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timofeeve [1]
2 years ago
7

Uranium hexafluoride is a solid at room temperature, but it boils at 56C. Determine the density of uranium hexafluoride at 60.C

and 745 torr.
Chemistry
1 answer:
Mandarinka [93]2 years ago
5 0

Answer: The density of Uranium Hexafluoride at 60 °C and 745 torr is 0.0127 g/ml

Explanation:

Uranium Hexafluoride is only present as gas at 60 °C and 745 torr.

So, we know PV = nRT, from that

n/V = P/ RT = (745 torr)/(62.36 L-torr/mol*K)(337 K) = 0.0354 mol/L.

Density = (352.02 g/mol)(0.0354 mol/L)/(1000 cm³/L) = 0.0127 g/cm³

To learn more about density of Uranium Hexafluoride from the given link

brainly.com/question/17987457

#SPJ4

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Describe the temperature changes that occur in ice as energy is added, starting in the frozen state and ending in the vapor stat
Ket [755]

Explanation:

When water is frozen then it is known as ice and its state is solid. So, its molecules will be held closer to each other as they are held by strong intermolecular forces of attraction.

As a result, its temperature will be minimum as its molecules have least kinetic energy.

It is known that kinetic energy of a substance is directly proportional to temperature.

As,            K.E = \frac{3}{2}kT

where            K.E = kinetic energy

                       T = temperature

                       k = boltzmann constant

When solid changes into liquid state then it means molecules of a substance has gained kinetic energy due to which there occurs more collisions between the molecules.

Hence, temperature of substance also increases.

Whereas when liquid state of a substance changes intro vapor state then it means that more kinetic energy has gained by the molecules due to which there will be much more collisions between the molecules.

Hence, temperature will be maximum in vapor state.

8 0
3 years ago
Calculate the percent yield when 500 grams of carbon dioxide react with an excess of water to produce 640 grams of carbonic acid
Elena-2011 [213]

Answer:

Percent yield =  90.5%

Explanation:

Given data:

Mass of carbon dioxide = 500 g

Mass of water = excess

Actual yield of carbonic acid = 640 g

Percent yield = ?

Solution:

Balanced chemical equation:

CO₂ + H₂O  → H₂CO₃

Number of moles of carbon dioxide

Number of moles  = Mass / molar mass

Number of moles = 500 g/ 44 g/mol

Number of moles = 11.4 mol

Now we will compare the moles of H₂CO₃ with CO₂.

                              CO₂          :              H₂CO₃

                                 1             :                  1

                               11.4           :                11.4

Mass of carbonic acid:

Mass = number of moles × molar mass

Mass = 11.4 mol × 62.03 g/mol

Mass = 707.14 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield =  640 g/ 707.14 g × 100

Percent yield =  90.5%

7 0
4 years ago
An unknown compound has the following chemical formula:
Softa [21]

Answer:

Formula of the compound = P_2O_5

Explanation:

Given,

No. of mole of O = 8.20 mol

No. of mole of P = 3.30 mol

Chemical formula = P_2O_x

Ratio of P and O = \frac{2}{x}

\frac{2}{x} =\frac{3.3}{8.20} \\x \times 3.3 = 8.20 \times 2\\x = \frac{8.20 \times 2}{3.3} \\x=4.9969\\ = 5

formula of the compound = P_2O_5

5 0
3 years ago
________ is a measure of the amount of matter in an object.
Zepler [3.9K]
Mass i think hope and this helps u
6 0
3 years ago
Read 2 more answers
An element has five isotopes. Calculate the atomic mass of this element using the information below. Show all your work. Using t
Katena32 [7]

Answer: Sol:-

Data provided in the question is :-

Atomic mass of isotope -1 = 64 amu

Atomic mass of isotope -2 = 66 amu

Atomic mass of isotope -3 = 67 amu

Atomic mass of isotope -4 = 68 amu

Atomic mass of isotope - 5 = 70 amu

Percentage abundace of isotope - 1 = 48.89 %

Percentage abundance of isotope -2 = 27.81 %

Percentage abundance of isotope - 3 = 4.11%

Percentage abundance of isotope-4 = 18.57%

Percentage abundance of isotope - 5 = 0.62 %

Formula used :-

Average atomic mass of an element =[ {(atomic mass of isotope-1 * percentage abundance of isotope-1) + ( atomic mass of isotope-2 * percentage abundance of isotope -2) + ( atomic mass of isotope -3 * percantege abundance of isotope-3 ) + ( atomic mass of isotope-4 * percentage abundance of isotope-4) + (atomic mass of isotope-5 * percentage abundance of isotope-5)} / 100]

Calculation :-

Put all the value in the formula :-

Average atomic mass of an element = [{(64 * 48.89) + (66 * 27.81) + (67 * 4.11) + (68 * 18.57) + (70 * 0.62)} / 100] amu

= [{(3128.96) + (1835.46) +(257.37) + (1262.76) + (43.4)} / 100] amu

= {(6528.04) / 100} amu

= 65.2804 amu

Average atomic mass of an element is = 65.2804 amu

Then this mass is approximatly equal to atomic mass of zinc so this element would be zinc

atomic mass of zinc = 65.38 \approx 65.2804 amu

5 0
3 years ago
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