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irinina [24]
3 years ago
9

The correct increasing order of Bond angle is H2O < NH3 < BCl3 < CCl4 H2O < NH3 < CCl4 < BCl3 NH3 < H2O &lt

; BCl3 < CCl4 NH3 < BCl3 < CCl4 < H2O
Chemistry
1 answer:
Step2247 [10]3 years ago
5 0

Answer:

H2O < NH3 < CCl4 < BCl3

Explanation:

Bond angle of a molecule or an ion can be explained by two concepts which are either by their valence shell electron pair repulsion (VSEPR) model or hybridization.

The VSEPR model  determines the total number of electron pairs surrounding the central atom of the species. All the electron pairs will orient themselves in such a way as to minimize the electrostatic repulsions between them. These repulsions hence determine the geometry of the covalent bond angles around the central atom.

Hence; as the number of lone pairs increases from zero to 2 the bond angles diminish progressively.

Hybridization is the mixing and blending of two or more pure atomic orbitals to form two or moe hybrid atomic orbitals which are identical in shape and energy. During the process of formation of these hybrid orbitals, The bonds formed i.e the sigma bond and the pi bond determines the bond angle of such compound.

From the given compounds;

H20 have a bond angle of 104.5°

NH3 have a bond angle of 107°

BCl3 have a bond angle of 120°

CCl4 have a bond angle of 109.5°

thus in an increasing order of bond angle:

H2O < NH3 < CCl4 < BCl3

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Energy is transferred from the sun to Earth mainly by
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Answer:

electromagnetic waves

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2 years ago
In writing a chemical equation that produces hydrogen gas, the correct representation of hydrogen gas is
Dennis_Churaev [7]

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h2

Explanation:

5 0
3 years ago
10) How many grams are there in 1.00 x 10^24 molecules of BC13?
lana66690 [7]

194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

Explanation:

In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.

It is known that 1 moles of any element has 6.022×10²³ molecules.

Then 1 molecule will have \frac{1}{6.022*10^{23} } moles.

So 1*10^{24} molecules = \frac{1*10^{24} }{6.022*10^{23} } =1.66 moles

Thus, 1.66 moles are included in BCl₃.

Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.

As it is known as 1 mole contains molecular mass of the compound.

As the molecular mass of BCl₃ will be

Molecular mass of BCl_{3}= Mass of Boron + (3*Mass of chlorine)

Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.

Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.

grams of BCl_{3}=Molar mass of BCl_{3}*Number of moles

grams of BCl_{3}=117.17*1.66=194.5 g

So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

3 0
3 years ago
PLZ HELP Question 14 of 25 What is the name for a representation of the physical world?
JulijaS [17]

Answer:

Model

Explanation:

A model of anything is something you make to represent it in it's physical world form

8 0
3 years ago
For this question, the "entropy term" refers to "-TΔS". Addition reactions are generally favorable at low temperatures because _
nata0808 [166]

Answer:

Lowering the temperature typically reduces the significance of the decrease in entropy. That makes the Gibbs Free energy of the reaction more negative. As a result, the reaction becomes more favorable overall.  

Explanation:

In an addition reaction there's a decrease in the number of particles. Consider the hydrogenation of ethene as an example.

\rm H_2C\text{=}CH_2\; (g) + H_2\; (g) \stackrel{\text{Ni}^\ast}{\to} H_3C\text{-}CH_3\; (g).

When \rm H_2 is added to \rm H_2C\text{=}CH_2 (ethene) under heat and with the presence of a catalyst, \rm H_3C\text{-}CH3 (ethane) would be produced.

Note that on the left-hand side of the equation, there are two gaseous molecules. However, on the right-hand side there's only one gaseous molecule. That's a significant decrease in entropy. In other words, \Delta S < 0.

The equation for the change in Gibbs Free Energy for a particular reaction is:

\Delta G = \Delta H + (\underbrace{- T \, \Delta S}_{\text{entropy}\atop \text{term}}).

For a particular reaction, the more negative \Delta G is, the more spontaneous ("favorable") the reaction would be.

Since typically \Delta S < 0 for addition reactions, the "entropy term" of it would be positive. That's not very helpful if the reaction needs to be favorable.

T (absolute temperature) is always nonnegative. However, lowering the temperature could help bring the value of

8 0
3 years ago
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