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My name is Ann [436]
2 years ago
7

How many grams of ammonia would be produced from the reaction of 2.64 moles of hydrogen gas reacting with excess nitrogen?

Chemistry
1 answer:
Anvisha [2.4K]2 years ago
5 0

Answer:

29.92g

Explanation:

tell me if my answer is wrong

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What is the answer to this chemistry question?
olga_2 [115]

Answer:

The answer to your question is remplacement double

Explanation:

Data

Lead (II) nitrate = Pb(NO₃)₂

Potassium iodide = KI

Process

1.- Write the balanced chemical reaction

             Pb(NO₃)₂  +  2KI  ⇒   PbI₂  +  2KNO₃

2.- Conclusion

This is a remplacement double reaction because there are two reactants that interchange cations and the products are a combination of the reactants.

5 0
3 years ago
2 C2H6 + 7 026 H2O + 4 CO2
Kay [80]

Answer:

Oxygen is limiting reactant

Explanation:

Based on the chemical reaction:

2C2H6 + 7O2 → 6H2O + 4CO2

<em>2 mole of ethane reacts with 7 moles of oxygen</em>

<em />

For a complete reaction of 5.25 moles of ethane are required:

5.25 moles Ethane * (7mol Oxygen / 2mol Ethane) = 18.38 moles of oxygen

As there are just 15.0 moles of oxygen

<h3>Oxygen is limiting reactant</h3>
3 0
2 years ago
an atmosphere is considered hazardous if it contains a hazardous gas in excess of 10 percent of the hazardous material's:
svlad2 [7]

Lower flammable limit means the lowest concentration of a material that will propagate a flame.

What is hazardous atmosphere?

It is an atmosphere that may expose employees to risk of death, incapacitation, impairment of ability to self-rescue, injury, or acute illness from one or more of following causes

  • Flammable gas, vapor, or mist in excess of 10 percent of lower flammable limit (LFL)
  • Airborne combustible dust at concentration that meets or exceeds its LFL

What is lower flammable limit?

  • It means the lowest concentration of a material that will propagate a flame.
  • The LFL is usually expressed as percent by volume of material in air (or other oxidant)
  • Atmospheres with concentration of flammable vapors at or above 10 percent of lower explosive limit (LEL) are considered hazardous when located in confined spaces.
  • However, atmospheres with flammable vapors below 10 percent of LEL are not necessarily safe. Such atmospheres are too lean to burn

Learn more about lower flammable limit at brainly.com/question/2456135

#SPJ4

6 0
1 year ago
Describe how you might determine the m/z and relative abundance of the ions contributing to the peak at 21.876 min
aliya0001 [1]

The m/z and relative abundance of the ions contributed to the peak at 21.876 min. The relative abundance will be 21.876%.

<h3>What is relative abundance?</h3>
  • The proportion of atoms with a particular atomic mass present in an element sample taken from a naturally occurring sample is known as the relative abundance of an isotope.
  • When the relative abundances of an element's isotopes are multiplied by their atomic masses and the results are added up, the result is the element's average atomic mass, which is a weighted average.
  • Chemists often divide the number of atoms in a particular isotope by the sum of the atoms in all the isotopes of that element, then multiply the result by 100 to determine the percent abundance of each isotope in a sample of that element.

To learn more about relative abundance with the given link

brainly.com/question/1594226

#SPJ4

8 0
2 years ago
A 17.6-g sample of ammonium carbonate contains ________ mol of ammonium ions.
postnew [5]
These problems are a bit interesting. :)

First let's write the molecular formula for ammonium carbonate. 

NH4CO3 (Note! The 4 and 3 are subscripts, and not coefficients)

17.6 gNH4CO3

Now to convert to mol of one of our substances we take the percent composition of that particular part of the molecule and multiply it by our starting mass. This is what it looks like using dimensional analyse. 

17.6 gNH4CO3 * (Molar Mass of NH4 / Molar Mass of NH4CO3)

Grab a periodic table (or look one up) and find the molar masses for these molecules! Well. In this case I'll do it for you. (Note: I round the molar masses off to two decimal places)

NH4 = 14.01 + 4*1.01 = 18.05 g/mol
NH4CO3 = 14.01 + 4*1.01 + 12.01 + 3*16.00 = 78.06 g/mol


17.6 gNH4CO3 * (18.05 molNH4 / 78.06 molNH4CO3)
= 4.07 gNH4

Now just take the molar mass we found to convert that amount into moles!

4.07 gNH4 * (1 molNH4 / 18.05 gNH4) = 0.225 molNH4

4 0
2 years ago
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