<span>Mass of nitrogen = 14.0067
</span>
Mass of oxygen = 15.9994
In this compound nitrogen = 36.86 /
14.0067 = 2.63
<span>And oxygen = 63.14 / 15.9994 = 3.95 <span>
now we have: N----- 2.63 and O----3.95
by dividing both with the smallest number we get
</span></span>
<span>N-------2.63/2.63 = 1<span>
<span>O-------3.95/2.63 = 1.5
To get whole numbers we multiply both by 2
</span></span></span>
N= 1 x 2 = 2
And O = 1.5 x 2= 3
<span>So, the empirical formula is N</span>₂O₃.
The amount of Mg(OH)2 present after the reaction is complete is 0.136 moles of Mg(OH)2.
The equation of the reaction is;
2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)
Number of moles of Mg(OH)2 = 8.00 g/58 g/mol = 0.138 moles
Number of moles of HNO3 = 0.205 M × 24.2 mL/1000 = 0.00496 moles
Given that;
2 moles of HNO3 reacts with 1 mole of Mg(OH)2
0.00496 moles of HNO3 reacts with 0.00496 moles × 1 mole /2 moles = 0.00248 moles of Mg(OH)2
Hence, Mg(OH)2 is the reactant in excess.
The amount of Mg(OH)2 remaining = Amount present - Amount reacted
Hence; 0.138 moles - 0.00248 moles = 0.136 moles of Mg(OH)2
Learn more: brainly.com/question/9743981
Trick question: you just stated everything in the beakers.
Answer: C
Explanation:
The one closest to the atomic center, there is a single 1s orbital that can hold 2 electrons. At the next energy level, there are four orbitals.
