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Bess [88]
3 years ago
6

What is the difference between active transport? Which one requires energy?

Chemistry
2 answers:
Kaylis [27]3 years ago
4 0
Off topic but you have the same name as my friend
erma4kov [3.2K]3 years ago
3 0

Answer:

I believe its active transport

Explanation:

I really hope this helps im really sorry if its wrong.

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jeka94
The answer is Heyyy friend in know because I took the test
4 0
2 years ago
In which of these diatomic molecules would you NOT find an octet of electrons
Semmy [17]

Answer:

The hydrogen molecule is the only one in which can not find an octet of electrons around each atom.

Explanation:

Let's evaluate each case.  

1. Nitrogen (N₂):

With Z = 7, nitrogen has the following electronic configuration

1s²

2s² 2p³  → valence electrons

Since its valence electrons are 5, in the molecule one nitrogen atom shares 3 electrons with the other one, and each remains with an electron pair, so <u>each atom has an octet of electrons.</u>

2. Hydrogen (H₂):

With Z = 1, its electronic configuration is:

1s¹  → valence electron

In the molecule, the hydrogen atoms share the only electron they have, so they will have only 2 electrons around. In this diatomic molecule, <em><u>we can not find an octet.</u></em>

3. Oxygen (O₂):

Z = 8. Electronic configuration:

1s²

2s² 2p⁴  → valence electrons

In the diatomic molecule, each oxygen atom shares 2 electrons with the other one and remains with 2 pairs of electrons, therefore, <u>each oxygen atom has an octet</u>.      

4. Fluorine (F₂)

Z = 9. Electronic configuration:

1s²

2s² 2p⁵  → valence electrons

In this molecule, each fluorine atom shares 1 electron with the other and remains with 3 pairs of electrons, hence, <u>each fluorine atom has an octet of electrons around</u>.

Finally, we can say that the hydrogen molecule is the only one in which can not find an octet of electrons around each atom.

I hope it helps you!  

8 0
2 years ago
Describe the structure of a typical metal such as iron ?<br>assaaap
dimaraw [331]

Answer:

Metals consist of giant structures of atoms arranged in a regular pattern. The electrons from the outer shells of the metal atoms are delocalised , and are free to move through the whole structure. This sharing of delocalised electrons results in strong metallic bonding .

Explanation:

mark me as brainliest

8 0
2 years ago
Read 2 more answers
When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing poin
LenaWriter [7]

The question is incomplete, the complete question is:

When 177. g of alanine (C_3H_7NO_2) are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is 5.9^oC lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is 7.2^oC lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.

<u>Answer:</u> The van't Hoff factor for potassium bromide in X is 1.63

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

\Delta T_f=i\times K_f\times m

OR

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}} ......(1)

  • <u>When alanine is dissolved in mystery liquid X:</u>

\Delta T_f=5.9^oC

i = Vant Hoff factor = 1 (for non-electrolytes)

K_f = freezing point depression constant

m_{solute} = Given mass of solute (alanine) = 177. g

M_{solute} = Molar mass of solute (alanine) = 89 g/mol

w_{solvent} = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

5.9=1\times K_f\times \frac{177\times 1000}{89\times 800}\\\\K_f=\frac{5.9\times 89\times 800}{1\times 177\times 1000}\\\\K_f=2.37^oC/m

  • <u>When KBr is dissolved in mystery liquid X:</u>

\Delta T_f=7.2^oC

i = Vant Hoff factor = ?

K_f = freezing point depression constant = 2.37^oC/m

m_{solute} = Given mass of solute (KBr) = 177. g

M_{solute} = Molar mass of solute (KBr) = 119 g/mol

w_{solvent} = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

7.2=i\times 2.37\times \frac{177\times 1000}{119\times 800}\\\\i=\frac{7.2\times 119\times 800}{2.37\times 177\times 1000}\\\\i=1.63

Hence, the van't Hoff factor for potassium bromide in X is 1.63

7 0
3 years ago
What is the ratio of ions of aluminium fluoride component i.e. Aluminium ions to fluorine ions
Ad libitum [116K]

Answer:

Al:F3​:: 3:19.

Explanation:

8 0
2 years ago
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