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Lelechka [254]
2 years ago
15

Natalia dipped two feathers in oil. Then she dipped one feather in cold water and the other feather in hot water. She swirled bo

th feathers around for 20 seconds, then removed them. Her results are in the table. Cold Water Hot Water Before Oil After Oil After Water Before Oil After Oil After Water Mass 1.2 g 4.5 g 4.2 g 1.2 g 4.4 g 3.5 g She also took a picture of each feather after swirling it in water. two feathers, each with oil and beads of water Was either the cold water or the hot water able to remove the oil? Explain your reasoning.
Chemistry
1 answer:
Alika [10]2 years ago
8 0

Answer:

The hot water was better for removing the oil.

Explanation:

You can see that because the mass went down more with the hot water. So, that means that more oil was taken out of the feather with hot water.

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See explanation

Explanation:

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For the acid HNO3, its conjugate base is NO3^- while for the acid H3O^+, its conjugate base is H2O.

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Stuck on question 22
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The most condensed state of matter is solid
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Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of
Brut [27]

<u>Answer:</u> The smallest temperature change is shown by water.

<u>Explanation:</u>

To calculate the heat absorbed or released, we use the equation:

q=mC\times \Delta T      ......(1)

where,

q = heat absorbed = 200.0 J

m = mass of the substance

C = specific heat of substance

\Delta T = change in temperature

As, the same amount of heat is getting absorbed in all the cases. So, the change in temperature will depend on the product of mass and specific heat.

For the given options:

  • <u>Option a:</u>  50.0 g Fe, C_{Fe}=0.449J/g^oC

We are given:

m=50.0g\\C_{Fe}=0.449J/g^oC

Putting values in equation 1, we get:

200.0J=50.0g\times 0.449J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 0.449}=8.99^oC

Change in temperature = 8.99°C

  • <u>Option b:</u>  50.0 g water, C_{water}=4.18J/g^oC

We are given:

m=50.0g\\C_{water}=4.18J/g^oC

Putting values in equation 1, we get:

200.0J=50.0g\times 4.18J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 4.18}=0.96^oC

Change in temperature = 0.96°C

  • <u>Option b:</u>  25.0 g Pb, C_{Pb}=0.128J/g^oC

We are given:

m=50.0g\\C_{Pb}=0.128J/g^oC

Putting values in equation 1, we get:

200.0J=25.0g\times 0.128J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.128}=62.5^oC

Change in temperature = 62.5°C

  • <u>Option d:</u>  25.0 g Ag, C_{Ag}=0.235J/g^oC

We are given:

m=25.0g\\C_{Ag}=0.235J/g^oC

Putting values in equation 1, we get:

200.0J=25.0g\times 0.235J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.235}=34.04^oC

Change in temperature = 34.04°C

  • <u>Option e:</u>  25.0 g granite, C_{granite}=0.79J/g^oC

We are given:

m=25.0g\\C_{Fe}=0.79J/g^oC

Putting values in equation 1, we get:

200.0J=25.0g\times 0.79J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.79}=10.13^oC

Change in temperature = 10.13°C

Hence, the smallest temperature change is shown by water.

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<h3><u>Answer;</u></h3>

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