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krek1111 [17]
2 years ago
12

Jeff has 10 grams of water and 10 grams of vegetable oil in separate containers. Both liquids have a temperature of 24°C. Jeff h

eats both liquids over a flame for five minutes. When he’s finished, he discovers that the temperature of the oil increased more than the temperature of the water. What can Jeff conclude from this experiment? Vegetable oil is a better conductor of heat than water. Water is a better conductor of heat than vegetable oil. Vegetable oil has a higher heat capacity than water. Vegetable oil has a lower heat capacity than water. Vegetable oil is a good insulator.
Chemistry
2 answers:
Anna11 [10]2 years ago
8 0
I think the answer would be that vegetable oil is a better conductor of heat than water.
Sonbull [250]2 years ago
6 0

Answer: Vegetable oil has a lower heat capacity than water.

Explanation:

Specific heat capacity is the heat required to raise the temperature of 1 g of a substance through 1^0C.

q=m\times c\times \Delta T

q = heat absorbed

m= mass of substance

c= specific heat

\Delta T = change in temperature

As both vegetable oils and water both have same mass and have same initial temperature and absorbed same heat.

As the temperature of the oil increased more than the temperature of the water, it means the specific heat of oil is less than that of water as the heat absorbed is same.

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Answer:2,4&5 A.

Explanation:

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The town of Natrium, West Virginia, derives its name from the sodium produced in the electrolysis of molten sodium chloride (NaC
Alona [7]

Explanation:

The reaction equation will be as follows.

           Na^{+} + e^{-} \rightarrow Na(s)

Hence, moles of Na = moles of electron used

Therefore, calculate the number of moles of sodium as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

                             = \frac{4500 g}{23 g/mol}    (as 1 kg = 1000 g)

                             = 195.65 mol

As,     Q = n \times F       where F = Faraday's constant

              = 195.65 mol \times 96500 C

              = 1.88 \times 10^{7} mol C

Relation between electrical energy and Q is as follows.

               E = Q \times V

Hence, putting the given values into the above formula and then calculate the value of electricity as follows.

              E = Q \times V

                 = 1.88 \times 10^{7} \times 5

                 = 9.4 \times 10^{7} J

As 1 J = 2.77 \times 10^{-7} kWh

Hence,      \frac{9.4 \times 10^{7}}{2.77 \times 10^{-7}} kWh

                = 3.39 kWh

Thus, we can conclude that 3.39 kilowatt-hours of electricity is required in the given situation.

7 0
2 years ago
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In the reaction of nitrogen gas with oxygen gas to produce nitrogen oxide, what is the effect of adding more oxygen gas to the i
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<h3>Answer:</h3>

The Equilibrium would shift to produce more NO

<h3>Explanation:</h3>

The reaction is;

N₂(g) + O₂(g) ⇆ 2NO(g)

  • When a reaction is at equilibrium then the forward reaction rate will be equivalent to the reverse reaction rate. Additionally, the concentration of the reactants and products are the same.
  • From Le Chatelier's principle, additional reactants favor the formation of more products while additional products favor the formation of more reactants.
  • For example, when more oxygen is added then more Nitrogen (II) oxide will be formed.
  • Oxygen is a reactant and when increased it favors forward reaction which leads to the formation of more NO which is the product.

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2 years ago
The density of gas particles in a section of Earth’s atmosphere decreases. Which of the following is the most likely explanation
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6 0
2 years ago
A 0.380 kg sample of aluminum (with a specific heat of 910.0 J/(kg x K)) is heated to 378 K and then placed in 2.40 kg of water
lbvjy [14]

Answer:

The equilibrium temperature of the system is 276.494 Kelvin.

Explanation:

Let consider the system formed by the sample of aluminium and water as a control mass, in which the sample is cooled and water is heated until thermal equilibrium is reached. The energy process is represented by First Law of Thermodynamics:

Q_{water} -Q_{sample} = 0

Q_{water} = Q_{sample}

Where:

Q_{water} - Heat received by water, measured in joules.

Q_{sample} - Heat released by the sample of aluminium, measured in joules.

Given that no mass is evaporated, the previous expression is expanded to:

m_{w}\cdot c_{p,w}\cdot (T-T_{w}) = m_{s}\cdot c_{p,s}\cdot (T_{s}-T)

Where:

m_{s}, m_{w} - Mass of water and the sample of aluminium, measured in kilograms.

c_{p,s}, c_{p,w} - Specific heats of the sample of aluminium and water, measured in joules per kilogram-Kelvin.

T_{s}, T_{w} - Initial temperatures of the sample of aluminium and water, measured in Kelvin.

T - Temperature which system reaches thermal equilibrium, measured in Kelvin.

The final temperature is now cleared:

(m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s})\cdot T = m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}

T = \frac{m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}}{m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s}}

Given that m_{s} = 0.380\,kg, m_{w} = 2.40\,kg, c_{p,s} = 910\,\frac{J}{kg\cdot K}, c_{p,w} = 4186\,\frac{J}{kg\cdot K}, T_{s} = 378\,K and T_{w} = 273\,K, the final temperature of the system is:

T = \frac{(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)\cdot (378\,K)+(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)\cdot (273\,K)}{(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)+(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)}

T = 276.494\,K

The equilibrium temperature of the system is 276.494 Kelvin.

7 0
2 years ago
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