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Kruka [31]
3 years ago
6

Name the type of chemical bonding in potassium bromide

Chemistry
1 answer:
Rzqust [24]3 years ago
5 0
The bond between Potassium and Bromide is considered iconic, Because of how they lose electrons and become ions when they bond together.
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Zn + H2SO4 ZnSO4 + H2<br><br> The mole ratio of zinc to zinc sulfate is
ValentinkaMS [17]

Answer:

Ratio is 1:1

Explanation:

I do not see any coefficients infront of the reactants and the products, therefore, we can automatically assume that every reactant and product is 1 mole. Don't get confused by the 4 off the O. It just means that 1 mole of sulfate has 1 zinc and 4 oxygens.

6 0
3 years ago
who reported four “element” classifications, but included some substances that were combinations of elements rather than true el
kozerog [31]
Answer: Antoine Lavoisier.

He classified the elements into four groups: elastic fluids, nonmetals, metals and earths. Some of the called elementes were not really elements (light and heat). Others were compounds, e.g. hydrochloric acid.






7 0
3 years ago
Read 2 more answers
Why is ocean water more dense than freshwater
zalisa [80]
A. Cuz it contains more salt
Hope this helps good luck on ur examsss :))))
8 0
3 years ago
Which of the following would you except to see in the death of a star that is less than 0.5 solar mass
Ket [755]

B. White Dwarf.

<h3>Explanation</h3>

The star would eventually run out of hydrogen fuel in the core. The core would shrink and heats up. As the temperature in the core increases, some of the helium in the core will undergo the triple-alpha process to produce elements such as Be, C, and O. The triple-alpha process will heat the outer layers of the star and blow them away from the core. This process will take a long time. Meanwhile, a planetary nebula will form.

As the outer layers of gas leave the core and cool down, they become no longer visible. The only thing left is the core of the star. Consider the Chandrasekhar Limit:

Chandrasekhar Limit: 1.4 \;M_\odot.

A star with core mass smaller than the Chandrasekhar Limit will not overcome electron degeneracy and end up as a white dwarf. Most of the outer layer of the star in question here will be blown away already. The core mass of this star will be only a fraction of its 0.5 \;M_\odot, which is much smaller than the Chandrasekhar Limit.

As the star completes the triple alpha process, its core continues to get smaller. Eventually, atoms will get so close that electrons from two nearby atoms will almost run into each other. By Pauli Exclusion Principle, that's not going to happen. Electron degeneracy will exert a strong outward force on the core. It would balance the inward gravitational pull and prevent the star from collapsing any further. The star will not go any smaller. Still, it will gain in temperature and glow on the blue end of the spectrum. It will end up as a white dwarf.

7 0
3 years ago
Which of the following is not the same as 12.5 millimeters? 0.0000125 km 0.00125 hm 0.0125 m 1.25 cm
erma4kov [3.2K]
1 mm (millimeter) = 0.000001 km (kilometer)
12.5 mm = <span>0.0000125 km

1 mm = </span><span>0.00001 hm (hectometer)
12.5 mm = </span><span>0.000125 hm

1 mm = </span>0.001 m (meter)
12.5 mm = 0.0125 m

1 mm = 0.1 cm (centimeter)
12.5 mm = 1.25 cm

So the only one of the answer choices that doesn't equal 12.5 mm is 0.00125 hm, since 12.5 mm is <span>0.000125 hm.

Answer:
</span><span>0.00125 hm
</span><span>
Hope this helps!</span>
5 0
3 years ago
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