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aleksley [76]
3 years ago
9

Suppose you start with 1.80 g of mgco3·5h2o, after complete dehydration how many grams of the sample do you have remaining?

Chemistry
1 answer:
elixir [45]3 years ago
5 0
The  number of  grams of the sample that remained  after  hydration is 0.87  grams


                calculation
calculate the molar  mass of MgCO3. 5 H2O  which is

24 + 12+ (3 x16) + ( 5  x18)  =  174 g/mol

we  well  know  after  hydration  water evaporate leaving behind   MgCO3
therefore the  mass  remained  is  for  MgCO3

find  the   molar  mass of  MgCO3 =  24 +12 +( 3 x16)  = 84  g/mol

the  mass  remained is therefore

 = 
molar  mass of MgCO3 /  molar mass of MgCO3.5H2O  x 1.80 g

=  84g/mol / 174 g/mol  x1.8g = 0.87  grams
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What is the volume in L of a 0.825 mole sample of Ar at 600 mm Hg and 300 K?
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Answer:

V = 25.7 L

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PV = nRT

In this formula,

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        > V = volume (L)

        > n = number of moles

        > R = constant (0.0821 L*atm/K*mol)

        > T = temperature (K)

While there is a different constant that can be used if you want to keep the pressure in mmHg, there is a more common constant used when the pressure is in atm. So, to find the volume, you need to (1) convert mmHg to atm (by dividing by 760) and then (2) calculate the volume (using Ideal Gas Law).

<u>(Step 1)</u>

600 mm Hg              1 atm
-------------------  x  ---------------------  =  0.789 atm
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<u>(Step 2)</u>

PV = nRT

(0.789 atm) x V = (0.825 mole)(0.0821 L*atm/K*mol)(300 K)

(0.789 atm) x V = 20.32

V = 25.7 L

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