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4 years ago

Can someone please explain to me the metallic bond?

Chemistry

1 answer:

Arlecino [84]4 years ago

6 0

Metallic bond is the electrostatic attraction between a positive ion and delocalised(free)electrons

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The answer to the multiple choice question

GenaCL600 [577]

The answer is A but i'm not sure .

3 0

3 years ago

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In class we derived the Gibbs energy of mixing for a binary mixture of perfect gases. We also discussed that the same result is

GaryK [48]

Answer:

Attached below

Explanation:

Free energy of mixing = ΔGmix = Gf - Gi

attached below is the required derivation of the

<u>a) Molar Gibbs energy of mixing</u>

ΔGmix = Gf - Gi

hence : ΔGmix = ∩RT ( X1 In X1 + X2 In X2 + X3 In X3 + ------- )

<u>b) molar excess Gibbs energy of mixing</u>

Ni = chemical potential of gas

fi = Fugacity

N°i = Chemical potential of gas when Fugacity = 1

ΔG = RT In ( a2 / a1 )

4 0

3 years ago

Equal but opposite forces acting on an object results in what?

forsale [732]

Answer:

Action given and reaction taken

Also known as

Newton's third law of motion

Explanation:

An action will be done such as bouncing a ball on the wall

- You throw the ball (Action)

- The ball bounces back (Reaction)

Hope this Helps

3 0

3 years ago

How many grams of Na2SO4 can be produced from 423.67 g of NaCl?

san4es73 [151]

Mass of Na2SO4= 514.18 grams

<h3>Further explanation</h3>

Given

423.67 g of NaCl

Required

mass of Na2SO4

Solution

Reaction

2NaCl + H2SO4 → Na2SO4 + 2HCl

mol NaCl :

= 423.67 g : 58.5 g/mol

= 7.24

From the equation, mol Na2SO4 :

= 1/2 x mol NaCl

= 1/2 x 7.24

= 3.62

Mass Na2SO4 :

= 3.62 mol x 142,04 g/mol

= 514.18 grams

4 0

3 years ago

At 517 mm Hg and 24 °C, a sample of gas occuples a volume of 95 ml. The gas is transferred to a 225-ml flask and the temperature

vodomira [7]

Answer:

$P_2=194.78mmHg$

Explanation:

Hello,

In this case, we employ the combined ideal gas law in order to understand the volume-gas-pressure behavior as shown below:

$\frac{P_1V_1}{T_1}= \frac{P_2V_2}{T_2}$

Hence, solving for the final pressure P2, we obtain (do not forget temperature must be absolute):

$P_2=\frac{P_1V_1T_2}{V_2T_1}=\frac{517mmHg*95mL*(-8.0+273.15)K}{(24+273.15)K*225mL}\\ \\P_2=194.78mmHg$

Best regards.

7 0

4 years ago

Read 2 more answers