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Mandarinka [93]
2 years ago
11

elect all statements that correctly describe the stability of eclipsed and staggered conformers. Multiple select question. Stagg

ered conformers are lower in energy than eclipsed conformers. Electron-electron repulsion between bonds in an eclipsed conformer may contribute to the low energy of this conformer. Staggered conformers possess torsional strain. Eclipsed conformers possess torsional strain.
Chemistry
1 answer:
alisha [4.7K]2 years ago
4 0

Staggered conformers are lower in energy than eclipsed conformers , Option A is the right answer.

<h3>What are Eclipsed and Staggered Confirmation ?</h3>

The arrangement of atoms in molecule such that  the dihedral angle formed is 60 degree is called Staggered Conformation and the arrangement of atoms in which the dihedral angle is 0 degree is called Eclipsed Confirmation

Among the given statements

The statement that best describe the stability of Staggered Conformers and eclipsed conformers is Staggered conformers are lower in energy than eclipsed conformers .

To know more about Eclipsed and Staggered Confirmation

brainly.com/question/27852131

#SPJ1

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Given 4.80g of ammonium carbonate, find:
V125BC [204]

Answer:

1) 0.05 mol.

2) 0.1 mol.

3) 0.05 mol.

4) 0.4 mol.

5) 2.4 x 10²³ molecules.

Explanation:

<em>1) Number of moles of the compound:</em>

no. of moles of ammonium carbonate = mass/molar mass = (4.80 g)/(96.09 g/mol) = 0.05 mol.

<em>2) Number of moles of ammonium ions :</em>

  • Ammonium carbonate is dissociated according to the balanced equation:

<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

<em>∴ The no. of moles of NH₄⁺ ions in 0.05 mol of (NH₄)₂CO₃ </em>= (2.0)(0.05 mol) =  <em>0.1 mol.</em>

<em>3) Number of moles of carbonate ions :</em>

  • Ammonium carbonate is dissociated according to the balanced equation:

<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of CO₃²⁻ ions in 0.05 mol of (NH₄)₂CO₃ = (1.0)(0.05 mol) = 0.05 mol.

<em>4) Number of moles of hydrogen atoms:</em>

  • Every 1.0 mol of (NH₄)₂CO₃  contains:

2.0 moles of N atoms, 8.0 moles of H atoms, 1.0 mole of C atoms, and 3.0 moles of O atoms.

<em>∴ The no. of moles of H atoms in 0.05 mol of (NH₄)₂CO</em>₃ = (8.0)(0.05 mol) = <em>0.4 mol.</em>

<em>5) Number of hydrogen atoms:</em>

  • It is known that every mole of a molecule or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.

<u><em>Using cross multiplication:</em></u>

1.0 mole of H atoms contains → 6.022 x 10²³ atoms.

0.4 mole of H atoms contains → ??? atoms.

<em>∴ The no. of atoms in  0.4 mol of H atoms</em> = (6.022 x 10²³ molecules)(0.4 mole)/(1.0 mole) = <em>2.4 x 10²³ molecules.</em>

8 0
3 years ago
At 170C, a sample of hydrogen gas occupies 125cm3. What will be the volume at 1000C? =161cm3 If the volume of a given mass of ga
pickupchik [31]

1. V= 161

2. V = 37.3

<h3 /><h3>Further explanation  </h3>

Charles's Law states that  

<em>When the gas pressure is kept constant, the gas volume is proportional to the temperature  </em>

\tt \dfrac{V1}{T1}=\dfrac{V2}{T2}

1.

\tt \dfrac{125}{17+273.15~K}=\dfrac{V2}{100+273.15}\\\\V2=160.7\rightarrow 161~cm^3

2.

\tt \dfrac{27.3}{0+273.15`K}=\dfrac{V2}{100+273.15}\\\\V2=37.3~cm^3

3 0
3 years ago
If an atom contains more <br> electrons than protons, it is
Ray Of Light [21]

Answer:

If an atom has the same number of electrons as protons, it is a neutral atom.

Explanation:

8 0
3 years ago
How many L of 3.0 M H2SO4 solution can be prepared by using 100.0 mL OF 18 M H2SO4?
max2010maxim [7]

Answer: A volume of 600 mL of 3.0 M H_{2}SO_{4} solution can be prepared by using 100.0 mL OF 18 M H_{2}SO_{4}.

Explanation:

Given: V_{1} = ?,        M_{1} = 3.0 M\\

V_{2} = 100.0 mL,       M_{2} = 18 M

Formula used to calculate the volume is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\3.0 M \times V_{1} = 18 M \times 100.0 mL\\V_{1} = \frac{18 M \times 100.0 mL}{3.0M}\\= 600 mL

Thus, we can conclude that a volume of 600 mL of 3.0 M H_{2}SO_{4} solution can be prepared by using 100.0 mL OF 18 M H_{2}SO_{4}.

6 0
3 years ago
Student follow up
sleet_krkn [62]

Answer:

Magnesium and calcium belong to the second group i. e. alkaline earth metals. They are known as earth metals because they are extracted from the earth. They are very reactive elements. Their reactivity increases when we go from top to bottom because the outermost electrons goes farther from the nucleus i. e. atomic radius increases so less energy is needed for its removal.

8 0
3 years ago
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