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allsm [11]
3 years ago
11

What is the partial pressure of argon, par, in the flask? express your answer to three significant figures and include the appro

priate unit?
Chemistry
1 answer:
monitta3 years ago
3 0
Given which are missing in your question:
the flask is filled with 1.45 g of argon at 25 C° 
So according to this formula (Partial pressure):
PV= nRT
first, we need n, and we can get by substitution by:
n =  1.45/mass weight of argon
   = 1.45 / 39.948 = 0.0363 mol of Ar
we have R constant = 0.0821
and T in kelvin = 25 + 273 = 298
and V = 1 L
∴ P * 1 = 0.0363* 0.0821 * 298 = 0.888 atm

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iogann1982 [59]

The molecules  of hydrogen gas that are  formed is  when 48.7 g  of sodium are added to water  is  6.375  x 10²³  molecules


  <u><em>calculation</em></u>

2 Na +2H₂O →  2 NaOH  +H₂

Step 1: find the moles  of sodium (Na)

moles =mass÷   molar mass

from periodic table the molar mass of Na = 23 g/mol

moles= 48.7 g÷ 23 g/mol =2.117 moles

Step 2:use the mole ratio to determine  the  moles of H₂

  from  given equation  Na:H₂ is 2:1

therefore the moles  of H₂ = 2.117 moles x 1/2=1.059 moles

Step 3: find the molecules of H₂ using the Avogadro's law

According  to  Avogadro's  law  1 mole = 6.02 x 10²³ molecules

                                                      1.059 moles = ?  molecules

by cross multiplication

= [(1.059 moles x 6.02 x10²³  molecules) /  1 mole] =6.375  x 10²³  molecules

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Fittoniya [83]
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