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Dima020 [189]
1 year ago
7

Boron’s chemistry is not typical of its group.(a) Cite three ways in which boron and its compounds differ significantly from the

other 3A(13) members and their compounds.
Chemistry
1 answer:
Mila [183]1 year ago
6 0

Boron’s chemistry is not typical of its group. is group 3A (13) shows the increasing metallic character from Al to Tl.

All Boron compounds are covalent whereas the other elements in group 3A (13) form mostly ionic compounds.

Except for Boron, the other elements of group 3A (13) show increasing metallic character from Al to Tl. But Boron is a metalloid.

Compared to the other elements in group 3A, boron has a lower reactivity in chemical terms (13)

The metalloid boron (B), as well as the metals aluminium (Al), gallium (Ga), indium (In), and thallium, are all part of group 3A (or IIIA) of the periodic table (Tl). In contrast to the other members of Group 3A, the element borax primarily forms covalent connections.

To learn more about group 3A (13) refer the link:

brainly.com/question/5489194

#SPJ4

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8 0
3 years ago
Which of the following solutions would have the highest pH? Assume that they are all 0.10 M in acid at 25°C. The acid is followe
Anastaziya [24]

Answer:

HCN~~Ka=4.9x10^-^1^0

Explanation:

In this case, we have to remember the <u>relationship between the Ka value and the pH</u>. We can use the general reaction for any acid with his Ka value expression:

HA~->~H^+~+~A^-    Ka=\frac{[H^+][A^-]}{[HA]}

In the Ka expression, we have a<u> proportional relationship</u> between Ka and the concentration of H^+. Therefore, if we have a higher Ka value we will have a smaller pH (lets keep in mind that with a higher

So, if we have to find the higher pH value we need to search the <u>smaller Ka value</u> in this case HCN~~Ka=4.9x10^-^1^0.

I hope helps!

8 0
3 years ago
What kind of weather does a stationary front bring?
Margaret [11]
Same as a warm front drizzle rain or light snow, I think
7 0
3 years ago
Read 2 more answers
What is the h+ of a solution with a ph of 5.6
7nadin3 [17]
Just have to do antilog

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6 0
3 years ago
A lead ball is added to a graduated cylinder containing 41.7 ml of water, causing the level of the water to increase to 96.0 mL.
rjkz [21]

Answer:

Vlead ball=54.3 mL

Explanation:

The graduated cylinder contains 41.7mL of water

mL is a volume unit.

Water volume = 41.7 mL

The lead ball caused an increase of volume from 41.7 mL to 96.0 mL

The new volume is the lead ball volume plus the original water volume :

Final volume = Vlead ball+ Water original volume

96.0mL=Vleadball +41.7mL

Vlead ball=96.0mL-41.7mL

Vleadball=54.3mL

This is actually true if we suppose that the lead ball is fully sunken in the water.

We always must consider that the volume difference is the volume that the sunken object is occupying in the water.

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3 years ago
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