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lutik1710 [3]
3 years ago
7

What is the maximum possible number of electrons in a ground-state carbon atom that have ml = 0?

Chemistry
1 answer:
juin [17]3 years ago
3 0
Ground state means 1s which can hold 2 electrons.
l for 1s is = 0
ml = 0 (given)
possible values of m = 0
so it can hold maximum of 2 electrons. One spin up and other spin down.
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Classify the carbohydrate tagatose by both the carbonyl group and the number of carbon atoms. D-tagatose is sugar with six carbo
Bingel [31]

Answer:

Hexose category and ketohexose category

Explanation:

The classification of the Carbohydrate tagatose by carbonyl group is that it is a monosaccharide and has a hexose structure hence it belongs to the Hexose category

Based on the number of carbon atoms the structure has a ketofunctionality hence it is classified under the ketohexose category

Attached below is the remaining part of the solution

6 0
3 years ago
What was the result of the atomic theory?
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Answer:

The result of the atomic theory was atomic theory proposed that all matter was composed of atoms, also postulated that chemical reactions resulted in the rearrangement of the reacting atoms.

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How many moles of carbon are there in a 0.70 g (3.5 carat) diamond?
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3.5 x 10^22 moles of carbon are there in a 0.70 g (3.5 carat) diamond.

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7 0
3 years ago
I need help solving this!
zmey [24]

Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

Explanation:

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As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol

The given reaction equation is as follows.

C + 2H_{2} \rightarrow CH_{4}

This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.

Moles of H_{2} = \frac{9.14}{2}\\= 4.57 mol

Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

5 0
3 years ago
A chemist must dilute 97.1 ml of aqueous magnesium fluoride solution until the concentration falls to 389 microMolarity . He'll
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Answer:

0.302L

Explanation:

<em>...97.1mL of 1.21m M aqueous magnesium fluoride solution</em>

<em />

In this problem the chemist is disolving a solution from 1.21mM = 1.21x10⁻³M, to 389μM = 389x10⁻⁶M. That means the solution must be diluted:

1.21x10⁻³M / 389x10⁻⁶M = 3.11 times

As the initial volume of the original concentration is 97.1mL, the final volume must be:

97.1mL * 3.11 = 302.0mL =

0.302L

6 0
3 years ago
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