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creativ13 [48]
3 years ago
12

A student has 7.05 g of zinc powder, 1.60 L of a 3.40 M calcium nitrate solution, and 1.50 L of a 1.60 M lead(II) nitrate soluti

on. Which of these two solutions will react with the zinc powder?
1) calcium nitrate
2) lead(II) nitrate

Chemistry
1 answer:
trapecia [35]3 years ago
6 0

<u>Answer:</u> Zinc will react with lead (II) nitrate solution.

<u>Explanation:</u>

Single displacement reaction is defined as the reaction in which more reactive element displaces a less reactive element from its chemical reaction.

The reactivity of metal is determined by a series known as reactivity series. The metals lying above in the series are more reactive than the metals which lie below in the series.

General equation for single displacement reaction follows:

A+BC\rightarrow AC+B

When zinc is reacted with calcium nitrate, the reaction does not take place as zinc is less reactive than calcium. Zinc lies below in the series than calcium.

Zn+Ca(NO_3)_2\rightarrow \text{No reaction}

But, when zinc is reacted with lead (II) nitrate, the reaction do take place as zinc is more reactive than lead. Zinc lies above in the series than lead.

The chemical equation for the reaction of zinc and lead (II) nitrate follows:

Zn+Pb(NO_3)_2\rightarrow Zn(NO_3)_2+Pb

Hence, zinc will react with lead (II) nitrate solution.

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All of the following statements concerning voltaic cells are true EXCEPT
MissTica

Answer:

oxidation occurs at the cathode.

Explanation:

In a voltaic cell electrons move from anode to cathode. At the anode, species give up electrons. This is an oxidation reaction depicted by the oxidation half equation. At the cathode, species accept electrons and become reduced. This is depicted by the reduction half equation. In summary; in a Voltaic cell, oxidation occurs at the anode while reduction occurs at the cathode.

3 0
3 years ago
MnO2 + 4HCl reacts to form MnCl2 + CL2 + 2H2O If .86 mol of MnO2 and 48.2 g of HCl react, which reactant is the limiting reagent
Katen [24]
HCl:
<span>
m=48,2g
M=36,5g/mol

n = m/M = 48,2g / 36,5g/mol = 1,32mol

1mol          : 4mol
MnO</span>₂        + 4HCl ⇒ MnCl₂ + Cl₂ + 2H₂O
0,86mol     :  1,32mol
                     limiting reagent 
0,33 will react

HCl is limiting reagent.
6 0
3 years ago
A saturated solution of Pb(IO3)2 in pure water has a lead ion concentration of 5.0 x 10-5 Molar. What is the Ksp value of Pb(IO3
Orlov [11]

Answer:

Option (E) is correct

Explanation:

Solubility equilibrium of Pb(IO_{3})_{2} is given as follows-

                   Pb(IO_{3})_{2}\rightleftharpoons Pb^{2+}+2IO_{3}^{-}

Hence, if solubility of Pb(IO_{3})_{2} is S (M) then-

                             [Pb^{2+}]=S(M) and [IO_{3}^{-}]=2S(M)

Where species under third bracket represent equilibrium concentrations

So, solubility product of Pb(IO_{3})_{2} , K_{sp}=[Pb^{2+}][IO_{3}^{-}]^{2}

Here, [Pb^{2+}]=S(M)=5.0\times 10^{-5}M

So, [IO_{3}^{-}]=2S(M)=(2\times 5.0\times 10^{-5})M=1.0\times 10^{-4}M

So, K_{sp}=(5.0\times 10^{-5})\times (1.0\times 10^{-4})^{2}=5.0\times 10^{-13}

Hence option (E) is correct

7 0
3 years ago
HELP ASAP!!!!!!
harina [27]

that seems very false but I believe its the second one

7 0
3 years ago
Read 2 more answers
How many particles would be found in a 1.224 g sample of K2O
lbvjy [14]

Answer:

9.96*10^21

Explanation:

Molar mass of K2O=29*2+16

= 74g per mol

number of moles in the sample= 1.224/ 74

=0.1654

Number of particles in 1 mole=6.0221409*10^23

Number of particles= 0.01654*6.0221409*10^23

=9.96*10^21

4 0
4 years ago
Read 2 more answers
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