When the process of condensation occurs, the kinetic energy of particles

F. If 20.0 moles of CO2 is exhaled, how many particles of H2O is produced?

erma4kov [3.2K]

Answer:

11 moles

Explanation:

1 mole of C12H22O11 molecules produces 12 moles of CO2 molecules and 11 moles of H2O molecules.

7 0

3 years ago

A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

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

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

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

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

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

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

3 years ago

If the substance in the diagram was heated to T2, what is happening if the substance is allowed to approach T1?

VMariaS [17]

Answer:

b

Explanation:

5 0

3 years ago

PLEASE HURRY WILL MARK BRAINLIEST

mr Goodwill [35]

Answer:

12 Neutrons

1 Valence Electron

23 ions [11 Protons and 12 Neutrons]

22 ions [11 Electrons and Protons]

Explanation:

To find the number of neutrons in an element, you simply take its atomic number and deduct that from the atomic mass [round if necessary]. There is one valence electron because according to the Periodic Table of Elements, the first three energy levels can fit 2 - 8 electrons:

1st Energy Level → 2 Electrons

2nd Energy Level → 8 Electrons

3rd Energy Level → 8 Electrons

4th Energy Level → 18 Electrons

5th Energy Level → 18 Electrons

6th Energy Level [Lanthanide Series] → 32 Electrons

7th Energy Level [Actinide Series] → 32 Electrons

So, as you can see, 2 + 8 gives you 10. So from there, we already filled up 2 energy levels. Now going to the third one will leave us with only one electron left over. That is where that valence electron is.

** I encourage you to write this down in your Chemistry notebook somewhere, so you can keep this stored in your memory at all times. There are some pointers in the answer as well. Copy that.

I am joyous to assist you anytime.

4 0

3 years ago

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01

m_a_m_a [10]

Answer:

The correct answer is - Positive charge occupies a very small volume in the atom.

Explanation:

Ernest Rutherford's experiments exhibited the presence of the nuclear core: a small region with the greater part of the mass of the atom and the positive charge.

Rutherford's gold foil analyze gave three conclusions:

- the particle is generally vacant space

- in it is a little, thick core or dense nucleus

- the core is positively charged.

6 0

3 years ago

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