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3 years ago

To double the average kinetic energy of helium

Chemistry

1 answer:

mixas84 [53]3 years ago

5 0

Answer:

54°C

Explanation:

Hello,

To find the temperature of the gas, we would need a relationship between kinetic energy and temperature.

K.E = 3/2RT/Na

Where K.E = kinetic energy

R = ideal gas constant

T = temperature of the ideal gas

Na = Avogadro's number

2K.E₁ = K.E₂ (required)

Since both constant (R and Na) are equal,

K.E = T (direct proportionality)

But T = 27°C

T2 = 2 × 27 = 54°C

To double the kinetic energy of a helium gas at 27°C the temperature of the gas should be increased to 54°C

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Magnesium reacts with a certain element to form a compound with the general formula MgX. What would the most likely formula be f

ziro4ka [17]

Answer:K2X

Explanation: Valency can be defined as the combining power of an element. It is the valency that dictates the value an element will have when writing a chemical formula for its compound.

MgX is a compound of magnesium and an element X. The valency of magnesium in most of its compound is +2. Now for the 2 to have been absent in the chemical formula, this shows that the element X itself have a valency if -2 for the valencies of both to have canceled out.

Now considering the element potassium, it is an alkaline metal belonging to group 1 of the periodic table. Hence, it is expected that it has a valency of +1

Forming a compound with element X means there would be an exchange of valencies between the two. We have established that x has a valency of -2. The formula of the compound thus formed by exchanging the valencies of both element would be K2X

7 0

2 years ago

Read 2 more answers

For all of the following questions 20.00 mL of 0.200 M HBr is titrated with 0.200 M KOH.

HACTEHA [7]

Answer :

The concentration of $H^+$ before any titrant added to our starting material is 0.200 M.

The pH based on this $H^+$ ion concentration is 0.698

Explanation :

First we have to calculate the concentration of $H^+$ before any titrant is added to our starting material.

As we are given:

Concentration of HBr = 0.200 M

As we know that the HBr is a strong acid that dissociates complete to give hydrogen ion $H^+$ and bromide ion $Br^-$ .

As, 1 M of HBr dissociates to give 1 M of $H^+$

So, 0.200 M of HBr dissociates to give 0.200 M of $H^+$

Thus, the concentration of $H^+$ before any titrant added to our starting material is 0.200 M.

Now we have to calculate the pH based on this $H^+$ ion concentration.

pH : It is defined as the negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pH=-\log (0.200)$

$pH=0.698$

Thus, the pH based on this $H^+$ ion concentration is 0.698

3 0

3 years ago

Erythrogenic acid, C18H26O2, is an acetylenic fatty acid that turns avivid red on exposure to light. On catalytic hydrogenation

lara31 [8.8K]

Answer:

use research

Explanation:

6 0

3 years ago

Molybdenum has a molar mass of 95.94g/mol. How many molecules of molybdenum are in 150.0 g of molybdenum

Kitty [74]

We are given the molar mass of Molybdenum as 95.94 g/mol. Also, the chemical symbol for Molybdenum is Mo. This question is asking for the amount of molecules of molybdenum in a 150.0 g sample. However, since molybdenum is a metal and it is in the form of solid molybdenum, Mo (s), it is not actual a molecule. A molecule has one or more atom bonded together. We will instead be finding the amount of atoms of Molybdenum present in the sample. To do this we use Avogadro's number, which is the amount of atoms/molecules of a substance in 1 mole of that substance.

150.0 g Mo/ 95.94 g/mol = 1.563 moles of Mo

1.563 moles Mo x 6.022 x 10²³ atoms/mole = 9.415 x 10²³ atoms Mo

Therefore, there are 9.415 x 10²³ atoms of Molybdenum in 150.0 g.

5 0

4 years ago

Ammonia and oxygen react to form nitrogen and water.

Nata [24]

Answer:

A. 19.2 g of O2.

B. 3.79 g of N2.

C. 54 g of H2O.

Explanation:

The balanced equation for the reaction is given below:

4NH3(g) + 3O2(g) → 2N2+ 6H2O(g)

Next, we shall determine the masses of NH3 and O2 that reacted and the masses of N2 and H2O produced from the balanced equation.

This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96 g

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 from the balanced equation = 2 x 28 = 56 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 6 x 18 = 108 g

Summary:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2 to produce 56 g of N2 and 108 g of H2O.

A. Determination of the mass of O2 needed to react with 13.6 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2.

Therefore, 13.6 g of NH3 will react with = (13.6 x 96)/68 = 19.2 g of O2.

Therefore, 19.2 g of O2 are needed for the reaction.

B. Determination of the mass of N2 produced when 6.50 g of O2 react.

This is illustrated below:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Therefore, 6.5 g of O2 will react to produce = (6.5 x 56)/96 = 3.79 g of N2.

Therefore, 3.79 g of N2 were produced from the reaction.

C. Determination of the mass of H2O formed from the reaction of 34 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted to 108 g of H2O.

Therefore, 34 g of NH3 will react to produce = (34 x 108)/68 = 54 g of H2O.

Therefore, 54 g of H2O were obtained from the reaction.

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3 years ago