Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
No. Of moles = mass/molar mass of (NH4)3PO4=2.38/149=0.02
No. Of particles= no. Of moles x 6.0x10^23 = 0.02 x 6.0x10^23 = 1.2x10^22
No. Of particles= no. Of moles x 6.0x10^23 = 0.02 x 6.0x10^23 = 1.2x10^22
The number of particles in 2.38 g ammonium phosphate are [tex]\rm \bold{9.0345\;\times\;10^2^1}[/tex].
The number of particles in a mole of sample is given by Avogadro law. The number of particles in a mole of a sample is equivalent to [tex]\rm 6.023\;\times\;10^2^3[/tex].
Computation for number of molecules in [tex]\rm (NH_4)_3PO_4[/tex]
The given mass of ammonium phosphate is 2.38 g.
The molar mass of ammonium phosphate is 149 g/mol
The moles of the sample is given as:
[tex]\rm Moles=\dfrac{Mass}{Molar\;mass} \\\\Moles\;(NH_4)_3PO_4=\dfrac{2.38\;g}{149\;g/mol}\\\\ Moles\;(NH_4)_3PO_4=0.015\;mol[/tex]
The available moles of ammonium phosphate is 0.015 mol.
According to Avogadro's law, the number of particles in the sample is given as:
[tex]\rm 1\;mol=6.023\;\times\;10^2^3\;particles\\\\0..015\;mol=\dfrac{6.023\;\times\;10^2^3\;particles}{1\;mol}\;\times\;0.015\;mol\\\\ 0.015\;mol=9.0345\;\times\;10^2^1\;particles[/tex]
The number of particles in 2.38 g ammonium phosphate are [tex]\rm \bold{9.0345\;\times\;10^2^1}[/tex].
Learn more about the number of particles, here:
https://brainly.com/question/1445383
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
