Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
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
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
