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Setler [38]
3 years ago
14

Click on the Delta H changes sign whan a process is reversed button within the activity and analyze the relationship between the

two reactions that are displayed. The reaction that was on the screen when you started and its derivative demonstrate that the reaction enthalpy, ΔH, changes sign when a process is reversed. Consider the reaction H2O(l)→H2O(g), ΔH =44.0kJ What will ΔH be for the reaction if it is reversed?
Chemistry
1 answer:
OleMash [197]3 years ago
6 0

Answer:

ΔH = - 44.0kJ

Explanation:

H2O(l)→H2O(g), ΔH =44.0kJ

In the reaction above, liquid water changes to gaseous water. This occurs through a process known as boiling. This process requires heat, hence the ΔH  is positive.

If he reaction is reversed, we have;

H2O(g)→H2O(l)

In this reaction, gaseous water changes to liquid water. This process is known as condensation. The water vapor loses heat in this reaction. Hence ΔH would be negative but still  have the same value.

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Which planets distance from the sun is 778,600,000
Nastasia [14]

Jupiter, I believe?

3 0
4 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
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alexgriva [62]

Answer:

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Explanation:

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6 0
3 years ago
How could you improve the accuracy of your measurements?
yaroslaw [1]

Explanation:

Accuracy is the nearness of observed values or measured values to the true value.

The difference between the measured and observed value gives the error incurred during the experimental procedure.

Here are some ways to improve the accuracy of measurements;

  • Use the right instrument to make recordings and to take measurements.
  • Employ the right hands and improve your technical knowledge.
  • Be very careful and observant while making measurement.
  • Ensure the instruments are properly calibrated.
  • Make sure the experiment follows the right design.
7 0
3 years ago
Picric acid has been used in the leather industry and in etching copper. However, its laboratory use has been restricted because
olchik [2.2K]

Answer:

0.3023 M

Explanation:

Let Picric acid = H_{picric}

So,  H_{picric}     +       H_2}O          ⇄      H_3}O^+     +     Picric^-

The ICE table can be given as:

                          H_{picric}     +       H_2}O          ⇄      H_3}O^+     +     Picric^-

Initial:                0.52                                               0                  0

Change:             - x                                                 + x                 + x

Equilibrium:      0.52 - x                                        + x                 + x

Given that;

acid dissociation constant  (K_a) = 0.42

K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}

0.42 = \frac{[x][x]}{0.52-x}}

0.42 = \frac{[x]^2}{0.52-x}}

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x²  + 0.42x - 0.2184 = 0                   -------------------- (quadratic equation)

Using the quadratic formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}    ;     ( where +/-  represent ± )

= \frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}

= \frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}

= \frac{-0.42+\sqrt {1.0496} }{2}     OR   \frac{-0.42-\sqrt {1.0496} }{2}

= \frac{-0.42+1.0245}{2}       OR    \frac{-0.42-1.0245}{2}

= \frac{0.6045}{2}                 OR    -\frac{1.4445}{2}

= 0.30225          OR     - 0.72225

So, we go by the +ve integer that says:

x =  0.30225

x = [ H_3}O^+ ] = [   Picric^- ] =  0.3023  M

∴  the value of  [H3O+] for an 0.52 M solution of picric acid  = 0.3023 M     (to 4 decimal places).

6 0
4 years ago
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