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Alex_Xolod [135]
3 years ago
10

Acetyl-CoA labeled with 14C in both of its acetate carbon atoms is incubated with unlabeled oxaloacetate and a crude tissue prep

aration capable of carrying out the reactions of the citric acid cycle. After one turn of the TCA cycle, oxaloacetate would have 14C in:
a. all four carbon atoms.
b. no pattern that is predictable from the information provided.
c. none of its carbon atoms.
d. the keto carbon and one of the carboxyl carbons.
e. the two carboxyl carbons.
Chemistry
1 answer:
Bumek [7]3 years ago
5 0

Answer:

a.all four carbon atoms.

Explanation:

Acetyl-CoA labeled with 14C in both of its acetate carbon atoms is incubated with unlabeled oxaloacetate and a crude tissue preparation capable of carrying out the reactions of the citric acid cycle. After one turn of the TCA cycle, oxaloacetate would have 14C in  all four carbon atoms.

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Alexxandr [17]

Answer:

u meant the answer to this ques

Explanation:

22

×12

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44

22 ×

----------

264

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or the answer can be 22×12=264

7 0
4 years ago
The quantity 44 liters expressed in cubic meters is ____.
Lerok [7]

D. 0.044 m3

hope this helped

7 0
4 years ago
A 4.36-g sample of an unknown alkali metal hydroxide is dissolved in 100.0 mL of water. An acid-base indicator is added, and the
BARSIC [14]

Answer:

Rb+

Explanation:

Since they are telling us that the equivalence point was reached after 17.0 mL of   2.5 M HCl were added , we can calculate the number of moles of HCl which neutralized our unknown hydroxide.

Now all the choices for the metal cation are monovalent, therefore the general formula for our unknown is XOH and  we know the reaction is 1 equivalent acid to 1 equivalent base. Thus we have the number of moles, n,  of XOH and from the relation n = M/MW we can calculate the molecular weight of XOH.

Thus our calculations are:

V = 17.0 mL x 1 L / 1000 mL = 0.017 L

2.5 M HCl x 0.017 L = 2.5 mol/ L x 0.017 L = 0.0425 mol

0.0425 mol = 4.36 g/ MW XOH

MW of XOH = (atomic weight of X + 16 + 1)

so solving the above equation we get:

0.0425 = 4.36 / (X + 17 )

0.7225 +0.0425X = 4.36

0.0425X = 4.36 -0.7225 = 3.6375

X = 3.6375/0.0425 = 85.59

The unknown alkali is Rb which has an atomic weight of 85.47 g/mol

6 0
4 years ago
A cork has a mass of 3 grams and a volume of 16 cms calculate the density
REY [17]

hey mate here is ur answer

solution

mass{m}=3 gram

             =3/1000

             

volume{v}=16cm

                =16/100

               

density=m/v

           =3/1000÷16/100

           =3/160

            =0.01875kg/m3

6 0
4 years ago
An unknown radioactive substance has a half-life of 3.20hours . If 46.2g of the substance is currently present, what mass A0 was
iogann1982 [59]

Answer:

a) a0 was 46.2 grams

b) It will take 259 years

c) The fossil is 1845 years old

Explanation:

<em>An unknown radioactive substance has a half-life of 3.20hours . If 46.2g of the substance is currently present, what mass A0 was present 8.00 hours ago?</em>

A = A0 * (1/2)^(t/h)

⇒ with A = the final amount = 46.2 grams

⇒ A0 = the original amount

⇒ t = time = 8 hours

⇒ h = half-life time = 3.2 hours

46.2 = Ao*(1/2)^(8/3.2)

Ao = 261.35 grams

<em>Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take in years for 34.0% of an Am-241 sample to decay?</em>  

t = (ln(0.66))-0.693) * 432 = 259 years

It will take 259 years

<em>A fossil was analyzed and determined to have a carbon-14 level that is 80% that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil?</em>

<em />

t = (ln(0.80))-0.693) * 5730 = 1845

The fossil is 1845 years old

7 0
3 years ago
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