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2 years ago

Which of the following is an empirical formula? H6O2 H402 H2O H2O2

Chemistry

2 answers:

Jobisdone [24]2 years ago

6 0

Answer:

hola responderé

Explanation:

el h2o es agua y el agua es mojada gracias por su atención

topjm [15]2 years ago

5 0

${\qquad\qquad\huge\underline{{\sf Answer}}}$

<h3>What does an empirical formula mean ? </h3>

- Empirical formula represents a molecule with its elements combined in simplest ratio, it's not necessarily same as the form in which the compound really exists in the nature. Hence we can say that empirical formula shows the simple ratios in which atoms combined to form a molecule.

For example : Hydrogen peroxide

- Hydrogen peroxide exists in nature as $\sf H_2O_2$ , which is its molecular formula.

but it's empirical formula will be H0 [ simplest ratio of 1 : 1 ]

Therefore, the correct choice is :

C. $\sf H_2O$

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An atom has the following electron configuration 1s2s2sp63s23p64s2 How many valance electrons does this atom have?

JulijaS [17]

Answer:

The electron configuration for this atom is Calcium, which has 2 valence electrons.

Explanation:

Following the periodic table and with the electron configuration, you will end up with calcium, which has 2 valence electrons. (Always follow the electron configuration from left to right! It begins at hydrogen, then to helium... and so on.)

1s2 -> He....

2s2 -> Be....

2p6 -> Ne...

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3 years ago

How does a balanced chemical equation demonstrate conservation of mass?

Mariulka [41]

Answer:

A balanced chemical equation will demonstrate the law of conservation of mass because the amount of elements on one side (reactants) will equal the amount of elements on the other side (products).

Explanation:

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1 C

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Products:

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6 0

2 years ago

Read 2 more answers

Select correct examples of linear molecules for five electron groups. Select correct examples of linear molecules for five elect

Shalnov [3]

Explanation:

Formula to calculate hybridization is as follows.

Hybridization = $\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

So, hybridization of $BeCl_{2}$ is as follows.

Hybridization = $\frac{1}{2}[V + N - C + A]$

= $\frac{1}{2}[2 + 2]$

= 2

Hybridization of $BeCl_{2}$ is sp. Therefore, $BeCl_{2}$ is a linear molecule. There will be only two electron groups through which Be is attached.

Similarly, hybridization of $XeF_{2}$ is calculated as follows.

Hybridization = $\frac{1}{2}[V + N - C + A]$

= $\frac{1}{2}[8 + 2]$

= 5

Therefore, hybridization of $XeF_{2}$ is $sp^{3}d. Therefore, [tex]XeF_{2}$ is also a linear molecule. Though there are three lone pair of electrons present on a xenon atom and it is further attached with fluorine atoms through two electron pairs. Hence, there are in total five electron groups.