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NNADVOKAT [17]
2 years ago
12

Which of the following has the greatest mass for one mole of the compound?

Chemistry
1 answer:
ddd [48]2 years ago
3 0
I think it would be c
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When 1.0g of diethyl ether is converted to vapor at its boiling point, about how much heat is absorbed? (C4H10O, Hvap= 15.7kj/mo
Aloiza [94]

Answer:

Option D = 0.2 Kj

Explanation:

Given data:

Mass of diethyl ether = 1.0 g

Hvap = 15.7 Kj / mol

Heat absorbed = ?

Solution:

Q = mass ×  Hvap /  molar mass

Q = 1.0 g × 15.7 Kj / mol / 74.12 g/mol

Q = 15.7 Kj / 74.12

Q =  0.212 KJ

8 0
2 years ago
What type of reaction is represented in the following example?
Nikitich [7]

Answer:

endothermic

Explanation:

the negative sign of the heat indicates that heat is being drawn into the system, thus endothermic

3 0
3 years ago
Read 2 more answers
If 0.092J of heat causes a 0.267 degree C temperature change, what mass of water is present?
Thepotemich [5.8K]

Answer:

0.082g

Explanation:

The following data were obtained from the question:

Heat (Q) = 0.092J

Change in temperature (ΔT) = 0.267°C

Specific heat capacity (C) of water = 4.184J/g°C

Mass (M) =..?

Thus, the mass of present can be obtained as follow:

Q = MCΔT

0.092 = M x 4.184 x 0.267

Divide both side by 4.184 x 0.267

M = 0.092 / (4.184 x 0.267)

M = 0.082g

Therefore, mass of water was present is 0.082.

6 0
2 years ago
A balloon filled with air has a volume of 6.50 L, a pressure of 0.900 atm, and a temperature of 25.0oC. If it is left out overni
kari74 [83]

Answer:

4.92 L

Explanation:

Rearrange ideal gas law and solve.

Change C to K.

- Hope that helps! Please let me know if you need further explanation.

8 0
3 years ago
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.

Learn more about calorimetry:

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7 0
1 year ago
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