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erma4kov [3.2K]
3 years ago
8

Which of the above solutions could be used to remove lead from lead (II) nitrate?

Chemistry
1 answer:
alisha [4.7K]3 years ago
8 0

Ni solution could be used to remove lead from lead (II) nitrate.

Explanation:

The removal of lead from lead nitrate will take place by displacement reaction.

In displacement reaction less reactive element is displaced by more reactive element form it compound.

The reactivity is decided by the placement of metal in the activity series.

A metal which is at higher position in the activity series will be able to displace the metal or element having lower position.

From the options given we will check their position in the activity series in comparison to Pb

Cu is lower in series than Pb hence cannot displace.

Hg is placed lower in the series than Pb hence cannot displace Pb.

Ag is placed lower than Pb in the series hence cannot displace Pb.

Ni is placed above the Pb in activity series hence can displace lead.

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A sample of 250 g of water are heated from 40°C to 121°C, calculate the amount of heat energy absorbed.
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Answer:

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Explanation:

H=MC(change of temp.)

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H= 250x4.186x81=84766.5J

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If a liquid substance is transferred to a different container, what can be predicted about the volume of the liquid in the new c
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The same

Explanation:

If a liquid substance is transferred to a different container, the volume of the liquid in the new container will remain the same.

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Microwave radiation has a wavelength on the order of 1.0 cm. Calculate the frequency and the energy of a single photon of this r
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Answer :

(1) The frequency of photon is, 3\times 10^{10}Hz

(2) The energy of a single photon of this radiation is 1.988\times 10^{-23}J/photon

(3) The energy of an Avogadro's number of photons of this radiation is, 11.97 J/mol

Explanation : Given,

Wavelength of photon = 1.0cm=0.01m     (1 m = 100 cm)

(1) Now we have to calculate the frequency of photon.

Formula used :

\nu=\frac{c}{\lambda}

where,

\nu = frequency of photon

\lambda = wavelength of photon

c = speed of light = 3\times 10^8m/s

Now put all the given values in the above formula, we get:

\nu=\frac{3\times 10^8m/s}{0.01m}

\nu=3\times 10^{10}s^{-1}=3\times 10^{10}Hz    (1Hz=1s^{-1})

The frequency of photon is, 3\times 10^{10}Hz

(2) Now we have to calculate the energy of photon.

Formula used :

E=h\times \nu

where,

\nu = frequency of photon

h = Planck's constant = 6.626\times 10^{-34}Js

Now put all the given values in the above formula, we get:

E=(6.626\times 10^{-34}Js)\times (3\times 10^{10}s^{-1})

E=1.988\times 10^{-23}J/photon

The energy of a single photon of this radiation is 1.988\times 10^{-23}J/photon

(3) Now we have to calculate the energy in J/mol.

E=1.988\times 10^{-23}J/photon

E=(1.988\times 10^{-23}J/photon)\times (6.022\times 10^{23}photon/mol)

E=11.97J/mol

The energy of an Avogadro's number of photons of this radiation is, 11.97 J/mol

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