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____ [38]
3 years ago
13

A 100.0-mL sample of 1.00 M NaOH is mixed with 50.0 mL of 1.00 M H2SO4 in a large Styrofoam coffee cup calorimeter fitted with a

lid through which a thermometer passes. The acid-base reaction is as follows:
2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)
The temperature of each solution before mixing is 22.3 °C. After mixing, the temperature of the solution mixture reaches a maximum temperature of 31.4 °C. Assume the density of the solution mixture is 1.00 g/mL, its specific heat is 4.18 J/g.°C, and no heat is lost to the surroundings. Calculate the enthalpy change, in kj, per mole of H2SO4 in the reaction.
a. +85.6 kJ/mol.
b. -85.6 kJ/mol.
c. +5.71 kJ/mol.
d. -5.71 kJ/mol.
e. -114 kJ/mol.
Chemistry
1 answer:
worty [1.4K]3 years ago
5 0

Answer:

THE ENTHALPY CHANGE IN KJ/MOLE IS +114 KJ/MOLE.

Explanation:

Heat = mass * specific heat capacity * temperature rise

Total volume = 100 + 50 = 150 mL

Total mass = density * volume

Total mass = 1 * 150 mL = 150 g

So therefore, the heat evolved during the reaction is:

Heat = 150 * 4.18 * ( 31.4 - 22.3)

Heat = 150 * 4.18 * 9.1

Heat = 5705.7 J

Equation for the reaction:

2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)  

From the equation, 2 moles of NaOH reacts with 1 mole of H2SO4 to produce 1 mole of Na2SO4 and 2 moles of water

50 mL of 1 M of H2SO4 contains

50 * 1 / 1000 mole of acid

= 0.05 mole of acid

The production of 1 mole of water evolved 5705.7 J of heat and hence the enthalpy changein kJ per mole will be:

0.05 mole of H2SO4 produces 5705.7 J of heat

1 mole of H2SO4 will produce 5705.7 / 0.05 J

= 114,114 J / mole

In kj/mole = 114 kJ/mole.

Hence, the enthalpy change of the reaction in kJ /mole is +114 kJ/mole.

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Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.
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Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

\%C_E = \frac{E}{T} * 100\%

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In (Ca(OH)_2)

The amount of all elements is: (i.e formula mass of (Ca(OH)_2))

T = 1 * Ca + 2 * H + 2 * O

T = 1 * 40 + 2 * 1 + 2 * 16

T = 74

The amount of calcium is: (i.e formula mass of calcium)

E = 1 * Ca

E = 1 * 40

E = 40

So, the percentage component of calcium is:

\%C_E = \frac{E}{T} * 100\%

\%C_E = \frac{40}{74} * 100\%

\%C_E = \frac{4000}{74}\%

\%C_E = 54.05\%

The amount of hydrogen is:

E = 2 * H

E = 2 * 1

E = 2

So, the percentage component of hydrogen is:

\%C_E = \frac{E}{T} * 100\%

\%C_E = \frac{2}{74} * 100\%

\%C_E = \frac{200}{74}\%

\%C_E = 2.70\%

Similarly, for oxygen:

The amount of oxygen is:

E = 2 * O

E = 2 * 16

E = 32

So, the percentage component of oxygen is:

\%C_E = \frac{E}{T} * 100\%

\%C_E = \frac{32}{74} * 100\%

\%C_E = \frac{3200}{74}\%

\%C_E = 43.24\%

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2 years ago
NO GUESSING!ONLY RIGHT ANSWER
lutik1710 [3]
<h3> <u>ANSWER</u></h3>

2. Neon, only

<h3><u>EXPLANATION</u></h3>

When Na and F combine they form an electron configuration of 2-8. Na electron configuration is 2-8-1 while F is 2-7, so when they form an ionic bond F will gain Na outermost electron to complete its valence shell due to having a higher negativity. Neon has an electron configuration of 2-8 but argon has 2-8-8.

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