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3 years ago

Identify whether longhand notation or noble-gas notation was used in each case below.

Chemistry

1 answer:

n200080 [17]3 years ago

6 0

Answer:

The given electronic configuration is long hand notation.

Explanation:

Long-hand notation of representing electronic configuration is defined as the arrangement of total number of electrons that are present in an element.

Noble-gas notation of representing electronic configuration is defined as the arrangement of valence electrons in the element. The core electrons are represented as the previous noble gas of the element that is considered.

The given electronic configuration of potassium (K):

The above configuration has all the electrons that are contained in the nucleus of an element. Thus, this configuration is a long-hand notation.

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Ppredict the identity of the precipitate in the below reaction:<br><br> BaCl2(aq) + K2SO4(aq) →

Annette [7]

Answer: The precipitate formed is $BaSO_4$

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid or precipitated form are represented by (s) after their chemical formulas.

A double displacement reaction in which one of the product is formed as a solid is called as precipitation reaction.

The balanced chemical equation is:

$K_2SO_4(aq)+BaCl_2(aq)\rightarrow BaSO_4(s)+2KCl(aq)$

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2 years ago

Draw structures from the following names, and determine which compounds are optically active:(c) 1,2-dibromo-2-methylbutane

dolphi86 [110]

The given compound 1,2-dibromo-2-methylbutane is an optically active compound .

Because this compound does not have plane of symmetry (POS) and center of symmetry (COS) i.e. does not have di-symmetry . And also forms non superimposable mirror image . the compound is optically active .

It has chiral center.

Here , the chiral carbon is attached to the 4 distinct groups such as : methyl , ethyl , bromine , bromomethane .

<h3>What is di-symmetry?</h3>

Di-symmetry is that which have no center of symmetry and plane of symmetry and alternate axis of symmetry .

<h3>Chiral center :</h3>

Have Sp3 hybridized center (4sigma bond ) .

4 distinct group is attached to the chiral atom. form non -superimposable mirror image .

<h3>What is optical isomerism ?</h3>

Same molecular formula and same structural formula . also have same physical and chemical properties .

They differ in their behavior towards plane polarized light (ppl) .

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brainly.com/question/9522537

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1 year ago

Aluminium +hydrochloride acid ________+_________

REY [17]

Answer:

Aluminium Chloride + Hydrogen

Please vote for Brainliest and I hope this helps!

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2 years ago

As a human being, we are considered to be a ..... because we rely on other organism for food

ryzh [129]

Heterotrophs i think :)

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3 years ago

Read 2 more answers

1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0

3 years ago