is:_____ **Any help would be greatly appreciated!**
frozen [14]

Answer:

The bond dissociation energy to break 4 bonds in 1 mol of CH is 1644 kJ

Explanation:

Since there are 4 C-H bonds in CH₄, the bond dissociation energy of 1 mol of CH₄ is 4 × bond dissociation energy of one C-H bond.

From the table one mole is C-H bond requires 411 kJ, that is 411 kJ/mol. Therefore, 4 C-H bonds would require 4 × 411 kJ = 1644 kJ

So, the bond dissociation energy to break 4 bonds in 1 mol of CH₄ is 1644 kJ

5 0
2 years ago
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