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3 years ago

Copper chloride burns bright green 510x10^9 m what is the frequency in Hz?

Chemistry

1 answer:

Katen [24]3 years ago

5 0

Answer:

5.88 x 10⁻⁴Hz

Explanation:

Given parameters:

Wavelength = 510 x 10⁹m

Unknown:

Frequency = ?

Solution:

To solve this expression, we use the equation below:

V = f∧

V is the velocity = 3 x 10⁸m/s

f is the frequency

∧ is the wavelength

3 x 10⁸ = f x 510 x 10⁹

f = 5.88 x 10⁻⁴Hz

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3. Examine the chemical formulas for the Arrhenius acids in Model 1. What feature do all the

Nostrana [21]

Answer:

They all have a hydrogen written separatly towards the beginning of the formula.

Explanation:

If you look at the Arrhenius acid formulas, you can see they all contain a hydrogen (H) at the beginning of them.

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4 years ago

Which substance in each of the following pairs would you expect to have the higher boiling point? Explain why.

satela [25.4K]

Explanation:

Boiling point is defined as the temperature when the vapor pressure of liquid becomes equal to the atmospheric pressure surrounding the liquid. And, during this temperature liquid state of substance changes into vapor state.

More stronger is the presence of intermolecular forces within the molecules of a compound more heat is required by it to break the bond. Hence, boiling point will also increase.

(a) As both Ne and Xe are noble gases and has intermolecular dispersion forces. But as the atomic size of Xe is more than the atomic size of Ne and we know that dispersion forces increases with increase in size.

Hence, boiling point of Xe will be more than the boiling point of Ne.

(b) Both $CO_{2}$ and $CS_{2}$ are non-polar in nature with intermolecular dispersion forces. So, more is the molecular weight of a compound more heat will be required by it in order to break the bonds.

As molecular weight of $CS_{2}$ is more than the molecular weight of $CO_{2}$ . Hence, $CS_{2}$ will have high boiling point.

(c) Molecular weight of $CH_{4}$ is 16 g/mol and molecular weight of $Cl_{2}$ is 35 g/mol.

Hence, $Cl_{2}$ will have high boiling point due to the presence of high molecular weight as compared to $CH_{4}$ and more strong dispersion forces.

(d) $F_{2}$ is a covalent compound so, it will also be non-polar in nature. Hence, it has weak intermolecular dispersion forces. On the other hand, LiF is an ionic compound and it has strong intermolecular forces of attraction due to the presence of opposite charges on the combining atoms.

Hence, LiF will have high boiling point as compared to $F_{2}$ .

(e) Both $NH_{3}$ and $PH_{3}$ are polar molecules in which dipole-dipole forces does exist. Since, nitrogen atom is more electronegative than phosphorous atom so, it will have stronger hydrogen bonding and strong intermolecular forces.

Hence, $NH_{3}$ will have high boiling point as compared to $PH_{3}$ .

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3 years ago

Giving two reasons,classify air as a mixture or compound.

ollegr [7]

1.) Air can be separated into its constituents such as oxygen, nitrogen etc.

2.) Air has a variable composition because at different places different amount of gases are present in air.

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4 years ago

Using the law of conservation of energy, what is the kinetic energy at C? K.E. =

Yanka [14]

Explanation:

It is also important to note that potential energy can be positive, zero or negative. ... Now that the kinetic energy and potential energy have been defined, we can now apply the Law of Conservation of Energy. In other words, the kinetic energy plus the potential energy equals a constant (KE+PE=Constant).

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3 years ago

A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate. Use 1.80 x 10−5 as Ka for ac

Anika [276]

Answer:

a. pH = 5.22

b. Acidic.

c. pH = 5.14

Explanation:

a. It is possible to find the pH of a buffer using Henderson-Hasselbalch equation (H-H equation):

pH = pKa + log₁₀ [A⁻] / [HA]

<em>Where pKa is -log Ka (For acetic acid = 4.74), [A⁻] is molar concentration of conjugate base (Acetate salt) and [HA] concentration of the weak acid (Acetic acid).</em>

Replacing:

pH = 4.74 + log₁₀ [0.600M] / [0.200M]

<em>You use the concentration of the acetic acid and sodium acetate because you're adding equal volumes, doing the ratio of the species the same</em>

<em />

b. As the solution has a pH lower that 7.0, it is considered as a <em>acidic solution.</em>

c. When you add HCl to the buffer, the reaction is:

CH₃COO⁻ + HCl → CH₃COOH + Cl⁻

<em>Where acetate ion reacts with the acid producing acetic acid.</em>

As you have 0.200L of the buffer, 0.100L are of the acetate ion and 0.100L of the acetic acid. Initial moles of both compounds and moles of HCl added are:

CH₃COO⁻: 0.100L ₓ (0.600mol / L) = 0.0600 moles

CH₃COOH: 0.100L ₓ (0.200mol / L) = 0.0200 moles

HCl: 3.0mL = 3x10⁻³L ₓ (0.034mol / L) = 0.00010 moles HCl

The moles added of HCl are the same moles you're consuming of acetate ion and producing of acetic acid. Thus, moles after the reaction are:

CH₃COO⁻: 0.0600 moles - 0.0001 moles = 0.0509 moles

CH₃COOH: 0.0200 moles + 0.0001 moles = 0.0201 moles

Replacing in H-H equation:

pH = 4.74 + log₁₀ [0.0509moles] / [0.0201moles]

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8 0

3 years ago