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TEA [102]
3 years ago
13

what has 4 valence electrons, shows properties of metals and non metals, and has 32 protons in the nucleus of each atom

Chemistry
1 answer:
12345 [234]3 years ago
6 0
Here is your answer

\bold{Germanium}

REASON:

Elements which have 4 valence electrons are generally metalloids.

The metalloids show the properties of both metals and non-metals.

We know that,

no. of protons= Atomic number

So,

Atomic no.= 32

Hence,

The element is Germanium which is a metalloid with 4 valence electrons and has 32 protons in nucleus of each atom because it has atomic no. 32

HOPE IT IS USEFUL
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<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)

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m = mass of the substance

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\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:

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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

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  • <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

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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

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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

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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

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