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AnnyKZ [126]
3 years ago
5

Explain why lithium fluoride is a solid at room temperature

Chemistry
1 answer:
SashulF [63]3 years ago
8 0
Lithium fluoride is a solid at room temperature because it is a salt that is held together by ionic bonds. Lithium. fluoride has a giant ionic structure.
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Which statement describe compound​
Katena32 [7]

Answer:

Explain more please

Explanation:

you have to at the statements

7 0
3 years ago
How many ATOMS of boron are present in 3.61 grams of boron trifluoride ?
Delvig [45]

Hi there!

\large\boxed{3.203 *  10^{22} atoms}

We can use the following conversions to solve:

Total mass --> amount of mols --> amount of atoms (Avogadro's number)

Begin by calculating the amount of boron trifluoride in 3.61 grams:

3.61 g * (1 mol BF₃ / 67.8 g) ≈ 0.0532 mol BF₃

Use avogadro's number to convert:

0.0532 mol * 6.02× 10²³atoms / 1 mol = 3.203 × 10²² atoms

3 0
3 years ago
The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in
astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}

For this, it is necessary to know the values in meters for any of these diameters:

\\ 1m = 10^{3}mm = 1e+3mm

\\ 1m = 10^{12}pm = 1e+12pm

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m

<h3>Diameter of a biscuit in meters</h3>

\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

6 0
3 years ago
Lab 2: paper chromatography of organic dyes<br> Picture of questions below.
Anika [276]

Answer:

The three primary colors used when mixing dyes or paints are red, yellow, and blue. Other colors are often a mixture of these three colors. Try running a chromatography test again with non-primary-color markers, like purple, brown, and orange.

Explanation:

<h3><em>Mixtures that are suitable for separation by chromatography include inks, dyes and colouring agents in food. ... As the solvent soaks up the paper, it carries the mixtures with it. Different components of the mixture will move at different rates. This separates the mixture out.</em></h3>

<em />

7 0
2 years ago
What is the steric number of H2O2?
chubhunter [2.5K]

Answer: 1(2)+6(2)=14

Explanation:

Not sure, but maybe.

8 0
3 years ago
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