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frosja888 [35]
3 years ago
5

The radius of a helium atom is a 31 PM. What is the radius in nanometers?

Chemistry
1 answer:
Paladinen [302]3 years ago
7 0

Answer:

Thus, the radius of the helium atom in nanometers is - 0.031 nm

Explanation:

Given that:-

The radius of the helium atom = 31 pm

Considering the conversion of length in pm to the length in nm as:-

1 pm = 0.001 nm

So,

Applying the above conversion factor in the radius of helium atom as:-

Radius = 31\times 0.001 nm = 0.031 nm

<u>Thus, the radius of the helium atom in nanometers is - 0.031 nm</u>

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Explanation:

Yes :)

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3 years ago
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In the laboratory, a volume of 100 mL of sulfuric acid (H2SO4) is recorded. How many g are there of the liquid if its density is
ser-zykov [4K]

Answer:

\large \boxed{\text{183 g}}  

Explanation:

\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\1.83 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{100 cm}^{3}}\\\\m & = & \text{183 g}\\\end{array}\\\text{There are $\large \boxed{\textbf{183 g}}$ of sulfuric acid.}

8 0
3 years ago
How many chlorine ions are required to bond with one aluminum ion and why??
antiseptic1488 [7]
3 Chlorine ions are required to bond with one aluminum ion.

In ionic bonds, metals atoms loses all its outermost shell electrons to form a cation. While, non metal atoms gains however many electrons in order to make its outermost electron shell be 8 (or 2 if there's only one shell).

Therefore, form the periodic table, we can see that aluminum has a atomic number of 13, which makes its electron arrangement be 2,8,3. So, in order to form a aluminum ion, an Al atom must lose 3 electrons. On the other hand, Chlorine has a atomic number of 17, which means it has the electron configuration of 2,8,7. It has to gain only 1 electron to have 8 outermost shell electron.

Thereofre, 3 Chlorine atom are required to gain all 3 electrons given out by just 1 aluminum ion.
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3 years ago
Using only the periodic table, determine the element that displays each given trend. Be sure to provide your reasoning.
RideAnS [48]

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

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3 years ago
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Answer: Bb you are loved and they may not be right for you and I’m sorry if there’s something going on you can message me but trust me you are loved!!

Explanation:

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