17) 8.4 / 20 x 100
18) 20 . 0.5150
19) 6,50% because (as you said) the law of definite proportions states that regardless of the amount, a compound is always composed of the same elements in the same proportion by mass
The molar enthalpy of combustion in kj/mol of magnesium is 620 kj/mol, Option D is the correct answer.
<h3>What is enthalpy of Combustion ?</h3>
The energy released when a fuel is oxidized by an oxidizing agent is called enthalpy of Combustion.
It is given that
a 1.0 g sample of magnesium is burned to form MgO. in doing so, 25.5 kj of energy are released.
Molecular weight of Magnesium = 24.35g
24.35 g makes 1 mole of Mg
1g = 1/24.35
For 0.04 moles 25.5 kJ is released
for 1 mole 25.5 *1/.04
= 620 kj/mol
Therefore the molar enthalpy of combustion in kj/mol of magnesium is 620 kj/mol.
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Sodium phosphide is the inorganic compound with the formula Na3P.
Answer:
Its phosphorus (P)
Explanation:
In writing the electron configuration for Phosphorus the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Phosphorous go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s. Since the 3s if now full we'll move to the 3p where we'll place the remaining three electrons. Therefore the Phosphorus electron configuration will be 1s22s22p63s23p3.