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Vladimir79 [104]
4 years ago
13

A student is performing an experiment to study thermodynamics. The student mixes citric acid and sodium bicarbonate in a calorim

eter to determine the energy released in the chemical reaction. In one paragraph, identify the system, the surroundings, and the reasons for using calorimetry in this type of experiment.
Chemistry
1 answer:
seraphim [82]4 years ago
5 0

Explanation :

In thermodynamics, a system is region or part of space which is being studied and observed while the surrounding is the region or space around the system which interacts with the system.

Here in the experiment ,system which is observed is reaction or changes when citric acid and sodium bicarbonate are mixed together. And the mixing is carried out in the calorimeter which serves as a surrounding around the system.

The reason behind the using the calorimeter is measure the energy change occurring during the reaction.

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What is the quantity of heat (in kJ) associated with cooling 185.5 g of water from 25.60°C to ice at -10.70°C?Heat Capacity of S
Cerrena [4.2K]

Taking into account the definition of calorimetry, sensible heat and latent heat,  the amount of heat required is 37.88 kJ.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

<h3>Sensible heat</h3>

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

<h3>Latent heat</h3>

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

  • <u><em>25.60 °C to 0 °C</em></u>

First of all, you should know that the freezing point of water is 0°C. That is, at 0°C, water freezes and turns into ice.

So, you must lower the temperature from 25.60°C (in liquid state) to 0°C, in order to supply heat without changing state (sensible heat).

The amount of heat a body receives or transmits is determined by:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

  • c= Heat Capacity of Liquid= 4.184 \frac{J}{gC}
  • m= 185.5 g
  • ΔT= Tfinal - Tinitial= 0 °C - 25.60 °C= - 25.6 °C

Replacing:

Q1= 4.184 \frac{J}{gC}× 185.5 g× (- 25.6 °C)

Solving:

<u><em>Q1= -19,868.98 J</em></u>

  • <u><em>Change of state</em></u>

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q = m×L

where L is called the latent heat of the substance and depends on the type of phase change.

In this case, you know:

n= 185.5 grams× \frac{1mol}{18 grams}= 10.30 moles, where 18 \frac{g}{mol} is the molar mass of water, that is, the amount of mass that a substance contains in one mole.

ΔHfus= 6.01 \frac{kJ}{mol}

Replacing:

Q2= 10.30 moles×6.01 \frac{kJ}{mol}

Solving:

<u><em>Q2=61.903 kJ= 61,903 J</em></u>

  • <u><em>0 °C to -10.70 °C</em></u>

Similar to sensible heat previously calculated, you know:

  • c = Heat Capacity of Solid = 2.092 \frac{J}{gC}
  • m= 185.5 g
  • ΔT= Tfinal - Tinitial= -10.70 °C - 0 °C= -10.70 °C

Replacing:

Q3= 2.092 \frac{J}{gC} × 185.5 g× (-10.70) °C

Solving:

<u><em>Q3= -4,152.3062 J</em></u>

<h3>Total heat required</h3>

The total heat required is calculated as:  

Total heat required= Q1 + Q2 +Q3

Total heat required=-19,868.98 J + 61,903 J -4,152.3062 J

<u><em>Total heat required= 37,881.7138 J= 37.8817138 kJ= 37.88 kJ</em></u>

In summary, the amount of heat required is 37.88 kJ.

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Solving part-1 only

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KMnO_4

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#2

Actually it's the oxidation number of Mn

Let's find how?

\\ \tt\Rrightarrow x+1+4(-2)=0

\\ \tt\Rrightarrow x+1-8=0

\\ \tt\Rrightarrow x-7=0

\\ \tt\Rrightarrow x=+7

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#3

  • Purple as per the color of potassium permanganate

#4

\boxed{\begin{array}{c|c|c}\boxed{\bf Tube} &\boxed{\bf Charge} &\boxed{\bf No\:of\; electrons\: loss}\\ \sf 2 &\sf +6 &\sf 6e^-\\ \sf 3& \sf +2 &\sf 2e- \\ \sf 4 &\sf 4 &\sf 4e^-\end{array}}

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2 years ago
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