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3 years ago

Determine whether a closed spiral notebook with no writing would be chiral or achiral.

1 answer:

4 0

A chiral object is one that is not superimposable on its mirror image. Conversely, an achiral object is one that is identical (superimposable) to its mirror image. I believe a closed spiral notebook is a chiral object. Hope this answers the question. Have a nice day.

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What are three ways of measuring the amount of substance?

Rashid [163]

Answer:

Measure its mass

Measure its volume

Measure its number of moles

Explanation:

3 0

2 years ago

CHEMISTRY HELP PLEASE *answer all questions please*

Lerok [7]

Answer:

Question 1:

A) 0.289 moles.

B) 1.74 x 10²³ atoms.

Question 2:

A) 0.30 moles.

B) it contains 0.3 moles of both Na and Cl.

C) it contains 6.023 x 10²³ atoms of both Na and Cl.

Question 3:

A) The number of moles of sucrose (C₁₂H₂₂O₁₁) ≅ 0.0228 moles.

B) The number of moles of C atoms in sucrose (C₁₂H₂₂O₁₁) = 0.2763 mole of C atoms.

The number of moles of H atoms in sucrose (C₁₂H₂₂O₁₁) = 0.5016 mole of H atoms.

The number of moles of O atoms in sucrose (C₁₂H₂₂O₁₁) = 0.2508 mole of O atoms.

C) The number of C atoms = 1.65 x 10²³ atoms.

The number of H atoms = 3.02 x 10²³ atoms.

The number of O atoms = 1.51 x 10²³ atoms.

Explanation:

Question 1:

A) The number of moles of Au in 57.01 g sample:

n = mass / molar mass,

mass = 57.01 g and molar mass = 196.966 g/mol e.

The number of moles of Au in the sample = (57.01 g) / (196.966 g/mole) = 0.289 moles.

B) The number of atoms of Au in the sample:

It is known that every mole of a substance contains Avogadro,s number (NA = 6.023 x 10²³) of molecules.

1.0 mole of Au → 6.023 x 10²³ atoms

0.289 mole of Au → ???? atoms

using cross multiplication:

The number of atoms of Au in the sample = (6.023 x 10²³ x 0.289 mole) / (1.0 mole) = 1.74 x 10²³ atoms.

Question 2:

A) The number of moles of 17.45 g of NaCl:

n = mass / molar mass,

mass = 17.45 g and molar mass = 58.44 g/mole.

The number of moles of NaCl = (17.45 g) / (58.44 g/mole) = 0.298 mole ≅ 0.30 moles.

B) The number of moles of each element in NaCl

NaCl → Na + Cl

Each mole of NaCl contains one mole of Na and one mole of Cl.

using cross multiplication:

1.0 mole NaCl → 1.0 mole Na

0.3 mole NaCl → ??? mole Na

The number of moles of Na atoms in NaCl = (1.0 mole Na x 0.3 mole NaCl) / (1.0 mole NaCl) = 0.3 mole of Na atoms.

by the same way; the number of moles of Cl atoms = (1.0 mole Cl x 0.3 mole NaCl) / (1.0 mole NaCl) = 0.3 mole of Cl atoms.

C) The number of atoms of each element in the sample:

It is known that every mole of a substance contains Avogadro,s number (NA = 6.023 x 10²³) of molecules.

1.0 mole of NaCl → 6.023 x 10²³ molecules

0.3 mole of NaCl → ???? molecules

using cross multiplication:

The number of molecules in 0.3 mole of NaCl = (6.023 x 10²³ x 0.3 mole) / (1.0 mole) = 1.8069 x 10²³ molecules.

Every molecule of NaCl contains one atom of Na and one atom of Cl.

So, it contains 6.023 x 10²³ atoms of both Na and Cl.

Question 3:

A) The number of moles of 7.801 g of sucrose (C₁₂H₂₂O₁₁):

n = mass / molar mass,

mass = 7.801 g and molar mass = 342.3 g/mole.

The number of moles of sucrose (C₁₂H₂₂O₁₁) = (7.801 g) / (342.3 g/mol) = 0.022789 mol ≅ 0.0228 moles.

B) The number of moles of each element in sucrose (C₁₂H₂₂O₁₁):

C₁₂H₂₂O₁₁ → 12C + 22H + 11O

Each mole of sucrose contains 12 moles of C, 22 moles of H, and 11 moles of O.

using cross multiplication:

1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 12.0 moles C

0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ??? moles C

The number of moles of C atoms in sucrose (C₁₂H₂₂O₁₁) = (12.0 moles C x 0.0228 moles of sucrose (C₁₂H₂₂O₁₁)) / (1.0 mole sucrose (C₁₂H₂₂O₁₁)) = 0.2763 mole of C atoms.

