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3 years ago

Briefly explain in your own words why the bond angle increases as the number of electron groups decreases

Chemistry

1 answer:

tekilochka [14]3 years ago

4 0

Answer:

i) The bond angle decreases due to the presence of lone pairs, which causes more repulsion on the bond pairs and as a result, the bond pairs tend to come closer. ii) The repulsion between electron pairs increases with an increase in electronegativity of the central atom and hence the bond angle increases.

Explanation:

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Which of the following pure compounds will exhibit hydrogen bonding?

zimovet [89]

Explanation:

For a compound to show hydrogen bonding it is necessary that the hydrogen atom of the compound should be attached to more electronegative atom like fluorine, oxygen or nitrogen.

For example, $CH_{3}CH_{2}OH$ , $CH_{3}NH_{2}$ and $NH_{3}$ all these compounds contain an electronegative atom attached to hydrogen atom.

Therefore, these pure compounds will exhibit hydrogen bonding.

Thus, we can conclude that out of the given options $CH_{3}CH_{2}OH$ , $CH_{3}NH_{2}$ and $NH_{3}$ are the pure compounds which will exhibit hydrogen bonding.

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3 years ago

Which description best characterizes the motion of particles in a solid?

Masja [62]

Answer:

D. Vibrating around fixed positions

Explanation:

It's right on edge

6 0

3 years ago

Can you give me more explanation in the picture?

PolarNik [594]

Read the paper more carefully if you don't understand it

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3 years ago

GIVING BRAINLIEST One mole of hydrogen gas (H2), reacts with one mole of bromine Br2(g) to produce 2 moles of hydrogen bromide g

JulsSmile [24]

Answer:

The equation to show the the correct form to show the standard molar enthalpy of formation:

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

Given, that 1 mole of $H_2$ gas and 1 mole of $Br_2$ liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.

$H_2(g) + Br_2(l)\rightarrow 2HBr(g) ,\Delta H_{f}^o= -72.58kJ$

Divide the equation by 2.

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

The equation to show the the correct form to show the standard molar enthalpy of formation:

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

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3 years ago

The process or method , by which a magnet loses its power , is called…………………… ..

andrey2020 [161]

Answer:

I think it is called demagnetisation.

3 0

3 years ago