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frutty [35]
2 years ago
15

g Select the irreversible reactions of glycolysis. conversion of glucose to glucose 6‑phosphate by hexokinase conversion of gluc

ose 6‑phosphate to fructose 6‑phosphate by phosphoglucose isomerase conversion of fructose 6‑phosphate to fructose 1,6‑bisphosphate by phosphofructokinase conversion of phosphoenolpyruvate to pyruvate by pyruvate kinase conversion of phosphoglycerate to phosphoenolpyruvate by enolase
Chemistry
1 answer:
alexdok [17]2 years ago
3 0

Answer:

1) Conversion of glucose to glucose 6-phosphate by hexokinase

2) Conversion of fructose 6-phosphate to fructose 1,6-biphosphate by phosphofructokinase

3) Conversion of phosphoenolpyruvate to pyruvate by pyruvate kinase

Explanation:

There are 10 steps in the glycolysis pathway, three of which are irreversible. The enzymes controlling these reactions have not only catalytic properties but the irreversibility of the reaction gives them regulatory properties as well.  These reactions serve as control points in the pathway.

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James was given a sample of sodium bicarbonate NaHCO3 in a weighing dish to analyze. His teacher asked him to find the percent o
gtnhenbr [62]
This question could be answered easily if the results of the abundance of the other elements are given. You will just have to subtract the sum of all their abundances to 100. Since it's not given, the solution would just be:

Na = 23 g/mol* 1 = 23 g
H = 1 g/mol * 1 = 1 g
C = 12 g/mol * 1 = 12 g
O = 16 g/mol * 3 = 48 g
Total = 84 g

% O = 48/84 * 100 = <em>57.14%</em>

5 0
2 years ago
Read 2 more answers
Consider isotopes ions protons and electrons. how many of these did dalton not discuss in his atomic theory?
MAVERICK [17]

Answer: He did not discuss about any of these.

Explanation: Dalton proposed some of the postulates for his atomic theory. They are:

1) Matter is made up of atoms which are not divisible.

2) Atoms of different elements combine in a fixed ratio to form compounds.

3) The atomic properties of given element are same including mass. This states that all the atoms of an element have same mass but the atoms of different elements have different masses.

4) No atoms are either created or destroyed during a chemical reaction.

5) Atoms of an element are identical in mass, size and other chemical and physical properties.

As it is visible from the postulates, he only discussed only about the atoms but not subatomic particles or isotopes.

8 0
2 years 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
Molecular bromine is 24 percent dissociated at 1600 k and 1.00 bar in the equilibrium br2 (
dlinn [17]
Br2 == 2Br

24% dissociated => n total moles, 0.24 mol*n of Br, and 0.76*n mol of Br2

=> partial pressure of Br, P Br = 0.24 bar, and
     partical pressure of Br2, P Br2 = 0.76 bar

kp = (P Br)^2 / P Br2 = (0.24)^2 / 0.76 = 0.0758


3 0
2 years ago
What volume of 0.210 M sulfuric acid is required to completely react with 2.14 g aluminium hydroxide?
Galina-37 [17]
3 H2SO4 + 2 Al(OH)3 → Al2(SO4)3 + 6 H2O

(2.14 g Al(OH)3) / (78.0036 g Al(OH)3/mol) x (3 mol H2SO4 / 2 mol Al(OH)3) / (0.210 mol/L H2SO4) =
0.19596 L = 196 mL H2SO4
3 0
1 year ago
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