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____ [38]
3 years ago
7

Hbbbbbbbbbbbbbbbhjjkk

Chemistry
2 answers:
sveticcg [70]3 years ago
8 0

Answer:

?????????

Explanation:

11Alexandr11 [23.1K]3 years ago
4 0

Answer:

hbbbbbbbbbbbbbbbhjjkk

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Calculate the [OH-]<br> [H+]=0.000025
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Explanation:

There are several ways to define acids and bases, but pH and pOH refer to hydrogen ion concentration and hydroxide ion concentration, respectively. The "p" in pH and pOH stands for "negative logarithm of" and is used to make it easier to work with extremely large or small values. pH and pOH are only meaningful when applied to aqueous (water-based) solutions. When water dissociates it yields a hydrogen ion and a hydroxide.

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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

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