All categories

1 year ago

show the step by step hybridization of hydroxonium ion H3O, draw the diagram showing the geometry and the bond angle

Chemistry

1 answer:

s344n2d4d5 [400]1 year ago

3 0

The central oxygen atom is a state of sp3 hybridization having only one available lone pair.

<h3>What is hybridization?</h3>

The term hybridization refers to the mixing of two or more atomic orbitals to yield hybrid orbitals that are suitable for bonding.

Looking at the ion H3O^+, we can see that the central oxygen atom is a state of sp3 hybridization having only one available lone pair. The image of the ion is shown here.

Learn more about hybridization in ions:brainly.com/question/16556024

#SPJ1

You might be interested in

Calcium Carbonate reacts with dilute hydrochloric acid. The equation for the reaction is shown. CaCo3 + 2Hcl = Cacl2 + H2O + Co2

Elena-2011 [213]

Answer:

Approximately $0.224\;\rm L$ , assuming that this reaction took place under standard temperature and pressure, and that $\rm CO_2$ behaves like an ideal gas. Also assume that the reaction went to completion.

Explanation:

The first step is to find out: which species is the limiting reactant?

Assume that $\rm CaCO_3$ is the limiting reactant. How many moles of $\rm CO_2$ would be produced?

Look up the relative atomic mass of $\rm Ca$ , $\rm C$ , and $\rm O$ on a modern periodic table:

$\rm Ca$ : $40.078$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm CaCO_3$ :

$\begin{aligned} & M(\rm CaCO_3) \\ &= 40.078 + 12.011 + 3 \times 15.999 \\&= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the number of moles of formula units in $1\; \rm g$ of $\rm CaCO_3$ using its formula mass:

$\begin{aligned}& n(\mathrm{CaCO_3})\\&= \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} \\ &= \frac{1\; \rm g}{100.086\; \rm g \cdot mol^{-1}} \approx 1.00\times 10^{-2}\; \rm mol\end{aligned}$ .

In the balanced chemical equation, the ratio between the coefficient of $\rm CaCO_3$ and that of $\rm CO_2$ is $\displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{CaCO_3})} = 1$ .

In other words, for each mole of $\rm CaCO_3$ formula units consumed, one mole of $\rm CO_2$ would be produced.

If $\rm CaCO_3$ is indeed the limiting reactant, all that approximately $1.00\times 10^{-2}\; \rm mol$ of $\rm CaCO_3\!$ formula would be consumed. That would produce approximately $1.00\times 10^{-2}\; \rm mol\!$ of $\rm CO_2$ .

On the other hand, assume that $\rm HCl$ is the limiting reactant.

Convert the volume of $\rm HCl$ to $\rm dm^{3}$ (so as to match the unit of concentration.)

$\begin{aligned}&V(\mathrm{HCl})\\ &= 50\; \rm cm^{3} \\ &= 50\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{10^{3}\; \rm cm^{3}} \\ &= 5.00\times 10^{-2}\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $5.00\times 10^{-2}\; \rm dm^{3}$ of this $\rm 0.05\; \rm mol \cdot dm^{-3}$

$\begin{aligned}& n(\mathrm{HCl}) \\ &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.05\; \rm mol \cdot dm^{-3}\\ &\quad\quad \times 5.00\times 10^{-2}\;\rm dm^{3} \\ &= 2.50 \times 10^{-3}\; \rm mol\end{aligned}$ .

Notice that in the balanced chemical reaction, the ratio between the coefficient of $\rm HCl$ and that of $\rm CO_2$ is $\displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{HCl})} = \frac{1}{2}$ .

In other words, each mole of $\rm HCl$ molecules consumed would produce only $0.5\;\rm mol$ of $\rm CO_2$ molecules.

Therefore, if $\rm HCl$ is the limiting reactant, that $2.50 \times 10^{-3}\; \rm mol$ of $\rm HCl\!$ molecules would produce only one-half as many (that is, $1.25\times 10^{-3}\; \rm mol$ ) of $\rm CO_2$ molecules.

If $\rm CaCO_3$ is the limiting reactant, $\rm 1.00\times 10^{-3}\; \rm mol$ of $\rm CO_2$ molecules would be produced. However, if $\rm HCl$ is the limiting reactant, $1.25\times 10^{-3}\; \rm mol$ of $\rm CO_2\!$ molecules would be produced.

In reality, no more than $\rm 1.00\times 10^{-3}\; \rm mol$ of $\rm CO_2$ molecules would be produced. The reason is that all $\rm CaCO_3$ would have been consumed before $\rm HCl$ was.

After finding the limiting reactant, approximate the volume of the $\rm CO_2\!$ produced.

Assume that this reaction took place under standard temperature and pressure (STP.) Under STP, the volume of one mole of ideal gas molecules would be approximately $22.4\; \rm L$ .

If $\rm CO_2$ behaves like an ideal gas, the volume of that $\rm 1.00\times 10^{-3}\; \rm mol$ of $\rm CO_2\!$ molecules would be approximately $\rm 1.00\times 10^{-3}\; \rm mol \times 22.4\; \rm L = 0.224\; \rm L$ .

3 0

3 years ago

Draw the structures of the starting materials needed to make 2-methylhept-3-yne in the spaces provided. The starting materials m

DIA [1.3K]

Answer: answered

Explanation:

4 0

3 years ago

Both the sweatshirt and the balloon are negatively charged. They are most

Makovka662 [10]

Answer:

B attract

Explanation:

There will be statication

8 0

2 years ago

When you look out the window you can see constantly moving air.Air is cycled in order to reach equilibrium

aleksley [76]

False, although the air is moving in an equilibrium us normal humans can not see the air actually moving. So it is false.

6 0

3 years ago

WILL GIVE BRAINLEST!!

qwelly [4]

Coastal flooding occurs when normally dry, low-lying land is flooded by seawater.[1] The extent of coastal flooding is a function of the elevation inland flood waters penetrate which is controlled by the topography of the coastal land exposed to flooding.The seawater can flood the land via from several different paths:

3 0

3 years ago

show the step by step hybridization of hydroxonium ion H3O, draw the diagram showing the geometry and the bond angle​

show the step by step hybridization of hydroxonium ion H3O, draw the diagram showing the geometry and the bond angle