Answer:
3Mg(s) +2P(s) -------> Mg3P2(s) + energy
Keq= [Mg3P2]/[Mg]^3 [P]^2
Explanation:
The equation for the formation of magnesium phosphide from its elements is;
3Mg(s) +2P(s) -------> Mg3P2(s) + energy
Hence we can see that three moles of magnesium atoms combines with two moles of phosphorus atoms to yield one mole of magnesium phosphide. The equation written above is the balanced chemical reaction equation for the formation of the magnesium phosphide.
The equilibrium expression for the reaction K(eq) will be given by;
Keq= [Mg3P2]/[Mg]^3 [P]^2
Answer:
I need help for this question
The molarity of a solution that contains 35.00 g of CuSO4 dissolved in 250.0 mL of water is 0.88M.
<h3>How to calculate molarity?</h3>
The molarity of a solution can be calculated using the following formula:
Molarity = no of moles/volume
According to this question, a solution consists of 35.00 g of CuSO4 dissolved in 250.0 mL of water.
no.of moles of CuSO4 = 35g ÷ 159.6g/mol
no. of moles of CuSO4 = 0.22 moles
Therefore; molarity of CuSO4 solution is calculated as follows:
M = 0.22 ÷ 0.25
M = 0.88M
Therefore, the molarity of a solution that contains 35.00 g of CuSO4 dissolved in 250.0 mL of water is 0.88M.
Learn more about molarity at: brainly.com/question/12127540
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
