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3 years ago

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The enthalpy of neutralization for the reaction of a strong acid with a strong base is −56 kJ/mol of water produced. How much en

ergy will be released when 174.4 mL of 0.500 M HNO3 is mixed with 153.6 mL of 0.416 M KOH?

1 answer:

denis23 [38]3 years ago

6 0

Answer:

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kskskw

Explanation:

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make

wokskwkwkwkwjjejesjsjjsjwjwj

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Predict whether the following reactions will be exothermic or endothermic.

Solnce55 [7]

Answer:

A. N₂(g) + 3H₂(g) -----> 2NH₃ exothermic

B. S(g) + O₂(g) --------> SO₂(g) exothermic

C. 2H₂O(g) --------> 2H₂(g) + O₂(g) endothermic

D. 2F(g) ---------> F₂(g) exothermic

Explanation:

The question says predict not calculate. So you have to use your chemistry knowledge, experience and intuition.

A. N₂(g) + 3H₂(g) -----> 2NH₃ is exothermic because the Haber process gives out energy

B. S(g) + O₂(g) --------> SO₂(g) is exothermic because it is a combustion. The majority, if not all, combustion give out energy.

C. 2H₂O(g) --------> 2H₂(g) + O₂(g) is endothermic because it is the reverse reaction of the combustion of hydrogen. If the reverse reaction is exothermic then the forward reaction is endothermic

D. 2F(g) ---------> F₂(g) is exothermic because the backward reaction is endothermic. Atomisation is always an endothermic reaction so the forward reaction is exothermic

5 0

3 years ago

Which of the following represents a physical change?

julsineya [31]

A. Phase changing. When phase changes nothing chemically changes about the substance, its still the same thing.

8 0

3 years ago

Read 2 more answers

Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

<u>For process 2 :</u>

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0

3 years ago

Read the following chemical equation.

elena55 [62]

Answer:

Iron gains three electrons.

5 0

3 years ago

Which of the following are cations? Check all that apply.

Lostsunrise [7]

Answer:

a

barium

b

calcium

e

aluminum

f

magnesium

g

copper

Forms cation

Explanation:

4 0

3 years ago