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MrMuchimi
3 years ago
12

Write chemical equation describing the complete combustion of liquid octane c8h18

Chemistry
1 answer:
Hunter-Best [27]3 years ago
4 0
Ljiskdhf;hsdhsoh=2 lol
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Which of the following elements would require the most energy to
umka2103 [35]

Option D:  Chromium would require the most energy to convert one mole of gaseous atoms into gaseous ions each carrying two positive charges.

<h3>What does the term “ionization energy” mean? </h3>

The ionization energy measures an element’s ability for participating in any chemical processes that calls up for the creation of ions or the donation of other electrons.

It is defined as the energy that any electron present in a gaseous atom or ion has in order to absorb so that it comes out of the influence of the nucleus and hence freely move

Ionization energy is also said as the minimum energy required to remove the most loosely bound electron that is present in an isolated gaseous atom or a positive ion or a molecule.

It can be easily connected to the type of chemical bonds that exist between the components in the compounds that they form.

<h3>Which element ionizes most energetically?</h3>

Helium, because it has the highest first ionization energy, whereas francium has one of the lowest.

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8 0
1 year ago
Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:
Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: Al_{2}S_{3} +H_{2}O  _\to  Al(OH)_{3} + H_{2}S

First, balance the chemical equation

Al_{2}S_{3} + 6H_{2}O  \to 2Al(OH)_{3} + 3H_{2}S

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

n_{Al_{2}S_{3}  } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3}  } = \frac{158g}{150.16g/mol}   \\n_{Al_{2}S_{3} }=1.05 mol

From the chemical reaction , the ratio of molar is 3mol H_{2}S/1 mol Al_{2}S_{3}. So, the moles of hydrogen sulfide are:

n_{H_{2} O} =\frac{131g}{18.02g/mol}

       = 7.26mol

From the chemical reaction, the molar ratio is 3 mol H_{2}S/6 mol H_{2}O. So, the moles of hydrogen sulfide are:

Moles of H_{2}S formed = 7.26 mol H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }

Th liming reactant isAl_{2}S_{3} beacuse the mass of Al_{2}S_{3} forms less product than water. Therefore, the maximum number of moles of H_{2}S is 3.15 mol.  We know that molar mass of H_{2}S is 34.10g/mol. So, the maximum mass of H_{2}S formed is,

m_{H_{2}S } = n_{H_{2}S } \times Molar mass of H_{2}S

         = 3.15 mol \times 34.10g/mol

         = 107.4g

Now, multiplying the number of moles of Al_{2}S_{3} by the molar ratio between Al_2S_3 and H_2O which is 6mol H_2O/1mol Al_2S_3 we get the number of moles of H_2O reacted.

Moles of H_2O reacted = 1.05 mol Al_{2}S_3 \times \frac{6 mol H_2O}{1 mol Al_2S_3}

                                     = 6.31 mol H_2O

The mass of H_2O is,

m_{H_{2} O} = 6.31 mol \times 18.02g/ mol

          = 114g

On subtracting, the mass of H_2O reacted from the given mass of H_2O is,

m_{H_2O} = (131-114)g

         = 17g

Hence, the excess remaining reactant is 17g

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8 0
2 years ago
A compound with the empirical formula CH 2 O has a formula mass of 180 g/mol. What is its molecular formula
olga nikolaevna [1]

Empirical formula mass

  • 12+2(1)+16
  • 28+2
  • 30g/mol

Molecular fornula mass:-180g/mol

  • n=Molecular formula mass/Empirical formula mass
  • m=180/30
  • n=6

Molecular formula:-

  • n×Empirical formula
  • 6(CH_2O
  • C_6H_12 O_6
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2 years ago
In which of the following are the symbol and name for the ion given correctly?
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3 years ago
How did planetesimals form?
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3 years ago
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