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polet [3.4K]
3 years ago
9

Which line (A or B) represents a faster speed?

Chemistry
1 answer:
BartSMP [9]3 years ago
3 0

Answer:

A

Explanation:

Because in 1 second, line A was faster than line B.

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If you have 10cm of snow with a volume of 40mL and a density of 0.5 g/mL how many inches of rain is this?
Lelu [443]

Density is the mass of compound divided by its volume can be shown as follows:

d = 0.5 g /mL = m /V = m /40 = 0.5

m = 20 g

40 mL of snow having 20 g of mass calculated from density.

Now, 10 cm of snow = 3.93 inches = 20 g  

As, 10 inches of rain will produce 11 inches of ice as the volume of ice is bigger than rain water.

10 inches rain = 11 inches snow

3.93 inches of snow produced by  

= (3.93 * 10) /11 = 3.57 inches rain.

Thus, 3.57 incehs of rain produces by 10 cm snow.

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Which of the following is accurate in describing the placement and classification of iodine?
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You can answer this question by only searching the element in the periodic table.

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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.

5 0
3 years ago
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