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3 years ago

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Chemistry

1 answer:

arlik [135]3 years ago

6 0

Given :

Chlorine's 3p sub level has 5 electrons in it.

To Find :

How many electron(s) more in the 3p sub-level to make it stable.

Solution :

Electronic configuration of Chlorine is :

[Ne] 3s² 3p⁵.

We know, p-orbit is stable with 6 electronic.

Therefore, to stable p-orbit 1 more electron is required in 3p sub-level.

Hence, this is the required solution.

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the density of glycerin is 1.26 g/cm3. How many pounds/foot3 is this ? use the conversion rates of 454g/1 pound and 28,317cm3/1f

lyudmila [28]

Answer:

78.6 lb/ft³

Step-by-step explanation:

Let's do this in steps.

1. Convert grams to pounds

D = (1.26 g/1 cm³) × (1 lb/454 g)

= 2.775 × 10⁻³ lb/cm³

2. Convert cubic centimetres to cubic feet

D = (2.775 × 10⁻³ lb/1 cm³) × (28 317cm³/1 ft³)

= 78.6 lb/ft³

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3 years ago

The expected o-c-h angle in this molecule is degrees. the expected hybridization at the central carbon is

WINSTONCH [101]

Each neutral carbon atom contains four valence electrons and may form up to four electron domains. Possible hybridizations include

$sp^{3}$ , four electron domains, as in ethane $\text{C}_2\text{H}_6$
$sp^{2}$ , three electron domains, as in ethene $\text{C}_2\text{H}_4$
$sp$ , two electron domains, as in ethyne $\text{C}_2\text{H}_2$

Molecules of each of the three hybridization demonstrate spatial configurations that would maximizes the separation between the electron domains.

Carbon atoms with a $sp^{3}$ hybridization would demonstrate a tetrahedral configuration with a bond angle of approximately $109.5\textdegree{}$
Carbon atoms with a $sp^{2}$ hybridization would demonstrate a triangular planar configuration with a bond angle of $120\textdegree{}$
Carbon atoms with a $sp$ hybridization would demonstrate a linear configuration with a bond angle of $180\textdegree{}$

Bond angles are characteristic of the spatial configuration of electron domains and identifies the hybridization of the central carbon atom.

Note that each hydrogen atom contains only one valence electron and would form only single bonds. It takes two valence electrons for oxygen atoms to achieve an octet such that each oxygen form only two bonds at a single time. Therefore given the fact that the carbon is bonded to both hydrogen and oxygen, only the following hybridizations are possible

$sp^{3}$ in which the oxygen atom forms a carbon-oxygen double bond with the central carbon atom;
$sp^{2}$ in which the oxygen atom forms a single bond with the central carbon atom and with a second atom.

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4 years ago

Household substances that contains acid

lidiya [134]

Answer:

milk, lemon juice, toothpaste and baking soda

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3 years ago

Problem PageQuestionSteam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas,

Serhud [2]

The question is incomplete. Her eis the complete question.

Steam reforming methane (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answer: $K_{c}$ = 2. $10^{-2}$

Explanation: The reaction for steam reforming methane is:

$CH_{4} + H_{2}O$ ⇒ $CO_{} + 3H_{2}$

To calculate the concentration equilibrium constant, first calculate the molarity ( $\frac{mol}{L}$ ) of each molecule of the reaction.

At 38°C: At the initial temperature, there no products yet

Molarity of CH4:

CH4 = $\frac{20}{125}$ = 0.16M

Molarity of H20:

H2O = $\frac{10}{125}$ = 0.08M

At final temperature:

Molarity of H2:

H2 = $\frac{18}{125}$ = 0.144M

According to the chemical reaction, the combination of 1 mol of each reagents produces 1 mol of CO and 3 mols of H2, so, for the products, the ratio is 1:3.

Molarity of CO:

CO = $\frac{0.144}{3}$ = 0.048M