The number of particles in one mole is given be Avagadro's number <span>6.022×10^23
Multiply by number of moles.
3 ×10^-21 mol * 6.022 ×10^23 molecules/mol = </span><span>1,807 molecules
(rounded to nearest whole number)
</span>
Answer:
Conclusion
Explanation:
I believe you were asking for the term that best matches with the description given. Typically the conclusion summarizes your experiment in a 1 to 2 paragraph format.
Answer:
Oxidation half reaction is written as follows when using using reduction potential chart
example when using copper it is written as follows
CU2+ +2e- --> c(s) +0.34v
oxidasation is the loos of electron hence copper oxidation potential is as follows
cu (s) --> CU2+ +2e -0.34v
Explanation:
Answer:
MgCO₃
Explanation:
From the question given above, we obtained:
MgF₂ + Li₂CO₃ —> __ + 2LiF
The missing part of the equation can be obtained by writing the ionic equation for the reaction between MgF₂ and Li₂CO₃. This is illustrated below:
MgF₂ (aq) —> Mg²⁺ + 2F¯
Li₂CO₃ (aq) —> 2Li⁺ + CO₃²¯
MgF₂ + Li₂CO₃ —>
Mg²⁺ + 2F¯ + 2Li⁺ + CO₃²¯ —> Mg²⁺CO₃²¯ + 2Li⁺F¯
MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF
Now, we share compare the above equation with the one given in the question above to obtain the missing part. This is illustrated below:
MgF₂ + Li₂CO₃ —> __ + 2LiF
MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF
Therefore, the missing part of the equation is MgCO₃
Given:
1.50 L
62.5 grams
and the MM of MgO: 40.31 g/mol
Molarity: mol/L
First, find mol.
62.5 g x 1mole ÷ 40.31 g = 1.55 mol
then divide mol and the given liters
1.55mol ÷ 1.50 L= 1.03 M
