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Schach [20]
3 years ago
15

If HCl were added to a solution containing lead(II) ions, sodium ions, nickel(II) ions, and potassium ions, which ions would pre

cipitate? (Select all that apply.). lead ions. nickel ions. potassium ions. sodium ions. none of them
Chemistry
1 answer:
Romashka-Z-Leto [24]3 years ago
8 0
When HCl is added to metal ions, metal chlorides are produced. In this problem, it is asked whether the given ions precipitate or not when added to HCl. According to the rule, all chlorides except Ag+, Pb 2+, Hg2 2+ are soluble. Hence the ion that would precipitate is only lead (II) ion.
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SpyIntel [72]

Answer:

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Explanation:

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8 0
3 years ago
Explain, in terms of electrons, why the bonding in strontium sulfide, SrS, is similar to the bonding in magnesium bromide, MgBr2
Aleksandr-060686 [28]

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Explanation: both Sr and Mg are earth alkaline metals and form ions Mg^2+

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5 0
3 years ago
What is the molarity of 1 mole of HCl in 5 liters of solution?
kicyunya [14]
Molarity = moles of solute(HCl)
                  ------------------------------------
                   volume of the solution
               
                =   1
                    ------
                     5 
               
                =  0.2M.

Hence option B is correct.
Hope this helps!!
5 0
3 years ago
Read 2 more answers
Which of the following is a homogeneous mixture? a. Na(s) b. H2O(l) c. Cl2(g) d. NaCl(aq)
iVinArrow [24]

Answer:

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7 0
3 years ago
Volume of HCl used 25.0mL 4 l
borishaifa [10]

Answer:

1.0 M

Explanation:

Reaction equation;

KOH(aq) + HCl(aq) -----> KCl(aq) + H2O(l)

Concentration of acid CA = ?

Concentration of base CB = 1.0 M

Volume of base VB = 25.60 - 0.50 = 25.1 ml

Volume of acid VB =  25.0 ml

Number of moles of acid NA = 1

Number of moles of base NB =2

CAVA/CBVB =NA/NB

CAVANB = CBVBNA

CA = CBVBNA/VANB

CA = 1 * 25.1 * 1/25.0 *1

CA = 1.0 M

8 0
3 years ago
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