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k0ka [10]
2 years ago
7

In a tissue that metabolizes glucose via the pentose phosphate pathway, C-1 of glucose would be expected to end up principally i

n:
A.
Carbon dioxide

B.
Glycogen

C.
Phosphoglycerate

D.
Pyruvate
Chemistry
1 answer:
Mnenie [13.5K]2 years ago
5 0

Answer:

A.

Carbon dioxide

Explanation:

In a tissue that metabolizes glucose via the pentose phosphate pathway, C-1 of glucose would be expected to end up principally in Carbon dioxide

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Four atoms and/or ions are sketched below in accordance with their relative atomic and/or ionic radii. Which of the following se
kotykmax [81]

Answer:

Hence the correct option is an option (b) Sr4, Cl,Br−,Na+.

Explanation:    

Bromine and chlorine belong to an equivalent group. As we go down the group the dimensions increases which too there's a charge on the bromine atom. therefore the size of the Br- is going to be larger in comparison to the chlorine atom.

Sr atom is within the second group, and also it's below the above-mentioned atoms.so Sr is going to be the larger one among all the atoms.

Sodium and chlorine belong to an equivalent period .size decrease from left to right. but due to the charge on sodium its size decreases and there's an opportunity that Na+ size could be adequate for Cl.      

Here we finally assume that two atoms are of an equivalent size (Na+ and Cl) which are less in size compared to the opposite two(Sr and Br-) during which one is greater (Sr)and the opposite is smaller(Br-).

4 0
2 years ago
Write a balanced nuclear equation for the formation of 28 Si 14 from beta-minus emission.
Fantom [35]

Answer:

Phosphorus-28 undergoes beta-minus decay to produce

  • an electron,
  • a Silicon-28 nuclei, and
  • an electron antineutrino.

\rm ^{28}_{13} Al \to ^{28}_{14} Si + ^{\phantom{-}0}_{-1}e + \bar{\mathnormal{v}}_{\rm e}

Explanation:

In simple words, when a nucleus undergoes beta-minus decay, a neutron is converted to a proton. An electron and an electron antineutrino will be released.

\rm ^{1}_{0}n^{0} \to ^{1}_{1}p^{+} + ^{\phantom{-}0}_{-1}e^{-} + \bar{\mathnormal{v}}_{\rm e}.

One way to tell whether a neutron is converted to a proton, but not vice versa, is to check the sum charges on the two sides of this equation.

  • Left-hand side: 0. Neutron is neutral.
  • Right-hand side: 1 + (-1) = 0. Each proton carries a charge of +1. Each electron (beta-minus particle) carries a charge of -1. Antineutrinos are neutral.

The charges on the two sides of this equation is the same. Hence this nuclear equation is possible (but not necessarily correct; however, if the proton and the neutron are in the wrong place the charge won't even be the same.)

Since the mass number of a proton and a neutron are both 1, the overall mass number of the atom will stay the same.

The atomic number is the number of protons in each atom. That number determines the symbol and the chemical properties of the atom. When one neutron in an atom is converted to a proton, the atomic number of the atom will increase by 1.

The atomic number of the daughter nucleus, silicon, is 14. It takes a parent nucleus with atomic number 14 - 1 = 13 to produce a silicon atom. Refer to a modern periodic table. Atomic number 13 corresponds to the element aluminum.

Also, the mass number of the daughter nucleus is 28. Since the mass number would stay the same in a beta decay, the mass number of the parent nucleus would also be 28. In other words, it takes an aluminum-28 atom to undergo beta-decay to produce a silicon-28 atom.

Complete the other details (electron and electron antineutrino) to obtain the equation

\rm ^{28}_{13} Al \to ^{28}_{14} Si + ^{\phantom{-}0}_{-1}e + \bar{\mathnormal{v}}_{\rm e}.

6 0
3 years ago
Which of the following molecules has one lone pair of electrons?
BigorU [14]
You can automatically rule out CH₄ since it has no lone pairs at all around the central atom. Water has 2. Ammonia is the only Lewis structure that contains one lone pair.
7 0
2 years ago
Read 2 more answers
Cho 4g CuO vào dung dịch axit clohidric 10% thì phản ứng vừa đủ.
Sholpan [36]

Answer:

Explanation:

a. CuO+ 2HCl⇒CuCl2+ H2O

b. n_{CuO}= \frac{4}{80}= 0,05 (mol)

⇒n_{CuCl2}= n_{CuO}=0,05 mol

⇒m_{CuCl2}= 0,05×135=6,75 (g)

c. n_{HCl}=2× n_{CuO}=0,1 (mol)

⇒m_{HCl}= 0,1×36,5= 3,65 (g)

⇒m_{dd HCl}= \frac{m_{HCl}}{10}×100=36,5 (g)

⇒ Nồng độ phần trăm dd sau phản ứng= Nồng độ % dd CuCl2=\frac{m_{CuCl2} }{m_{dd HCl+ m_{CuO} } }×100=\frac{6,75}{36,5+4} ×100≈ 16,67%

8 0
3 years ago
I need to know the answers to this to make sure I am right ​
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The answer is C............
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