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Nataliya [291]
2 years ago
15

For the reaction H2O) =H% + OH", if x moles of H and y moles of OH are formed K. is y A. *?/(1 – x) C. 1/xy D. xy ​

Chemistry
1 answer:
hjlf2 years ago
5 0

Answer: Correct option is B)

The standard enthalpy change of formation of a compound is the enthalpy change which occurs when one mole of the compound is formed from its elements under standard conditions. The equation showing the standard enthalpy change of formation of water is

H
2
(g)+

2
1

​ O

2

​ (g)⟶H

2                                                                   ↑

​ O(l)                                           1 mole of water formed.

∴   Enthalpy of formation is −X2kJ/mol.

________________________________________________________

Explanation:I hope this helped

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What<br>your events made Gautam Buddha to<br>go in search of the truth?​
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The Buddha was born into an aristocratic family in the Shakya clan but eventually renounced lay life. According to Buddhist tradition, after several years of mendicancy, meditation, and asceticism, he awakened to understand the mechanism which keeps people trapped in the cycle of rebirth. The Buddha then traveled throughout the Ganges plain teaching and building a religious community. The Buddha taught a middle way between sensual indulgence and the severe asceticism found in the Indian śramaṇa movement.[12] He taught a spiritual path that included ethical training and meditative practices such as jhana and mindfulness. The Buddha also critiqued the practices of Brahmin priests, such as animal sacrifice.

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Explanation:

8 0
3 years ago
Problem Page Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.2 g o
stich3 [128]

Answer: 0.0 grams

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of butane

\text{Number of moles}=\frac{5.2g}{58.12g/mol}=0.09moles

b) moles of oxygen

\text{Number of moles}=\frac{32.6g}{32g/mol}=1.02moles

2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O

According to stoichiometry :

2 moles of butane require 13 moles of O_2

Thus 0.09 moles of butane will require =\frac{13}{2}\times 0.09=0.585moles  of O_2

Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.

Thus all the butane will be consumed and 0.0 grams of butane will be left.

7 0
3 years ago
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