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4 years ago

What are the parts of an atom, and how are they used to determine an atom’s mass and identity?

Chemistry

1 answer:

Annette [7]4 years ago

4 0

The majority of an atoms' mass comes from the protons and neutrons that make up its nucleus.

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At 20 âˆ˜C the vapor pressure of benzene C6H6 is 75 torr and that of toluene C7H8 is 22 torr. Assume that benzene and toluene fo

Olegator [25]

Answer:

a) Xbenzene = 0.283

b) Xtoluene = 0.717

Explanation:

At T = 20°C:

⇒ vapor pressure of benzene (P*b) = 75 torr

⇒ vapor pressure toluene (P*t) = 22 torr

Raoult's law:

Pi = Xi.P*i

∴ Pi: partial pressure of i

∴ Xi: mole fraction

∴ P*i: vapor pressure at T

a) solution: benzene (b) + toluene (t)

∴ Psln = 37 torr; at T=20°C

⇒ Psln = Pb + Pt

∴ Pb = (Xb)*(P*b)

∴ Pt = (Xt)*(P*t)

∴ Xb + Xt = 1

⇒ Psln = 37 torr = (Xb)(75 torr) + (1 - Xb)(22 torr)

⇒ 37 torr - 22 torr = (75 torr)Xb - (22 torr)Xb

⇒ 15 torr = 53 torrXb

⇒ Xb = 15 torr / 53 torr

⇒ Xb = 0.283

b) Xb + Xt = 1

⇒ Xt = 1 - Xb

⇒ Xt = 1 - 0.283

⇒ Xt = 0.717

6 0

4 years ago

The electrons which are primarily responsible for chemical bonds are:

earnstyle [38]

Is there any choices because i can answer if you give me choices

4 0

3 years ago

Read 2 more answers

DO TODAY, PLS HURRY I DONT HAVE TIME :C (10 points, this is science, and its a writeing Question) some animals survive winter by

Lerok [7]

Answer: Beavers

Explanation: they do not hibernate

hope it helps

7 0

3 years ago

Please help me its due today thanks :D

nirvana33 [79]

Answer:

true

Explanation:

6 0

3 years ago

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When 106 g of water at a temperature of 21.4 °C is mixed with 64.3 g of water at an unknown temperature, the final temperature o

MissTica

Answer:

THE INITIAL TEMPERATURE OF THE SECOND SAMPLE IS 4.93 C OR 277.93 K

Explanation:

Mass of first sample of water = 106 g

Initial temp of first sample = 21.4 °C = 21.4 + 273 K = 294.4 K

Mass of second sample = 64.3 g

Final temp of theresulting mixture = 46.8 °C = 46.8 + 273 K = 319.8 K

Specific heat capacity of water = 4.184 J/g K

It is worthy to note that;

Heat gained by the first sample = Heat lost by the second sample

Since heat = mass * specific heat capacity * change in temperature, we have

Mass * specific heat * change in temp of the first sample = Mass * specific heat * change in temp. of the second sample

MC (T2 - T1) = MC (T2-T1)

106 * 4.184 * ( 319.8 - 294.4) = 64.3 * 4.184 * ( 319.8 - T1)

106 * 4.184 * 25.4 = 269.0312 ( 319.8 - T1)

11 265.0016 = 269.0312 (319.8 - T1)

Since the change in temperature = 319.8 -T1

Change in temperature =11265.0016 / 269.0312

Change in temperature = 41.87

Change in temperature = 319.8 -T1

41.87 = 319.8 - T1

T1 = 319.8 - 41.87

T1 = 277.93 K

T1 = 4.93 °C

So therefore, the initial temperature of the sacond sample is 4.73 °C or 277.93 K

7 0

3 years ago