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grandymaker [24]
3 years ago
10

What do you think would happen if a solid were exposed to extremely cold temperatures, or if a gas were exposed to extremely hot

temperatures?
Chemistry
1 answer:
grandymaker [24]3 years ago
8 0

At high temperature the gas would diffuse out as the pressure increases and at extremely low temperature the solid becomes compact.

Explanation:

The states of matter largely depends on the temperature. Any substance when crosses the threshold temperature its phase changes.

When temperature is low the motion of molecules is also low and internal energy also gets low. Solid have tendency of settling in low energy level and have highly compact molecules. At low temperature the solid would compress as molecules would be highly condensed.

Gas in the nature has its molecules quite far apart in matter. According to Kinetic theory of gases the increase in temperature causes rapid collisions of the gas molecule as the kinetic energy of molecules increases. The greater force of collision would cause increase in pressure of the container and increased diffusion rate.

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Assuming that the bath contains 250.0 g of water and that the calorimeter itself absorbs a negligible amount of heat, calculate
Anvisha [2.4K]

Answer:

-3272     kJ/mol

Explanation:

Given and known facts

Mass of Benzene = 0.187 grams

Mass of water = 250 grams

Standard heat capacity of water = 4.18 J/g∙°C

Change in temperature ΔT = 7.48°C

Heat

=250 * 4.18 * 7.48\\=7816.6 \\=7.82

Heat released by benzine is - 7.82 kJ

Now, we know that

0.187 grams of benzene release = -7.82  kJ heat

So, 1 g benzine releases

\frac{ -7.82 }{0.187}\\= -41.8

kJ/g

0.187 * \frac{1}{78.108}=0.00239 mol C6H6

Heat released

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4 0
3 years ago
Please help make sure its correct thanks
Wewaii [24]

The empirical formula of metal iodide : CoI₃(Cobalt(III) Iodide)

<h3>Further explanation</h3>

13.02 g sample of Cobalt , then mol Co(MW=58.933 g/mol) :

\tt mol=\dfrac{mass}{Ar}=\\\\mol=\dfrac{13.02~g}{58.933}\\\\mol=0.221

Mass of metal iodide formed : 97.12 g, so mass of Iodine :

\tt =mass~metal~iodide-mass~Cobalt\\\\=97.12-13.02\\\\=84.1~g

Then mol iodine (MW=126.9045 g/mol) :

\tt \dfrac{84.1}{126.9045}=0.663

mol ratio of Cobalt and Iodine in the compound :

\tt 0.221\div 0.663=1\div 3

5 0
2 years ago
Predict the empirical formula of these compounds. a. C2N2. b. P4O10. c. N2O5. d. NaCl. e. C9H20. f. B2H6. g. K2Cr2O7. h. Al2Br6.
Stella [2.4K]
The empirical formula is the simplest form of the formula expressed in the lowest ratio. In this case, we just have to divide each subscript by the greatest common factor. Hence.
a. CN
b. P2O5
c.N2O5
d.NaCl
e. C9H20
f. BH3
g.K2Cr2O7
h.AlB3
i.CH
j.SiCl4
5 0
3 years ago
Someone please help me with number 5. no idea what it is.
KATRIN_1 [288]
Answer is Sodium Hydroxide.
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These two elements are most likely good conductors of heat and electricity and shiny when solid.
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Copper and Aluminium..........
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