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
I'm gonna go ahead and guess, and say its B.
A conversion factor is ALWAYS equal to 1 :)
Answer:
SeF4 is a polar molecule
Explanation:
SeF4 is a polar molecule because a polar molecule is any molecule that have lone pairs of electrons in the central atom or have atoms that are electronegative and the electrons between that are covalently bonded are not evenly distributed.
The electronegative atoms of flourine in SeF4 are not evenly distributed and kind pairs of electrons are on the central atom.
Answer:
My guess is b or c but its robably wrong
Explanation:
I just also need points sorry <3
