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3 years ago

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The transition metals (with the exception of Zn, Cd and Hg) are very much hard and have low volatility. Why Zn ,Cd and Hg has lo

w melting points. (a) These have all their electrons paired. (b) These have maximum number of unpaired electrons. (c) These have very strong metallic bonding. (d) These have more metal-metal bonding.

1 answer:

lana66690 [7]3 years ago

6 0

Answer:

Zn ,Cd and Hg has low melting points because - <u>These have all their electrons paired.</u>

Explanation:

<u>These have all their electrons paired -: </u> All the electrons in the d-subshell are combined with Zn , Cd and Hg. The metallic bonds present in them are, therefore, weak. This is why they have low points of melting and bolting. therefore , this statement is true.
<u> These have maximum number of unpaired electrons</u>. Since , these transition metals have all electrons paired , so this statement is incorrect.
<u>These have very strong metallic bonding -:</u> Zn, Cd and Hg d-orbital are completely filled. They have poor metallic bonding and less compact packing due to their fully filled d-orbitals, so all of them are volatile in nature. <u>Therefore, this statement is also not val</u>id .
<u>These have more metal-metal bonding. -: </u> Since , their metallic bond is weak , so this statement is not true .

Hence , the correct option is A (These have all their electrons paired).

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3 years ago

A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.

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Answer:

$Ag_2SO_4$

Explanation:

Formula for the calculation of no. of Mol is as follows:

$mol=\frac{mass\ (g)}{molecular\ mass}$

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

$mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol$

Molecular mass of S = 32 g/mol

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Therefore, divide by 0.02657

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4 years ago

Read 2 more answers

Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =

Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Wherem

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314 J / K mol

Thus, given that, $E_a$ = ?

$k_2=4.0\times 10^3s^{-1}$

$k_1=1.5\times 10^3s^{-1}$

$T_1=5\ ^0C$

$T_2=25\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (5 + 273.15) K = 278.15 K

T = (25 + 273.15) K = 298.15 K

$T_1=278.15\ K$

$T_2=298.15\ K$

So,

$\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)$

$E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}$

$E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}$

$E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}$

$E_a=33813.28\ J/mol$

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

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3 years ago