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HACTEHA [7]
1 year ago
7

What is the energy of an electron in the first energy level of hydrogen?

Chemistry
1 answer:
dybincka [34]1 year ago
5 0

Answer:

B

Explanation:

hydrogen it's atomic number (Z) is

recall from the formular of Energy

E = -2.178 x 10-18/Z

= -2.178 x 10-18/1

E = -2.178 x 10-18

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EXTRA CREDIT
marin [14]

Answer:

\huge\boxed{\sf 36\ H\ atoms}

Explanation:

<u>Molecular formula from Glucose:</u>

C₆H₁₂O₆

<u>3 moles of Glucose:</u>

3C₆H₁₂O₆

In 1 mole of Glucose, there are 12 hydrogen atoms.

<u>In 3 moles:</u>

= 12 × 3

= 36 H atoms

\rule[225]{225}{2}

5 0
2 years ago
How can you find a tutor??​
yanalaym [24]

Depending on what you are doing, virtual, or in school, you can contact your school and you ask them, or you can have your teacher, based on the subject, tutor you.

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7 0
3 years ago
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What is water clarity and why is it important?
il63 [147K]

Answer:

it's a measure of how far down light can penetrate through the water column. ... Because water clarity is closely related to light penetration, it has important implications for the diversity and productivity of aquatic life that a system can support

6 0
3 years ago
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What is the maximum number of moles of Al2O3 that can be produced by the reaction of .4 mol of Al with .4 mol of O2
Darya [45]

<u>Given:</u>

Moles of Al = 0.4

Moles of O2 = 0.4

<u>To determine:</u>

Moles of Al2O3 produced

<u>Explanation:</u>

4Al + 3O2 → 2Al2O3

Based on the reaction stoichiometry:

4 moles of Al produces 2 moles of Al2O3

Therefore, 0.4 moles of Al will produce:

0.4 moles Al * 2 moles Al2O3/4 moles Al = 0.2 moles Al2O3

Similarly;

3 moles O2 produces 2 moles Al2O3

0.4 moles of O2 will yield: 0.4 *2/3 = 0.267 moles

Thus Al will be the limiting reactant.

Ans: Maximum moles of Al2O3 = 0.2 moles

4 0
3 years ago
Read 2 more answers
If 57.0 g of b2o3 is added to 44.7 g of cl2 and 68.8 g of c, what is the theoretical yield of boron trichloride?
sdas [7]
(57.0 g B2O3 / (69.6202 g B2O3/mol) x (4mol BCI3 / 2 mol B2O3) = 1.64 mol BC13

(44.7 g C12) / (70.9064 g C12/mol) x (4mol BCI3 / 6mol C12) = 0.42027 mol BC13

(68.8 g C) / (12.01078 G C/mol) x (4mol BCI3 / 3 mol C) = 7.63 mol BCI3

C12 is the limiting reactant.

(0.42027 mol BCI3) X (117 . 170 g BCI3/mol) = 49.2 g BCI3 in theory.
6 0
3 years ago
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