How many Mg2+ ions would be present in 3.0 L of seawater that is 0.052 M...

Question

Question

How many Mg2+ ions would be present in 3.0 L of seawater that is 0.052 M in Mg2+?

Answer 1

Answer #1

Molarity = moles/ V in mL

Mg²⁺ = 0.052 M

Volume = 3.0 L

moles = M*V

= 0.052*3.0

= 0.156 moles

1 mole of Mg²⁺ contains -------> 6.023*10^23 ions

0.156 moles of Mg²⁺ contains ------> 6.023*10^23*0.156

= 9.39*10^22 ions

Number of Mg²⁺ ions would be present in 3.0 L of seawater that is 0.052 M in Mg²⁺ = 9.39*10^22 ions

Answer 2

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