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3 years ago

An aluminum cation is smaller than the aluminum atom because it has fewer occupied energy levels. True false

Chemistry

2 answers:

Mars2501 [29]3 years ago

7 0

Answer: The given statement is true.

Explanation:

Atomic number of aluminium is 13 and in order to attain stability aluminium loses 3 electrons. As a result, neutral aluminium atom converts into $Al^{3+}$ ion.

Thus, due to loss of electrons aluminium cation is smaller than neutral aluminium atom.

Therefore, the statement an aluminum cation is smaller than the aluminum atom because it has fewer occupied energy levels, is true.

butalik [34]3 years ago

3 0

The molecules or atoms that are formed by gain or loss of one or more valence electrons are said to be ions.

When atom loss one or more valence electrons, results in formation of cation whereas when atom gain one or more valence electrons, then formation of anion occurs. Cations carry positive charge and anions carry negative charge.

In general, cations are smaller than the neutral atoms from which they are formed and anions are larger than the neutral atoms.

As cations are smaller than the related neutral atoms because the valence electrons are lost which are farthest away from the nucleus. After that, taking more electrons distant from the cation results in reduction of radius of the ion.

Thus, aluminium cation consist of few electrons which results in fewer occupied energy levels by the electrons further results in reduction of radius i.e. smaller size.

Hence, given statement is true i.e. aluminium atom is larger than the aluminium cation as cation has fewer occupied energy levels.

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3 years ago

What is the molar mass of an unknown gas with a density of 4.95 g/L at 1.00 atm and 25.0 °C?

mestny [16]

Answer:

121 g/mol

Explanation:

To find the molar mass, you first need to calculate the number of moles. For this, you need to use the Ideal Gas Law. The equation looks like this:

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = constant (0.0821 L*atm/mol*K)

-----> T = temperature (K)

Because density is comparing the mass per 1 liter, I am assuming that the system has a volume of 1 L. Before you can plug the given values into the equation, you first need to convert Celsius to Kelvin.

P = 1.00 atm R = 0.0821 L*atm/mol*K

V = 1.00 L T = 25.0. °C + 273.15 = 298.15 K

n = ? moles

PV = nRT

(1.00 atm)(1.00L) = n(0.0821 L*atm/mol*K)(298.15 K)

1.00 = n(0.0821 L*atm/mol*K)(298.15 K)

1.00 = (24.478115)n

0.0409 = n

Now, we need to find the molar mass using the number of moles per liter (calculated) and the density.

0.0409 moles ? grams 4.95 grams
---------------------- x ------------------ = ------------------
1 L 1 mole 1 L

? g/mol = 121 g/mol

**note: I am not 100% confident on this answer

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IRISSAK [1]

Answer:

0.2g

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A = A₀e^-kt

A => Activity at time (t)

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T => time in units that match the decay constant

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3 years ago

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3 years ago

The solubility of a gas is 0.890 g/L at a pressure of 120 kPa. What is the solubility of the gas if the pressure is changed to 1

IgorC [24]

The solubility of the gas if the pressure is changed to 100 kPa is 0.742 g/L

<h3>Effect of Pressure on Solubility </h3>

As the pressure of a gas increases, the solubility increases, and as the pressure of a gas decreases, the solubility decreases.

Thus, Solubility varies directly with Pressure

If S represents Solubility and P represents Pressure,

Then we can write that

S ∝ P

Introducing proportionality constant, k

S = kP

S/P = k

∴ We can write that

$\frac{S_{1} }{P_{1} } = \frac{S_{2} }{P_{2} }$

Where $S_{1}$ is the initial solubility

$P_{1}$ is the initial pressure

$S_{2}$ is the final solubility

$P_{2}$ is the final pressure

From the given information

$S_{1} = 0.890 \ g/L$

$P_{1} = 120 \ kPa$

$P_{2} = 100 \ kPa$

Putting the parameters into the formula, we get

$\frac{0.890}{120}=\frac{S_{2}}{100}$

$S_{2}= \frac{0.890 \times 100}{120}$

$S_{2}= 0.742 \ g/L$

Hence, the solubility of the gas if the pressure is changed to 100 kPa is 0.742 g/L

Learn more on Solubility here: brainly.com/question/4529762

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2 years ago