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olchik [2.2K]
4 years ago
14

Which equation has both a liquid and a gas as products? 2 upper H g upper O (s) right arrow 2 upper H g (l) plus upper O subscri

pt 2 (g). 2 upper M g (s) plus upper O subscript 2 (g) right arrow 2 upper M g upper O (s). 1 UJpper K upper C l upper O subscript 3 (s) right arrow 2 upper K upper C l (s) plus 3 upper O subscript 2 (g). Upper M g (s) plus 2 upper H upper C l (a q) right arrow upper H subscript 2 (g) plus upper M g upper C l subscript 2 (a q).
Chemistry
2 answers:
Mandarinka [93]4 years ago
8 0
<h2>Let us predict the gas and liquid products in option </h2>

Explanation:

option  1 :  2 upper H g upper O (s) right arrow 2 upper H g (l) plus upper O subscript 2 (g).

Chemical reaction

It is the reactants react to form products .

Chemical equation

It is the method of representing reaction in terms of moles , specifying states , symbols , molecular formulas etc .

This actually gives the complete info about the reaction .

In the above asked question or any other question we can specify the states by writting :

Solid as "S".

Liquid as "L"

Gas as (g).

inessss [21]4 years ago
8 0

Answer:

A

Explanation:

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Explanation:

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Calculate how many grams of hydrogen can be burned if 40. liters of oxygen at 200. k and 1.0 atm. show work pls.
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The grams of hydrogen gas can be burned if 40. liters of oxygen at 200. k and 1.0 atm is 4.88 grams.

<h3>How do we calculate grams from moles?</h3>

Grams (W) of any substance will be calculated by using their moles (n) through the following equation:

  • n = W/M, where

M = molar mass

And moles of the gas will be calculated by using the ideal gas equation as:

  • PV = nRT, where

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n = moles = ?

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On putting these values on the above equation, we get

n = (1)(40) / (0.082)(200) = 2.439 = 2.44 moles

  • Now grams of hydrogen gas will be calculated by using the first equation as:

W = (2.44mol)(2g/mol) = 4.88g

Hence required mass of hydrogen gas is 4.88g.

To know more about ideal gas equation, visit the below link:

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