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3 years ago

Aluminum boils at 2467°C. Aluminum’s boiling point in Kelvin is 2194. True or False.

Chemistry

2 answers:

Sergeeva-Olga [200]3 years ago

6 0

Answer:

The given statement is false.

Explanation:

Given that aluminum boils at 2,467°C

Boiling point of aluminum in kelvins is 2,194 K

Temperature in degree Celsius can be converted into Kelvins by relation:

$(T)^oC=273 +(T) K$

So, boiling point of aluminium in Kelvins will be:

$2,467 ^oC=273 +(2,467) K=2,740 K$

2,740 K ≠ 2,194 K

But the given value in Kelvins is not coming equal to our calculated value, Hence, the given statement is false.

IceJOKER [234]3 years ago

3 0

False. To get Kelvin add 273 to <span>°C to get your temperature into Kelvin. The correct boiling point of Aluminum in Kelvin is 2740 K.</span>

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Which pair of elements form an ionic bond?

Anastasy [175]

Answer:

As an example I can say sodium (Na) and chlorine (Cl).

Explanation:

An ionic bond occurs when a metal element reacts with a nonmetal element. Therefore in the answer given above the Na is metal and Cl is nonmetal and they form a molecule through ionic bonding.

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3 years ago

Living organisms make their own carbon

san4es73 [151]

Where’s the question is this a true or false question

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3 years ago

The force of attraction between a divalent cation and a divalent anion is 1.64 x 10-8 N. If the ionic radius of the cation is 0.

Elden [556K]

The radius of the anion is 7.413 nm

<h3>How to calculate the force of attraction between charges</h3>

The force of attraction (F) is given by the formula:

F = (1/4π∈r²)(Zc*e)(Za*e)

where:

∈ = permittivity of free space = 8.85*10⁻¹⁵ F/m

Zc = charge on the cation = +2

Zc = charge on the anion = -2

e = charge on an electron = 1.602 * 10⁻¹⁹ C

r = interionic distance

r = rc + ra

where rc and ra are the radius of the cation and anion respectively

F = 1.64 * 10⁻⁸ N

Therefore based on the equation of force of attraction:

1.64 *10⁻⁸ = [1/4π(8.85*10⁻¹⁵)r²](2 * 1.602*10⁻¹⁹)²

r² = 5.63 * 10⁻¹⁷

r = 7.50 nm

Since r = rc + ra

where rc = 0.087 nm

thus, ra = r - rc = 7.50 - 0.087

ra = 7.413 nm

Therefore, the radius of the anion is 7.413 nm

Learn more about ionic radius at: brainly.com/question/2279609

6 0

2 years ago

A gas occupies 4.31 liters at a pressure of 0.755 atm. Determine the volume if the pressure is increased to 1.25 atm. (Boyle)

vlabodo [156]

Answer:

Explanation:

The new volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we are finding the new volume

$V_2 = \frac{P_1V_1}{P_2} \\$

From the question we have

$V_2 = \frac{4.31 \times 0.755}{1.25} = \frac{3.25405}{1.25} \\ = 2.60324$

We have the final answer as

Hope this helps you

6 0

3 years ago

An unknown amount of Al203 decomposed producing 215 g of solid aluminum. 2Al2O3=4Al+3O2 How many grams of oxygen gas should be p

natulia [17]

Answer:

191.11 grams of oxygen gas should be produced.

Explanation:

The balanced reaction is:

2 Al₂O₃ → 4 Al + 3 O₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Al₂O₃: 2 moles
Al: 4 moles
O₂: 3 moles

Being the molar mass of each compound:

Al₂O₃: 102 g/mole
Al: 27 g/mole
O₂: 32 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Al₂O₃: 2 moles* 102 g/mole= 204 grams
Al: 4 moles* 27 g/mole= 108 grams
O₂: 3 moles* 32 g/mole= 96 grams

Then you can apply the following rule of three: if by stoichiometry 108 grams of aluminum are produced along with 96 grams of oxygen, 215 grams of aluminum are produced along with how much mass of oxygen?

$mass of oxygen=\frac{215 grams of aluminum*96 grams of oxygen}{108grams of aluminum}$

mass of oxygen= 191.11 grams

<u><em>191.11 grams of oxygen gas should be produced.</em></u>

6 0

3 years ago