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levacccp [35]
3 years ago
10

Please answer

Chemistry
1 answer:
lutik1710 [3]3 years ago
8 0

Answer:

See Explanation

Explanation:

8.  3, 1, -1, +1/2 => 3pₓ¹ => Aluminum

9.  4, 2, +1, +1/2 => 4d₁¹ => Chromium

10. 6, 1, 0, -1/2 => 6p₀² => Argon

11.  4, 3, +3, -1/2 => 4f₊₃² => Lutetium

12. 2, 1, +1, -1/2  => 2p₊₁² => Neon

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Is The atoms of the elements have different masses but the same volume true?
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Among the factors that affect the solubility of a substance are type of solvent and temperature.
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An isotope of potassium with a half-life of roughly one billion years is often used in radioactive dating. This isotope decays t
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Answer : The chemical symbol for the element that results from this process is, (Ar) for argon.

Explanation :

Electron capture : In this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. Simply, a proton and an electron combines together to form a neutron. Mass number does not change in this process.

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7 0
3 years ago
A compound is found to have 55.7% hafnium+and+44.3%+chlorine.+what+is+the+empirical+formula?
uranmaximum [27]

The empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

<h3>How to calculate empirical formula?</h3>

The empirical formula of a compound is a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.

The empirical formula of the given compound can be calculated as follows:

  • Hafnium = 55.7% = 55.7g
  • Chlorine = 44.3% = 44.3g

First, we convert mass values to moles by dividing by the molar mass of each element

  • Hafnium = 55.7g ÷ 178.49g/mol = 0.312mol
  • Chlorine = 44.3g ÷ 35.5g/mol = 1.25mol

Next, we divide each mole value by the smallest

  • Hafnium = 0.312 ÷ 0.312 = 1
  • Chlorine = 1.25 ÷ 0.312 = 4

Therefore, the empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

Learn more about empirical formula at: brainly.com/question/14044066

#SPJ1

7 0
1 year ago
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