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Katen [24]
3 years ago
10

Calculate the energy required to heat of ethanol from to . Assume the specific heat capacity of ethanol under these conditions i

s . Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
Marianna [84]3 years ago
5 0

Answer:

17 kJ

Explanation:

Calculation for the Calculate the energy required to heat 0.60kg of ethanol from 2.2°C to 13.7°C.

Using this formula

q = mC∆T

Where,

q represent Energy

m represent Mass of substance=0.60kg=600g

C represent Specific heat capacity=2.44J·g−1K−1.

∆T represent change in Temperature=2.2°C to 13.7°C.

Let plug in the formula

q=(0.60 kg x 1000 g/kg)(2.44 J/gº)(13.7°C-2.2°C)

q = (600g)(2.44 J/gº)(11.5º)

q=16.836 kJ

q= 17 kJ (Approximately)

Therefore the energy required to heat 0.60kg of ethanol from 2.2°C to 13.7°C will be 17 kJ

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Explanation:Photosynthesis is important because the earth and all of the creatures in the sea and land need the plants to be able to feed themselves with the help of the sun and with cellular respiration, those plants are able to take in the carbon dioxide that we give and make into oxygen with we need. there fore without them we would not be able to exist.

4 0
3 years ago
Which law relates to the ideal gas law?
Kamila [148]

The equation of state for a hypothetical ideal gas is known as the ideal gas law, sometimes known as the general gas equation. i.e. PV = nRT or P1V1 = P2V2.

  • According to the ideal gas law, the sum of the absolute temperature of the gas and the universal gas constant is equal to the product of the pressure and volume of one gram of an ideal gas.
  • Robert Boyle, Gay-Lussac, and Amedeo Avogadro's observational work served as the basis for the ideal gas law. The Ideal gas equation, which simultaneously describes every relationship, is obtained by combining all of their observations into a single statement.
  • When applying the gas constant R = 0.082 L.atm/K.mol, pressure, volume, and temperature should all be expressed in units of atmospheres (atm), litres (L), and kelvin (K).
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Learn more about ideal gas law here:

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3 0
1 year ago
A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f
a_sh-v [17]

Answer:

m_{H_2O}=39.0g

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g

Best regards.

8 0
3 years ago
What is the total number of liters of SO3 produced when 32.0 of SO2 reacts completely
ahrayia [7]
Write  the   chemical  equation   for  reaction
that  is  
2SO2+O2  --->2SO2

find  the   moles  of  SO2   used  =  moles=mass/molar mass  of  so2

=  32g/80g/mol=0.4 moles

by  use  of  reacting  ratio  between  SO2   and  SO3   which   is   2:2  therefore  the  moles  of  so3  is  also =  0.4  moles

STP  1 mole  =  22.4L.
what  about  0.4moles

= 0.4 /1 x22.4=8.96 liters
4 0
3 years ago
A 0.1326 g sample of magnesium was burned in an oxygen bomb calorimeter. the total heat capacity of the calorimeter plus water w
Sladkaya [172]

Answer: Th enthalpy of combustion for the given reaction is 594.244 kJ/mol

Explanation: Enthalpy of combustion is defined as the decomposition of a substance in the presence of oxygen gas.

W are given a chemical reaction:

Mg(s)+\frac{1}{2}O_2(g)\rightarrow MgO(s)

c=5760J/^oC

\Delta T=0.570^oC

To calculate the enthalpy change, we use the formula:

\Delta H=c\Delta T\\\\\Delta H=5760J/^oC\times 0.570^oC=3283.2J

This is the amount of energy released when 0.1326 grams of sample was burned.

So, energy released when 1 gram of sample was burned is = \frac{3283.2J}{0.1326g}=24760.181J/g

Energy 1 mole of magnesium is being combusted, so to calculate the energy released when 1 mole of magnesium ( that is 24 g/mol of magnesium) is being combusted will be:

\Delta H=24760.181J/g\times 24g/mol\\\\\Delta H=594244.3J/mol\\\\\Delta H=594.244kJ/mol

4 0
3 years ago
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