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worty [1.4K]
3 years ago
9

How many moles of magnesium chloride (mgcl2) are in 1.31 l of 0.38 m mgcl2?

Chemistry
1 answer:
Mumz [18]3 years ago
5 0
Molarity can be defined as the number of moles dissolved in 1 L of solution.
The molarity of MgCl₂ is 0.38 mol/L
In 1 L of MgCl₂ solution, amount go 0.38 moles is dissolved.
Therefore in 1.31 L of MgCl₂  - 0.38 mol /1 L x 1.31 L = 0.4978 mol rounded off  is 0.50 mol
In 1.31 L of solution, amount of 0.50 mol of MgCl₂ is present.
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An atom of 125Sn has a mass of 124.907785 amu. Calculate the binding energy in MeV per atom. Use the masses: mass of 1H atom = 1
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1.06×10³ MeV/atom

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