Answer:
No of Moles in excess at the end of the reaction is 0.25 moles
Explanation:
AgNO3 + Mg3P2 → Ag3P + Mg(NO3)2
Balancing the equation we get
6AgNO3 + Mg3P2 → 2Ag3P + 3Mg(NO3)2
6 moles of AgNO3 needs 1 mole of Mg3P2
using unitary method
AgNO3 = 
1.5 AgNO3 = 
= 1/4 = 0.25moles of Mg3P2
So 1.5 Moles of AgNO3 requires 0.25Mg3P2 for complete reaction but we have 0.5Moles of Mg3P2 available Therefore Mg3P2 is in excess
No of Moles in excess at the end of the reaction = 0.5 - 0.25 = 0.25moles
Maybe propelled like moving forward
boron, carbon, nitrogen, oxygen and flourine families in addition to the noble gases.
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A Bronsted-Lowry acid-base is a molecule or ion that donates a hydrogen ion in a reaction.
<em>Brainliest Please?</em>
Answer:
0.056moles HF and 0.70M
Explanation:
When a strong acid is added to a buffer, the acid reacts with the conjugate base.
In the system, NaF and HF, weak acid is HF and conjugate base is NaF. The reaction of NaF with HCl (Strong acid) is:
NaF + HCl → HF + NaCl
Initial moles of NaF and HF in 60.0mL of solution are:
NaF:
0.0600L × (0.80mol / L)= 0.048 moles NaF
HF:
0.0600L × (0.80mol / L)= 0.048 moles HF
Then, the added moles of HCl are:
0.0200L × (0.40mol / L) = 0.008 moles HCl.
Thus, after the reaction, moles of HF produced are 0.008 moles + the initial 0.048moles of HF, moles of HF are:
<em>0.056moles HF</em>
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In 20.0mL + 60.0mL = 80.0mL = 0.0800L, molarity of HF is:
0.056mol HF / 0.0800L = <em>0.70M</em>
