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Georgia [21]
3 years ago
7

What is an energy level

Chemistry
1 answer:
Helen [10]3 years ago
8 0

Answer: The fixed amount of energy that a system described by quantum mechanics

Explanation: Quantum mechanics are molecules, atom, electron, or nucleus.

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Answer:

it is A mark me big brain pls

Explanation:

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3 years ago
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What is the charge of an electron<br> A. -1<br> B. 0<br> C. +1<br> D. +2
Gelneren [198K]

Answer:

A

Explanation:

I believe it is negative 1.

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Is rain chemical or mechanical weathering?
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4 years ago
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The insulating and packing material Styrofoam is a dating and packing material Styrofoam is a polymer of styrene. Find the molec
Nataly [62]

Explanation:

In order to find the molecular formula, we have to find the empirical formula first of all.

It is known that is a styrofoam, elements present are carbon and hydrogen atoms.

Element     %         Atomic mass       Molar ratio           Simple ratio

    C        92.25          12.01              \frac{92.25}{12.01} = 7.68                 \frac{7.68}{7.68} = 1

     H        7.75            1.008            \frac{7.75}{1.008} = 7.68                 \frac{7.68}{7.68} = 1

As empirical formula of styrofoam is C_{x}H_{x}.

Hence, empirical mass = (12.01 + 1.008) g/mol = 13.018 g/mol = 13.0 g/mol (approx)

Molar mass given is 104 g/mol.

So,     \frac{\text{molar mass}}{\text{empirical formula mass}} = \frac{104}{13.0} = 8

Thus, we can conclude that molecular formula of the given styrene is C_{8}H_{8}.

8 0
4 years ago
A compound is found to contain 64.80 % carbon, 13.62 % hydrogen, and 21.58 % oxygen by weight. What is the empirical formula for
insens350 [35]

Answer:

Empirical formula is C₄H₁₀O

Explanation:

Values for C, H and O are determined as centesimal composition.

64.80 g of C in 100g of compound

13.62g of H in 100 g of compound

21.58 g of O in 100 g of compound.

We convert the mass to moles:

64.80 g . 1mol/ 12g = 5.4 moles of C

13.62 g . 1 mol /1g = 13.62 moles of H

21.58 g . 1 mol/16g = 1.35 moles of O

We pick the lowest value and we divide:

5.4 moles of C / 1.35 = 4 C

13.62 moles of H / 1.35 = 10 H

1.35 moles of O / 1.35 = 1 O

Empirical formula is C₄H₁₀O, it can be the diethyl ether.

We confirm, the excersise is well done.

Molar mass = 74g/mol

74 g of compound we have (12 . 4)g of C

In 100 g of compound we may have (100 . 48) / 74 = 64.8 g

5 0
3 years ago
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