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Lapatulllka [165]
4 years ago
11

Properties of metals

Chemistry
1 answer:
Nitella [24]4 years ago
3 0

Answer:

Physical Properties of Metals :

Metals are lustrous, malleable, ductile, good conductors of heat and electricity. Other properties include: State: Metals are solids at room temperature with the exception of mercury, which is liquid at room temperature (Gallium is liquid on hot days).

Hope this helps! :D

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When solid Ni metal is put into an aqueous solution of Pb(NO3)2, solid Pb metal and a solution of Ni(NO3)2 result. Write the net
Sonja [21]

Answer:

Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)

Explanation:

Hello there!

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Pb(NO_3)_2(aq)+Ni(s)\rightarrow Ni(NO_3)_2(aq)+Pb(s)

Now, we can separate the nitrates in ions as they are aqueous to obtain:

Pb^{2+}(aq)+2(NO_3)^-(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+2(NO_3)^-(aq)+Pb(s)

And then, we cancel out the nitrate ions as the spectator ones, for us to obtain the net ionic equation:

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Best regards!

6 0
3 years ago
) In the Deacon process for the manufacture of chlorine, HCl and O2 react to form Cl2 and H2O. Sufficient air (21 mole% O2, 79%
LiRa [457]

Answer:

Here's what I get.

Explanation:

1. Write the chemical equation

\rm 4HCl + O$_{2} \longrightarrow \,$ 2Cl$_{2}$ + 2H$_{2}$O

Assume that we start with 4 L of HCl

2. Calculate the theoretical volume of oxygen

\text{V}_{\text{O}_{2}}= \text{4 L HCl} \times \dfrac{\text{1 L O}_{2}}{\text{4 L HCl}} = \text{1 L O}_{2}}

3. Add 35% excess

\text{V}_{\text{O}_{2}}= \text{1 L O}_{2}} \times 1.35 = \text{1.35 L O}_{2}}

4. Calculate the theoretical volume of nitrogen

\text{V}_{\text{N}_{2}} = \text{1.35 L O}_{2}} \times \dfrac{\text{79 L N}_{2}}{\text{21 L O}_{2}}} = \text{5.08 L N}_{2}}

4. Calculate volumes of reactant used up

Only 85 % of the HCl is converted.

We can summarize the volumes in an ICE table

           4HCl     +       O₂    +    N₂   →    2Cl₂   +   2H₂O

I/L:          4               1.35         5.08         0              0

C/L:  -0.85(4)        -0.85(1)        0      +0.85(2)   +0.85(2)

E/L:     0.60             0.50        5.08       1.70          1.70

5. Calculate the mole fractions of each gas in the product stream

Total volume = (0.60 + 0.50 + 5.08 + 1.70 + 1.70) L = 9.58 L

\chi = \dfrac{\text{V}_{\text{component}}}{\text{V}_\text{total}} = \dfrac{\text{ V}_{\text{component}}}{\text{9.58}} = \text{0.1044V}_{\text{component}}\\\\\chi_{\text{HCl}} = 0.1044\times 0.60 = 0.063\\\\\chi_{\text{O}_{2}} = 0.1044\times 0.50 = 0.052\\\\\chi_{\text{N}_{2}} = 0.1044\times 5.08 = 0.530\\\\\chi_{\text{Cl}_{2}} = 0.1044\times 1.70 = 0.177\\\\\chi_{\text{H$_{2}${O}}} = 0.1044\times 1.70 = 0.177\\\\

8 0
3 years ago
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