Na₂O is an ionic compound. Na is in group 1 which means it has one electron in the outer shell. O is in group 6 so it has 6 electrons in the outer shell.
To become stable atoms need to either lose the electrons in the outer shell or gain electrons to gain a complete outer shell configuration.
To become stable Na loses its outer electron and becomes positively charged. It only loses one electron, so net charge is +1. O to become stable gains 2 electrons to complete its outer shell and becomes negatively charged. since it gains 2 electrons, the net charge is -2.
cation - Na⁺
anion - O²⁻
O gains 2 electrons but Na can give only one electron, therefore 2 Na⁺ ions are required.
the compound can be written by exchanging the charges
ions Na⁺ O²⁻
charge +1 -2
exchange Na₂O
Correct answer is <span>Na has +1 charge and O has -2 charge</span>
Answer:
39.6 g
Explanation:
The equation of the reaction is;
2Mg(s) + O2(g) --------> 2MgO(s)
To obtain the limiting reactant;
Number of moles in 26.4 g of Mg = 26.4g/24 g/mol = 1.1 moles
If 2 moles of Mg yields 2 moles of MgO
1.1 moles of Mg yields 1.1 * 2/2 = 1.1 moles of MgO
Number of moles in 26.4 g of O2 = 26.4 g/32g/mol = 0.825 moles
If 1 mole of O2 yields 2 moles of MgO
0.825 moles of O2 yields 0.825 moles * 2/1 = 1.65 moles of MgO
Hence Mg is the limiting reactant.
Theoretical yield of MgO = 1.1 moles of MgO * 40 g/mol = 44 g
Percent yield = 90%
Percent yield = actual yield/theoretical yield * 100
Actual yield = Percent yield * theoretical yield/100
Actual yield = 90 * 44/100
Actual yield = 39.6 g
4NH3+5O2 <=>4NO + 6H2O
Using the definition of Kp, we have
Kp=(Pno^4*Ph2o^6)/(Pnh3^4*Po2^5)
where Pno=partial pressure of NO, etc.
The numerical value for a given temperature can be evaluated when the actual partial pressures are known.
K gives 1 electron, and Br takes 1 electron.
