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stealth61 [152]
4 years ago
11

Using only the periodic table, determine the element that displays each given trend. Be sure to provide your reasoning.

Chemistry
1 answer:
RideAnS [48]4 years ago
3 0

Choose the element that:

- is more reactive: oxygen or argon.

- has the larger atomic radius: francium or sodium.

- has the smaller atomic radius: scandium or selenium.

- has more ionization energy: bismuth or mercury.

- has less ionization energy: nitrogen or phosphorus.

- has more electronegativity: xenon or silver.

- has less electronegativity: titanium or hafnium.

Explanation:

Choose the element that is more reactive: oxygen or argon.

Argon have a complete electronic shell which makes him inert from chemically point of view while oxygen is able to form compounds with the majority of the elements.

Choose the element that has the larger atomic radius: francium or sodium.

As you go down in a group the atomic radius is increasing.

Choose the element that has the smaller atomic radius: scandium or selenium.

As a general trend the atomic radius of elements decrease in periods from right to left for this elements .

Choose the element that has more ionization energy: bismuth or mercury.

Mercury have a complete electronic shell (5d¹⁰) while bismuth have 3 electrons in the last shell (6p³). It is harder to extract an electron from mercury compared to bismuth.

Choose the element that has less ionization energy: nitrogen or phosphorus.

Ionization energy decrease as you go down in a group.

Choose the element that has more electronegativity: xenon or silver.

Silver have a higher electronegativity than Xenon. Xenon is an inert gas and it has a complete electronic shell which explains his inert chemical reactivity.

Choose the element that has less electronegativity: titanium or hafnium.

As you go down a grup electronegativity is decreasing.

Learn more about:

trends in periodic table:

brainly.com/question/2293240

brainly.com/question/4433739

brainly.com/question/3266672

#learnwithBrainly

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Answer:

* x_{CH_3OH}=0.0425

* \%m/m_{CH_3OH}=7.31\%

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Explanation:

Hello,

In this case, for the mole fraction of methanol we use the formula:

x_{CH_3OH}=\frac{n_{CH_3OH}}{n_{CH_3OH}+n_{water}}

Thus, we compute the moles of both water (molar mass 18 g/mol) and methanol (molar mass 32 g/mol):

n_{CH_3OH}}=14.6g*\frac{1mol}{32g}=0.456molCH_3OH \\\\n_{water}}=185g*\frac{1mol}{18g}=10.3molH_2O

Hence, mole fraction is:

x_{CH_3OH}=\frac{0.456mol}{0.456mol+10.3mol}\\\\x_{CH_3OH}=0.0425

Next, mass percent is:

\%m/m_{CH_3OH}=\frac{m_{CH_3OH}}{m_{CH_3OH}+m_{water}}*100\%\\\\\%m/m_{CH_3OH}=\frac{14.6g}{14.6g+185g}*100\%\\\\\%m/m_{CH_3OH}=7.31\%

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m=\frac{n_{CH_3OH}}{m_{water}} =\frac{0.456mol}{0.185kg}\\ \\m=2.46m

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Explanation:

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