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strojnjashka [21]
3 years ago
5

Daltons idea that atoms cannot be divided into smaller parts was disproved by the discovery of the.

Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
4 0
It was disproved by the discovery of the electrons
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Identify the dominant intermolecular attraction in bh3.
k0ka [10]
The molecule BH3 is trigonal planar, with B in the center and H in the three vertices. Ther are no free electrons. All the valence electrons are paired in and forming bonds.

There are four kind of intermolecular attractions: ionic, hydrogen bonds, polar and dispersion forces.

B and H have very similar electronegativities, Boron's electronegativity is 2.0 and Hydrogen's electronegativity is 2.0.

The basis of ionic compounds are ions and the basis of polar compounds are dipoles.

The very similar electronegativities means that B and H will not form either ions or dipoles. So, that discards the possibility of finding ionic or polar interactions.

 Regarding, hydrogen bonds, that only happens when hydrogen bonds to O, N or F atoms. This is not the case, so you are sure that there are not hydrogen bonds.

When this is the case, the only intermolecular force is dispersion interaction, which present in all molecules.

Then, the answer is dispersion interaction.
6 0
3 years ago
A piece of unknown metal weighs 348g. When the metal piece absorbs 6.64kj of heat , its temperature increases from 24.4C to 43.6
Morgarella [4.7K]

Answer:

This metal could be the aluminium with a specific heat of H = 993 \frac{J}{kgC}

Explanation:

A pie of unknown metal presents a mass (M) of 348 g. This metal is heated using energy (E) of 6.64 kJ and the temperature increases from T1 =24.4 to T2 =43.6°C. We can calculate the specific heat (H) of this metal as follows

H = \frac{E}{M*(T2-T1)}

We can replace previously presented data in this equation. After simplifying and converting to adequated units, we found that

H = \frac{6640 J}{0.348 Kg*(43.6-24.4) C} =\frac{6640 J}{6.686 KgC}

Finally, the specific heat of this metal is

H = 993 \frac{J}{kgC}

The aluminium could be the metal, its specific heat is similar to that found in this problem.

Finally,  we can conclude that this metal could be the aluminium with a specific heat of H = 993 \frac{J}{kgC}

7 0
3 years ago
Carrying capacity is _____.
Alinara [238K]
<span>Carrying capacity is the number of organisms an ecosystem can support. It is the maximum size of a population that can survive in the ecosystem. If the animals reach the carrying capacity, the population may crash. As the consequence, the number of animals will decrease due to predators or diseases.</span>
5 0
3 years ago
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18. Which metal is capable of forming more than one cation? <br> Li<br> Ba<br> Al <br> Sn
maw [93]

Answer:

Li

Explanation:

3 0
3 years ago
2.50 g CuCl2 equals how many moles
Vitek1552 [10]
The molar mass of CuCl2 is 134.45 g/mol; therefore, you divide 2.5 g of CuCl2 by 134.45 g of CuCl2 leaving you with 0.019 moles
4 0
3 years ago
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