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mote1985 [20]
3 years ago
14

Selenium (Se) has 30 protons and 20 neutrons in each atom, therefore its atomic

Chemistry
1 answer:
allochka39001 [22]3 years ago
8 0

Answer:

30, 50

Explanation:

Hello,

In this since an element's atomic number is equal to the number of protons in its atom, we can infer that selenium's atomic number is 30. Moreover, due to the fact the the neutrons are equal to the atomic mass minus the atomic number or the number of protons, by knowing the number of neutrons we compute the atomic as follows:

neutrons=mass-protons\\\\mass=neutrons+protons\\\\mass=30+20\\\\mass=50a.m.u

Thus, answer is 30, 50.

Best regards.

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). The molar mass of an organic acid, a compound composed of carbon, hydrogen, and oxygen, is 194.14 g/mol. Combustion of a 1.50
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Answer:

The empirical formula is C₆H₁₀O₇.

Step-by-step explanation:

1. Calculate the masses of C, H, and O from the masses given.

Mass of C =  2.0402 g CO₂ × (12.01 g C/44.01 g CO₂) = 0.5568  g C

Mass of H = 0.6955 g H₂O  × (2.016 H/18.02 g H₂O)  = 0.077 81 g H

Mass of O = Mass of compound - Mass of C - Mass of H = (1.500 – 0.5568 – 0.077 81) g = 0.8654 g O

=====

2. Convert these masses to moles.

Moles  C = 0.5568  × 1/12.01  = 0.046 36

Moles H = 0.077 81 × 1/1.008 = 0.077 19

Moles O = 0.8654   × 1/16.00 = 0.054 09

=====

3. Find the molar ratios.

Moles  C = 0.046 36/0.046 36 = 1

Moles H = 0.077 19/0.046 36   = 1.665

Moles O = 0.054 09/0.046 36 = 1.167

======

4. Multiply the ratios by a number to make them close to integers

C  = 1        × 6 = 6

H = 1.665 × 6 = 9.991

O = 1.167 × 6  = 7.001

=====

5. Round the ratios to integers

C:H:O =6:10:7

=====

6. Write the empirical formula

The empirical formula is C₆H₁₀O₇.

=======

7. Calculate the empirical formula mass

C₆H₁₀O₇ = 6×12.01 + 10×1.008 + 7×16.00

C₆H₁₀O₇ = 72.01 + 10.08+ 112.0

C₆H₁₀O₇ = 194.09

=====

8. Divide the molecular mass by the empirical formula mass.  

MM/EFM = 194.14/194.09 = 1.000 ≈ 1

=====

9. Determine the molecular formula

MF = (EF)ₙ = (C₆H₁₀O₇)₁ = C₆H₁₀O₇

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