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Hoochie [10]
3 years ago
5

What is the empirical formula for a compound that is 31.9% potassium, 28.9% chlorine, and 39.2% oxygen?

Chemistry
1 answer:
notsponge [240]3 years ago
7 0
The  empirical formula  for a compound   is KClO3

     Explanation 
 
find  the  moles of each  element
moles  = % composition/molar  mass

molar  mass of  of  potassium =39g/mol ,chlorine = 35.5 g/mol, oxygen =16 g/mol

moles  of potassium  =  31.9 / 39 = 0.818  moles
moles of  chlorine    = 28.9/35.5 = 0.814 moles
moles of   oxygen  =  39.2/  16 =  2.45  moles


find the  mole ratio  by  dividing   with  the smallest  mole = 0.814  moles

potassium = 0.818/0.814  =1  
chlorine  =  0.814/0.814 = 1
oxygen =  2.45 /0.814 =3 


the empirical  formula   is therefore = KClO3
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Answer:

We need 7.5 mL of the 1M stock of NaCl

Explanation:

Data given:

Stock = 1M this means 1 mol/ L

A 0.15 M solution of 50 mL has 0.0075 moles NaCl per 50 mL

Step 2: Calculate the volume of stock we need

The moles of solute will be constant

and n = M*V  

M1*V1 = M2*V2

⇒ with M1 = the initial molair concentration = 1M

⇒ with V1 = the volume we need of the stock

⇒ with V2 = the volume we want to make of the new solution = 50 mL = 0.05 L

⇒ with M2 = the concentration of the new solution = 0.15 M

1*V1 = 0.15*(50)

V1 = 7.5 mL

Since 0.0075 L of 1M solution contains 0.0075 moles

50 mL solution will contain also 0.0075 moles but will have a molair concentration of 0.0075 moles / 0.05 L =0.15 M

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<h3>Answer:</h3>

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<h3>Explanation:</h3>

We are given;

Volume of NaOH solution = 2.5 Liters

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We are required to calculate the moles of NaOH

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