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tester [92]
3 years ago
9

Part iv. Is the neutralization reaction enthalpy favored?

Chemistry
1 answer:
Burka [1]3 years ago
5 0

Yes, it is a special case of enthalpy of neutralization.  

The enthalpy of neutralization (ΔHn) is the change in enthalpy that occurs when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to form water and a salt.

The standard enthalpy change of neutralization is the enthalpy change when solutions of an acid and an alkali react together under standard conditions to produce 1 mole of water.

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Which of the following are examples of types of mixtures?
aev [14]
I think it would be homogeneous and heterogeneous
6 0
3 years ago
(a) the temperature on a warm summer day is 87 °f. what is the temperature in °c? (b) many scientific data are reported at 25
olchik [2.2K]

Answers:

              (a)  30.55 °C

              (b) 298 K and 77°F

              (c)  204.44 °C and 477.44 K

              (d)  -320.8 °F and -196 °C

Explanation:

Converting °C into °F;

                                   °F  =  °C × 1.8 + 32

Converting °F into °C;

                                   °C  =  °F - 32 ÷ 1,8

Converting °C into K;

                                   K  =  °C + 273

Converting K into °C;

                                   °C  =  K - 273

8 0
3 years ago
Explain why 0.76 * 103 would not be considered a correct answer for a scientific notation problem? What should be done to fix th
Valentin [98]
The decimal should be after the 7 and before the 6.
7 0
2 years ago
Read 2 more answers
KMnO4 + KCI + H2SO4<br>MnSO4 + K2S04+ Cl₂ + H₂O​
yKpoI14uk [10]

Answer:

Balancing the equation

2KMnO₂+10KCl+8H₂SO₄⇒2MnSO₄+6K₂SO₄+8H₂O+5Cl₂

7 0
3 years ago
Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 500.
yaroslaw [1]

Answer:

12. is the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.

Explanation:

2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

initially

3.0 atm      0              0

At equilibrium

(3.0-2p)     p               3p

Equilibrium partial pressure of nitrogen gas = p = 0.90 atm

The expression of a pressure equilibrium constant will be given by :

K_p=\frac{p_{N_2}\times (p_{H_2})^3}{(p_{NH_3})^2}

K_p=\frac{p\times (3p)^3}{(3.0-2p)^2}

=\frac{0.90 atm\times (3\times 0.90 atm)^3}{(3.0-2\times 0.90 atm)^2}

K_p=12.30\approx 12.

12. is the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.

3 0
3 years ago
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