What is the product when NH3 decomposes

1 answer:

6 0

The decomposition reaction of ammonia is:

2NH3 ➡️ N2 + 3H2

The reverse formation reaction is known as the Haber process, which converts nitrogen and helium into ammonia.

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Which of the examples is potassium? es ) A) B) B C)

Nikitich [7]

Answer:

examples of things which contain potassium are:

green vegetables

root vegetables

fruits

potassium chloride

potassium sulphate

Explanation:

if you need a specific answer please send the options

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3 years ago

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3 years ago

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An element forms an oxide, E₂O₃, and a fluoride, EF₃.(a) Of which two groups might E be a member?

Fittoniya [83]

a) The E might belong to group 13.

As the formula of a chemical compound is derived by the cross multiplication of the valency of the atoms. As formula of the given oxide is and valency of O atom is -2, therefore valency of element E must be +3 in order to obtain E2O3.

Also, in EF3, the valency of E will be +3 because there are three atoms of fluorine who has an individual valency of -1. Thus, e will have the valency of +3.

The Group 13 is the boron group which has the following elements:

Boron
Aluminium
Gallium
Indium
Thallium

All these elements have the valency of +3.

To know more about Valency, refer to this link:

brainly.com/question/12717954

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1 year ago

A chemist must prepare of sodium hydroxide solution with a pH of at . He will do this in three steps: Fill a volumetric flask ab

77julia77 [94]

Answer:

0.0400 g for the example given below.

Explanation:

pH value is not provided, so we'll solve this problem in a general case and then we will use an example to justify it.

By definition, $pH = -log[H_3O^+]$ .
NaOH is a strong base, as it's a hydroxide formed with a group 1A metal, so it dissociates fully in water by the equation: $NaOH (aq)\rightarrow Na^+ (aq) + OH^- (aq)$ .
From the equation above, using stoichiometry we can tell that the molarity of hydroxide is equal to the molarity of NaOH: $[NaOH] = [OH^-]$ .
Concentration of hydroxide is then equal to the ratio of moles of NaOH and the volume of the given solution. Moles themselves are equal to mass over molar mass, so we obtain: $[OH^-] = [NaOH] = \frac{n_{NaOH}}{V} = \frac{m_{NaOH}}{M_{NaOH}V}$ .
We also know that $pOH = 14.00 - pH = -log[NaOH]$ . Take the antilog of both sides: $10^{-pOH} = 10^{pH - 14.00} = [NaOH] = \frac{m_{NaOH}}{M_{NaOH}V}$ .
Solve for the mass of NaOH: $m_{NaOH} = 10^{pH - 14.00}\cdot M_{NaOH}\cdot V$ .