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2 years ago

What happens to the glucose molecule during the process of cellular respiration?

Chemistry

2 answers:

Alex787 [66]2 years ago

8 0

The answer is A. It gets broken down.

Eduardwww [97]2 years ago

5 0

Answer:

it's broken down into carbon dioxide and water

Explanation:

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A positively charged substance is brought near the ion. What will most likely happen and why

d1i1m1o1n [39]

Answer: It repels positive ion and attracts negative ion.

Explanation:

There are 2 types of ions:

1. Cations: These ions are positively charged ions which are formed when a substance looses electrons.

2. Anions: These ions are negatively charged ions which are formed when a substance gains electrons.

It is known that, like charges repel each other and unlike charges attract each other.

As, it is given that the substance is positively charged, so it will attract anion and repel cation.

4 0

3 years ago

Which pair of samples contains the same number of oxygen atoms in each compound?0.20 mol Ba(OH)2 and 0.20 mol H2SO40.20 mol Br2O

Whitepunk [10]

Answer:

0.20 mol Br2O and 0.20 mol HBrO have the same number of oxygen atoms

Explanation:

0.20 mol Ba(OH)2 and 0.20 mol H2SO4

In Ba(OH)2 there are 2 moles of O atoms in every mol of Ba(OH)2.

Number of O atoms in 0.20 moles Ba(OH)2 = 2*0.20 = 0.40 moles O atom

In H2SO4 there are 4 moles of O atoms for every mol of H2SO4.

Number of O atoms in 0.20 moles H2SO4 = 4*0.20 = 0.80 moles O atom

⇒ 0.20 mol Ba(OH)2 and 0.20 mol H2SO4 do not have the same number of oxygen atoms.

0.20 mol Br2O and 0.20 mol HBrO

In Br2O there is 1 mol of O atoms in every mol Br2O

Number of O atoms in 0.20 moles Br2O = 0.20*1 = 0.20 moles O atom

In HBrO there is 1 mol of O atom in every mol HBrO

Number of O atoms in 0.20 moles HBrO = 0.20 *1 = 0.20 moles O atom

⇒ in 0.20 moles Br2O and 0.20 moles HBrO we have the same number of oxygen atoms

0.20 moles of Oygen contains: = 0.20 * 6.022*10^23 = 1.2 *10^23 O atoms

0.10 mol Fe2O3 and 0.50 mol BaO

In Fe2O3 there are 3 moles of O atoms in every mol of Fe2O3

Number of O atoms in 0.10 moles Fe2O3 = 0.10 * 3 = 0.30 moles O atom

In BaO there is 1 mol of O atoms in every mol BaO

Number of O atoms in 0.50 mol BaO = 1*0.50 = 0.50 moles O atom

⇒ 0.10 mol Fe2O3 and 0.50 mol BaO do not have the same number of oxygen atoms.

0.10 mol Na2O and 0.10 mol Na2SO4

In Na2O there is 1 mol of O atoms in every mol of Na2O

Number of O atoms in 0.10 mol Na2O = 1*0.10 = 0.10 moles O atom

In Na2sO4 there are 4 moles of O atoms in every mol of Na2SO4

Number of O atoms in 0.10 mol Na2SO4 = 4*0.10 = 0.40 moles O atom

⇒ 0.10 mol Na2O and 0.10 mol Na2SO4 do not have the same number of oxygen atoms.

4 0

3 years ago

In which orbitals would the valence electrons for selenium (Se) be placed?

meriva

Answer:

Valence electrons of selenium will be placed in s and p-orbitals.

Explanation:

Hope this helps.

3 0

2 years ago

Please help me in answering these questions

ki77a [65]

Answer:

which answer questions hi to nhi hai

7 0

3 years ago

Why is a terminal alkyne favored when sodium amide (NaNH2) is used in an elimination reaction with 2,3-dichlorohexane

mihalych1998 [28]

The question is incomplete, the complete question is;

Why is a terminal alkyne favored when sodium amide (NaNH2) is used in an elimination reaction with 2,3-dichlorohexane? product. A) The terminal alkyne is more stable than the internal alkyne and is naturally the favored B) The terminal alkyne is not favored in this reaction. C) The resonance favors the formation of the terminal rather than internal alkyne. D) The strong base deprotonates the terminal alkyne and removes it from the equilibrium.

E) The positions of the Cl atoms induce the net formation of the terminal alkyne.

Answer:

E) The positions of the Cl atoms induce the net formation of the terminal alkyne.

Explanation:

In this reaction, sterric hindrance plays a very important role. We know that sodamide is a strong base, it tends to attack at the most accessible position.

The first deprotonation yields an alkene. The strong base attacks at the terminal position again and yields the terminal alkyne. Thus the structure of the dihalide makes the terminal hydrogen atoms most accessible to the base. Hence the answer.

7 0

3 years ago