Which element often shares property's

Two adaptations to plants are described below:

Elza [17]

Answer:

D

Explanation: this was easy for me

7 0

3 years ago

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Draw the Lewis dot structure for RbIO2. Include all hydrogen atoms and nonbonding electrons. Show the formal charges of all atom

Rzqust [24]

Answer : The Lewis-dot structure of $RbIO_2$ is shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $RbIO_2$

As we know that rubidium has '1' valence electrons, iodine has '7' valence electrons and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in $RbIO_2$ = 1 + 7 + 2(6) = 20

As we know that $RbIO_2$ is an ionic compound because it is formed by the transfer of electron takes place from metal to non-metal element.

7 0

3 years ago

What fraction of the original atoms of the radioactive sample will remain after the given number of half-lives has passed?

meriva

Answer:

1/8

Explanation:

$\frac{1}{ {2}^{3} } = \frac{1}{8}$

Mark brainliest please

5 0

2 years ago

Is buoyancy a physical or chemical property?

gizmo_the_mogwai [7]

Buoyancy is considered a physical property. It is a type of physical property because it is related to the density and weight of the item, which are both physical.

7 0

3 years ago

Electrophilic addition of hbr to 3-methyl-2-hexene creates an asymmetric center at c-2. what is the product of this reaction?

AleksandrR [38]

This reaction would give rise to two products.

2-bromo-3-methylhexane, and
3-bromo-3-methylhexane.

However, 2-bromo-3-methylhexane would be more common than 3-bromo-3-methylhexane among the products.

The hydrogen atom in a hydrogen bromide molecule carries a partial positive charge. It is attracted to the double bond region with a high electron density. The hydrogen-bromine bond breaks when HBr gets too close to a double bond to produces a proton $\text{H}^{+}$ and a bromide ion $\text{Br}^{-}$ .

The proton would attack the double bond to produce a carbocation. It could attach itself to either the second or the third carbon atom.

$\phantom{\text{H}_3\text{C}-\text{CH}-\;}+\\\text{H}_3\text{C}-\text{CH}-\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3\\\phantom{\text{H}_3\text{C}-\;\;}|\phantom{\text{CH}-}\;}|\\\phantom{\text{H}_3\text{C}-\;} \text{H}\phantom{\text{CH}}\;\;}\text{CH}_3$
$\phantom{\text{H}_3\text{C}-\text{CH}-}\;\;\text{H}\\\phantom{\text{H}_3\text{C}-\;}+\phantom{\text{H}-\;}|\\\text{H}_3\text{C}-\text{CH}-\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3\\\phantom{\text{H}_3\text{C}-\;\;\;}\phantom{\text{CH}-}\;}|\\\phantom{\text{H}_3\text{C}-\;} \phantom{\text{H}}\phantom{\text{CH}}\;\;}\text{CH}_3$

Carbocations are unstable and might decompose over time. The first carbocation is more stable than the second for having three alkyl groups- i.e., straight carbon chains- attached to the center of the positive charge. Alkyl groups have stabilizing positive induction effect on positively-charged carbon. The second carbocation has only two, and is therefore not as stable. The first carbocation thus has the greatest chance to react with a bromide ion to produce a stable halocarbon.

Bromide ions are negatively charged. They attach themselves to carbocations at the center of positive charge. Adding a bromide ion to the first carbocation would produce 3-bromo-3-methylhexane whereas adding to the second produces 2-bromo-3-methylhexane.

The <em>most likely</em> product of this reaction is therefore 3-bromo-3-methylhexane.

6 0

3 years ago