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2 years ago

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A molecule with the formula ax3 uses __________ to form its bonds. multiple choice sp2 hybrid orbitals sp hybrid orbitals sp3d2

hybrid orbitals sp3d hybrid orbitals sp3 hybrid orbitals

2 answers:

Ivanshal [37]2 years ago

8 0

A molecule with formula AX3E uses SP3 to form its bonds.
Where A is the metal
X is any halogen which is bonded to it.
E is the lone pair
∴ A requires four electron pairs and tetrahedral.
Where tetrahedral symmetry is SP3 hybridized

GenaCL600 [577]2 years ago

4 0

A molecule with formula AX3E uses SP3 to form its bonds.

Where A is the metal

X is any halogen which is bonded to it.

E is the lone pair

A requires four electron pairs and tetrahedral.

Where tetrahedral symmetry is SP3 hybridized

Read more on Brainly.com - brainly.com/question/9810726#readmore

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Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

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2 years ago

Arc AB is One-sixth of the circumference of a circle. What is the radian measure of the central angle?

soldi70 [24.7K]

Answer:

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Explanation:

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2 years ago

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In the reaction 2H2O (1)+ 2Cl^- (aq)= H2(g)+Cl2 (g)+ 2OH^-(aq), which substance is reduced?

Oksanka [162]

Answer:- C. H

Explanations:- Reduction is gain of electron. In other words we could say that decrease in oxidation number is reduction.

As per the rules, oxidation number of hydrogen in its compounds is +1(except metal hydrides) and the oxidation number of oxygen in its compounds is -2.

The oxidation number in elemental form is zero.

In $H_2O$ , the oxidation number of H is +1 and oxidation number of O is -2. Oxidation number of Cl in $Cl^-$ is -1. On product side, the oxidation number of hydrogen in $H_2$ is zero and in $OH^-$ the oxidation number of H is +1 and that of O is -2. Oxidation number of Cl in $Cl_2$ is 0.

From above data, Oxidation number of O is -2 on both sides so it is not reduced.

Oxidation number of Cl is changing from -1 to 0 which is oxidation.

Oxidation number of H is changing from +1 to 0 which is reduction.

So, the right choice is C.H

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The majority of the mass of the atom is located in the _______. A) nucleus B) core electrons C) pions and quarks D) valence elec

NeX [460]

The majority of the mass of the atom is located in the nucleus. Remember that the nucleus contains both protons and neutrons and therefore, most of the mass of the atom.

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Answer:

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