By the same way; the number of moles of H atoms:

1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 22.0 moles H

0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ??? moles H

The number of moles of H atoms in sucrose (C₁₂H₂₂O₁₁) = (22.0 moles H x 0.0228 moles of sucrose (C₁₂H₂₂O₁₁)) / (1.0 mole sucrose (C₁₂H₂₂O₁₁)) = 0.5016 mole of H atoms.

Also; the number of moles of O atoms:

1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 11.0 moles O

0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ??? moles O

The number of moles of O atoms in sucrose (C₁₂H₂₂O₁₁) = (11.0 moles H x 0.0228 moles of sucrose (C₁₂H₂₂O₁₁)) / (1.0 mole sucrose (C₁₂H₂₂O₁₁)) = 0.2508 mole of O atoms.

C) The number of atoms of each element in the sucrose (C₁₂H₂₂O₁₁) sample:

It is known that every mole of a substance contains Avogadro,s number (NA = 6.023 x 10²³) of molecules.

1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 6.023 x 10²³ molecules

0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ???? molecules

using cross multiplication:

The number of molecules in 0.0228 mole of sucrose (C₁₂H₂₂O₁₁) = (6.023 x 10²³ x 0.0228 mole) / (1.0 mole) = 1.273 x 10²² molecules.

Each molecule of sucrose contains 12 atoms of C, 22 atoms of H, and 11 atoms of O.

So, the number of each atom that the sucrose (C₁₂H₂₂O₁₁) sample contains are:

The number of C atoms = (12 x 1.273 x 10²² molecules) = 1.65 x 10²³ atoms.

The number of H atoms = (22 x 1.273 x 10²² molecules) = 3.02 x 10²³ atoms.

The number of O atoms = (11 x 1.273 x 10²² molecules) = 1.51 x 10²³ atoms.

6 0

2 years ago

When 18.5 g of HgO(s) is decomposed to form Hg(l) and O2(g), 7.75 kJ of heat is absorbed at standard-state conditions. What is t

taurus [48]

Answer:

The standard enthalpy of formation of HgO is -90.7 kJ/mol.

Explanation:

The reaction between Hg and oxygen is as follows.

$\text{Hg(l)}+\frac{1}{2}{O_{2}\rightarrow \text{HgO(s)}$

From the given,

Molar mass of HgO = 216.59 g/mol

Mass of HgO decomposed = 18.5 g

Amount of heat absorbed = 7.75 kJ

From the reaction,

The standard enthalpy of formation = $+7.75\times\frac{kJ}{18.5 g}\frac{216.59}{1mol} \,\,= +90.7 kJ/mol$

During the decomposition of 1 mol of HgO , 90.7 kJ of energy absorbed.

For the formation of 1 mol of HgO , 90.7 kJ of energy is release

Therefore, the enthalpy of formation of mercury(II)Oxide is -90.7 kJ/mol

5 0

3 years ago

Solid sodium azide (NaN3) produces solid sodium and nitrogen gas. How many grams of sodium azide are needed to yield a volume of

ch4aika [34]

Answer:

52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P*V = n*R*T

In this case, the balanced reaction is:

2 NaN₃ → 2 Na + 3 N₂

You know the following about N₂:

P= 1.10 atm
V= 26.5 L
n=?

$R=0.082057 \frac{atm*L}{mol*K}$

T= 295 K

Replacing in the equation for ideal gas:

1.10 atm* 26.5 L= n* $0.082057 \frac{atm*L}{mol*K}$ *295 K

Solving:

$n=\frac{1.10 atm*26.5 L}{0.082057 \frac{atm*L}{mol*K} *295K}$

n= 1.2 moles

Now, the following rule of three can be applied: if 3 moles of N₂ are produced by stoichiometry of the reaction from 2 moles of NaN₃, 1.2 moles of N₂ are produced from how many moles of NaN₃?

$moles of NaN_{3}=\frac{1.2 molesofN_{2} *2 molesofNaN_{3} }{3 molesofN_{2} }$

moles of NaN₃= 0.8

Since the molar mass of sodium azide is 65.01 g / mol, then one last rule of three applies: if 1 mol has 65.01 grams of NaN₃, 0.8 mol how much mass does it have?

$mass of NaN_{3} =\frac{0.8 mol*65.01 grams}{1 mol}$

mass of NaN₃=52.008 grams

52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.

6 0

3 years ago

Read 2 more answers

What is the approximate value of the ccc bond angle in ch3ch2ch2oh

Fynjy0 [20]

Answer:

The value of the carbon bond angles are 109.5 °

Explanation:

CH3CH2CH2OH = propanol . This is an alcohol.

All bonds here are single bonds.

Single bonds are sp³- hybdridization type. To be sp3 hybridized, it has an s orbital and three p orbitals : sp³. This refers to the mixing character of one 2s-orbital and three 2p-orbitals. This will create four hybrid orbitals with similar characteristics.

Sp3- types have angles of 109.5 ° between the carbon - atoms.

This means that the value of the carbon bond angles are 109.5 °

6 0

3 years ago