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4 years ago

As electrical energy is converted into heat energy, the total amount in the system

Chemistry

2 answers:

schepotkina [342]4 years ago

5 0

Answer: Option (c) is the correct answer.

Explanation:

According to law of conservation of energy, energy can neither be created nor it can be destroyed as it can only be converted from one form to another.

For example, in the given situation electrical energy is converted into heat energy.

This means that the total amount of the electrical energy is actually equal to the total amount of heat energy. As only change in the form of energy has taken place.

Thus, we can conclude that as electrical energy is converted into heat energy, the total amount in the system remains the same.

Aleks04 [339]4 years ago

3 0

c. remains the same
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As electrical energy is converted into heat energy, the total amount in the system remains the same
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according to law of conservation of energy

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PLEASE HELP I WILL MARK BRAINLIEST!!! Reactions In Our World Lab Report

Cerrena [4.2K]

Answer:

Explanation:

6 0

3 years ago

A lead mass is heated and placed in a foam cup calorimeter containing 40.0 mL of water at 17.0°C. The water reaches a temperatur

lbvjy [14]

Answer: 502 Joules

Explanation:

To calculate the mass of water, we use the equation:

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

Density of water = 1 g/mL

Volume of water = 40.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{40.0mL}\\\\\text{Mass of water}=(1g/mL\times 40.0mL)=40.0g$

When metal is dipped in water, the amount of heat released by lead will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$q=m\times c\times \Delta T$

q = heat absorbed by water

$m$ = mass of water = 40.0 g

$T_{final}$ = final temperature of water = 20.0°C

$T_{initial$ = initial temperature of water = 17.0°C

$c$ = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

$q=40.0\times 4.186\times (20.0-17.0)]$

$q=502J$

Hence, the joules of heat were re-leased by the lead is 502

5 0

3 years ago

How do I do letter D? Someone help me please

Minchanka [31]

So for D you have to find energy right
from c you get wavelength Lambda
so on D use this
E = Hc / lambda
c is given 1.5 x 10 ^20
h = 6.624 x 10^-34
and then you get answer for energy

8 0

3 years ago

Why is it better to conduct an experiment more than once?

lilavasa [31]

The first reason to repeat experiments is simply to verify results. Different science disciplines have different criteria for determining what good results are. Biological assays, for example must be done in at least triplicate to generate acceptable data. Science is built on the assumption that published experimental protocols are repeatable.

2) The next reason to repeat experiments is to develop skills necessary to extend established methods and develop new experiments. “Practice make perfect” is true for the concert hall and the chemical laboratory.

3) Refining experimental observations is another reason to repeat. Maybe you did not follow the progress of the reaction like you should have.

4) Another reason to repeat experiments is to study and/or improve them in way. In the synthetic chemistry laboratory, for example, there is always a desire to improve the yield of a synthetic step. Will certain changes in the experimental conditions lead to a better yield? The only way to find out is to try it! The scientific method informs us that it is best to only make one change at a time.

5) The final reason to repeat an extraction, chromatographic or synthetic protocol is to produce more of your target substance. This is sometimes referred to scale-up.

8 0

3 years ago

If E=mc^2,solve for both m and c. Also, if m=80 and c=0.40 what is the value of E?

Pie

Answer:

$m = \frac{E}{c^2}$

$c = \sqrt{\frac{E}{m} }$

$E = 12.8 J$

Explanation:

Part 1: Solving for m

We are given that:

E = mc²

To solve for m, we will need to isolate the m on one side of the equation

This means that we will simply divide both sides by c²

$m = \frac{E}{c^2}$

Part 2: Solving for c

We are given that:

E = mc²

To solve for c, we will need to isolate the m on one side of the equation

This means that first we will divide both sides by m and then take square root for both sides to get the value of c

$c^2 = \frac{E}{m}\\ \\ c=\sqrt{\frac{E}{m}}$

Part 3: Solving for E

We are given that:

m = 80 and c = 0.4

To get the value of E, we will simply substitute in the given equation:

E = mc²

E = (80) × (0.4)²

E = 12.8 J

Hope this helps :)

4 0

3 years ago