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3 years ago

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

1 answer:

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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Why is it important that a hypothesis be stated so that it can be modified?

ANEK [815]

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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3 years ago

What coefficient should be used to make the following equation balanced? N2+O2--> _NO

IgorLugansk [536]

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Your answer should be:

N2 + O2 → 2NO

I hope this answered your question!

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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