Ca(OH)₂ ==> Ca²⁺ + 2 OH<span>-
Ca(OH)</span>₂ is <span>strong Bases</span><span>
</span>Therefore, the [OH-] equals 5 x 10⁻⁴ M. For every Ca(OH)₂ you produce 2 OH⁻<span>.
</span>
pOH = - log[ OH⁻]
pOH = - log [ <span>5 x 10⁻⁴ ]
pOH = 3.30
pH + pOH = 14
pH + 3.30 = 14
pH = 14 - 3.30
pH = 10.7
hope this helps!</span>
M(P)=3.72 g
M(P)=31 g/mol
m(Cl)=21.28 g
M(Cl)=35.5 g/mol
n(P)=m(P)/M(P)
n(P)=3.72/31=0.12 mol
n(Cl)=m(Cl)/M(Cl)
n(Cl)=21.28/35.5=0.60 mol
P : Cl = 0.12 : 0.60 = 1 : 5
PCl₅ - is the empirical formula
Answer:
c is the answer then check it out
Explanation:
Lithium diisopropylamide (LDA) is used in many organic synthesis and is a strong base. It is prepared by the acid base reaction of N,N-diisopropylamine ( [(CH₃)₂CH]₂NH ) and butyllithium ( Li⁺⁻CH₂CH₂CH₂CH₃ ).
The equation is show below as:
[(CH₃)₂CH]₂NH + Li⁺⁻CH₂CH₂CH₂CH₃ ⇒ [(CH₃)₂CH]₂N⁻Li⁺ + CH₃CH₂CH₂CH₃
N,N-diisopropylamine ( [(CH₃)₂CH]₂NH ) is a weaker acid and hence, LDA ( [(CH₃)₂CH]₂N⁻Li⁺ ) is stronger base. (Weaker acid has stronger conjugate base)
Butyllithium ( Li⁺⁻CH₂CH₂CH₂CH₃ ) is a very strong base and hence, butane ( CH₃CH₂CH₂CH₃ ) is a very weak acid. (Strong base has weaker conjugate acid)