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TEA [102]
3 years ago
15

There are two beakers that contain the same liquid substance at the same temperature. The larger beaker is 1,000ml and the small

er beaker is 100ml. How does the thermal energy of the liquid in the larger beaker compare to the thermal energy of the liquid in the smaller beaker?
A. The liquid in the larger beaker has the same amount of thermal energy as the liquid in thesmaller beaker.
B. The exact volume of liquid in each beaker must be known to compare the thermal energyof the liquids.
C. The liquid in the larger beaker has less thermal energy than the liquid in the smaller beaker.
D. The liquid in the larger beaker has more thermal energy than the liquid in the smaller beaker.
Chemistry
1 answer:
azamat3 years ago
6 0

Answer:

Answer choice B

Explanation:

Since you do not know the volume of the liquid in each beaker, the one in the smaller beaker could have more substance and therefore more thermal energy. If they had the same amount of substance, then the more voluminous one would radiate faster. However, since you do not know this, there is no way to tell. PM me if you have more questions. Hope this helps!

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How does the law of conservation of matter apply to chemical equations?
Tasya [4]

Answer: option D.

The total number of atoms of each element on both sides of the

equation must be the same.

Explanation:

3 0
3 years ago
What is the smallest particle of a covalent compound that has the properties of that compound?
Shtirlitz [24]
The smallest particle is the white blood cells because they are the ones that help you breathe
5 0
3 years ago
One gram of liquid benzene is burned in a bomb calorimeter. The temperature before ignition was 20.826 C, and the temperature af
Licemer1 [7]

Answer:

fH = - 3,255.7 kJ/mol

Explanation:

Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.

Qsystem + Qcombustion = 0

Qsystem = heat capacity*ΔT

10000*(25.000 - 20.826) + Qc = 0

Qcombustion = - 41,740 J = - 41.74 kJ

So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:

n = mass/molar mass = 1/ 78

n = 0.01282 mol

fH = -41.74/0.01282

fH = - 3,255.7 kJ/mol

4 0
3 years ago
How is impure copper refined by electrolysis
Gelneren [198K]

Answer:

By giving electricity to copper compound solution.

Explanation:

Electrolysis is one of the major way of refined copper. The copper containing solution has two electrodes.i) positive electrodes called anode. ii) negative electrodes called cathode. When electricity is pass into the copper containing solution electrolysis process is starts and impure copper is formed in anode and pure copper is formed in cathode.

So, We can get pure copper in cathode through electrolysis.

I Hope this will be helpful for you.

If this is helpful for you .Then choose this answer as brainliest answer.

6 0
2 years ago
Jason measures three values of density for his unknown. The values he obtains are: 1.019 g/mL 1.498 g/mL 1.572 g/mL What is the
jok3333 [9.3K]

<u>Answer:</u> The average of the densities of the given measurements is 1.363 g/mL

<u>Explanation:</u>

The equation used to calculate density of a substance is given by:

\text{Density of a substance}=\frac{\text{Mass of a substance}}{\text{Volume of a substance}}

We are given:

First measured value of density, d_1 = 1.019 g/mL

Second measured value of density, d_2 = 1.498 g/mL

Third measured value of density, d_3 = 1.572 g/mL

Putting values in above equation, we get:

\rho_{mix}=\frac{d_1+d_2+d_3}{3}

\rho=\frac{1.019+1.498+1.572}{3}\\\\\rho=1.363g/mL

Hence, the average of the densities of the given measurements is 1.363 g/mL

7 0
2 years ago
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