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Inga [223]
3 years ago
13

Consider this reaction involving an unknown element X. F2+2XBr⟶Br2+2XF When 1.300 g of XBr reacts, 0.7241 g of Br2 is produced.

Calculate the molar mass of X, and then identify its element symbol.
Chemistry
1 answer:
scZoUnD [109]3 years ago
6 0

The molar mass of X would be 63.55 g/mol and the chemical symbol of X would be Cu (copper).

From the equation of the reaction: F2+2XBr⟶Br2+2XF

The mole ratio of XBr to Br2 is 2:1

Mole of Br2 = mass/molar mass

                      = 0.7241/159.808

                       = 0.004531 mole

Mole of XBr = 2 x mole of Br2

                      = 2 x 0.004531

                        = 0.009062 mole

Molar mass of XBr = mass/mole

                            = 1.300/0.009062

                                 = 143.456 g/mol

Since the molar mass of Br is 79.904, the molar of X would be:

                             143.456 - 79.904

                                   = 63.55 g/mol

Hence, X would be Cu.

More on molar mass calculation can be found here: brainly.com/question/20691135

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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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