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icang [17]
1 year ago
9

Acetylene gas (ethyne; HC = CH) burns in an oxyacetylene torch to produce carbon dioxide and water vapor. The heat of reaction f

or the combustion of acetylene is 1259 kJ/mol.
(c) How many grams ofCO₂ form?
Chemistry
1 answer:
Yanka [14]1 year ago
6 0

The mass of CO2 produced by 26g of acetylene is 88g.

Given ,

In an oxyacetylene torch, acetylene gas (ethyne; HCCH) burns to produce carbon dioxide and water vapour.

The acetylene combustion reaction is given by,

H2O + HCCH + 5/2 O=O 2CO2

Heat of reaction for acetylene combustion = 1259kj/mol

CO2 has a molecular mass of 44g/mol.

2 moles of CO2 have a molecular mass of 88g.

On combustion, 1 mole of acetylene yields 2 moles of CO2.

Thus, 26g of acetylene produces 88g of CO2 when burned.

As a result, the mass of carbon dioxide produced by 26g of acetylene is 88g.

Learn more about acetylene here :

brainly.com/question/15346128

#SPJ4

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Vlad1618 [11]

Across a period I.E increases progressively from left to right

Explanation:

The trend of the first ionization energy is such that across a period I.E increases from left to right due to the decreasing  atomic radii caused by the increasing nuclear charge. This not compensated for by successive electronic shells.

  • Ionization energy is a measure of the readiness of an atom to lose an electron.
  • The lower the value, the easier it is for an atom to lose an electron.
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Ionization energy brainly.com/question/6324347

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3 0
3 years ago
You perform a combustion analysis on a 255 mg sample of a substance that contains only C, H, and O, and you find that 561 mg of
zhuklara [117]
The combustion of an organic compound is mostly written as,
                        CaHbOc + O2 --> CO2 + H2O
where a, b, and c are supposed to be the subscripts of the elements C, H, and O in the compound. Determining the number of moles of C and H in the product which is the same as that in the compound,
   (Carbon, C)   :   (561 mg) x (12/44) = 153 mg x (1 mmole/12 mg) = 12.75
   (Hydrogen, H) :  (306 mg) x (2/18)  = 34 mg x (1 mmole/1 mg) = 34 
   Calculating for amount of O in the sample,
(oxygen, O) = 255 - 153 mg - 34 mg = 68 mg x (1mmole/16 mg) = 4.25 
The empirical formula is therefore,
                        C(51/4)H34O17/4
                           C3H8O1
The molar mass of the empirical formula is 60. Therefore, the molecular formula of the compound is,
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