<span>What is the maximum number of electrons in the following energy level? n = 2
2
</span>
Since 1mL=1cm^3 the wood would sink due to it being more dense. I.e. 0.95>0.88
Explanation:
An object can possess energy in tow ways by it's motion or position
<u>Given:</u>
Mass of MgBr2 = 0.500 g
<u>To determine:</u>
Number of anions in 0.500 g MgBr2
<u>Explanation:</u>
Molar mass of MgBr2 = 24 + 2 (80) = 184 g/mol
Moles of MgBr2 = 0.500 g/184 g.mol-1 = 0.00271 moles
Based on stoichiometry-
1 mole of MgBr2 has 1 mole of Mg2+ cations and 2 moles of Br- anions
Therefore, 0.00271 moles of MgBr2 will have: 2 * 0.00271 = 0.00542 moles of Br-
Now,
1 mole of Br- contains 6.023 * 10²³ anions
0.00542 moles of Br- contain: 0.00542 * 6.023*10²³ = 3.264*10²¹ anions
Ans: There are 3.264*10²¹ anions in 0.5 g of MgBr2
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
