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katen-ka-za [31]
3 years ago
12

What are two properties of diamonds that make them different from other forms of carbon?

Chemistry
1 answer:
Airida [17]3 years ago
6 0
They are tetrahedral in shape
they carbon atom share all four of its outer electrons with other carbon atoms to form 3D shape
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Because you need to know what you are looking for before actually trying something so you can prevent any accidents by doing stuff at random
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What coefficient should be used to make the following equation balanced? N2+O2--> _NO
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<h2>                  ║∧║······→Hello.←······║∧║</h2>

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                                                N2 + O2 → 2NO

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I hope this answered your question!

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(<em>A brainliest would be appreciated!)</em>

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7 0
3 years ago
The volume of a given mass of an ideal gas at constant pressure is
zmey [24]

Answer:

inversely proportional to the kelvin temperature

Explanation:

PV=nRT

8 0
2 years ago
Read 2 more answers
Which of the following statements are FALSE of models? (select all that applies)
ira [324]

Answer:

A B and C

Explanation:

Hopefully this helps

7 0
3 years ago
Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr
VikaD [51]

Answer : The value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

Explanation :

The balanced cell reaction will be,

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

The half-cell reactions are:

Oxidation reaction (anode) : Pb(s)\rightarrow Pb^{2+}(aq)+2e^-

Reduction reaction (cathode) : 2Ag^+(aq)+2e^-\rightarrow 2Ag(g)

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

where,

\Delta G^o = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential of the cell = 0.93 V

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

\Delta G^o=-2\times 96500\times 0.93

\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol

Now we have to calculate the value of 'K'.

\Delta G^o=-RT\ln K

where,

\Delta G_^o =  standard Gibbs free energy  = -180 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K

K=3.6\times 10^{31}

Therefore, the value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

5 0
3 years ago
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