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sukhopar [10]
3 years ago
13

A divalent metal ion dissolved in dilute hydrochloric acid forms a precipitate when H2S is bubbled through the solution. Which i

on is is?
A. Ca2+ B. Mn2+ C. Zn2+. D. Cd2+
Chemistry
1 answer:
faltersainse [42]3 years ago
7 0

Answer:

D. Cd2+

Explanation:

The metal ions are classified into four groups which is based on their solubility.

Group II cations are the cations which are based on the solubility of the Sulphides.

Group II (Cu^{2+}, Bi^{3+}, Cd^{2+}, Hg^{2+}, As^{3+}, Sb^{3+}, Sn^{4+}) cations are the cations which produce very insoluble sulfides.

So, among the given options,  Cd^{2+} is the only cation which comes in Group II and thus, it will precipitate.

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Catalyst
Hoochie [10]

a. volume of NO : 41.785 L

b. mass of H2O : 18 g

c. volume of O2 : 9.52 L

<h3>Further explanation</h3>

Given

Reaction

4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Required

a. volume of NO

b. mass of H2O

c. volume of O2

Solution

Assume reactants at STP(0 C, 1 atm)

Products at 1000 C (1273 K)and 1 atm

a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

\tt \dfrac{4}{5}\times 0.5=0.4

volume NO at 1273 K and 1 atm

\tt V=\dfrac{nRT}{P}=\dfrac{0.4\times 0.08206\times 1273}{1}=41.785~L

b. 15 L NH3 at STP ( 1mol = 22.4 L)

\tt \dfrac{15}{22.4}=0.67~mol

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

\tt \dfrac{6}{4}\times 0.67=1

mass H2O(MW = 18 g/mol) :

\tt mass=mol\times MW=1\times 18=18~g

c. mol NO at 1273 K and 1 atm :

\tt n=\dfrac{PV}{RT}=\dfrac{1\times 35.5}{0.08206\times 1273}=0.34

mol ratio of NO : O2 = 4 : 5, so mol O2 :

\tt \dfrac{5}{4}\times 0.34=0.425

Volume O2 at STP :

\tt 0.425\times 22.4=9.52~L

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3 years ago
Determine the number of molecules in 43.9 g of carbon tetrachloride
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Explanation:

7 0
2 years ago
A chemist performs the same tests on two white solids, A and B. The results are shown in the table below.
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Answer:

i'm not sure but mabey A?

Explanation:

4 0
3 years ago
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