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seropon [69]
3 years ago
10

A gas sample occupies a volume of 1.264 L when the temperature is 168.0 °C and the pressure is 946.6 torr. How many molecules ar

e in sample?
Chemistry
1 answer:
alukav5142 [94]3 years ago
5 0

Answer:

0.26×10²³ molecules

Explanation:

Given data:

Volume of gas = 1.264 L

Temperature = 168°C

Pressure = 946.6 torr

Number of molecules of gas = ?

Solution:

Temperature = 168°C (168+273= 441 K)

Pressure = 946.6 torr (946.6/760 = 1.25 atm)

Now we will determine the number of moles.

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

n = PV/RT

n = 1.25  atm ×1.264 L / 0.0821 atm.L/ mol.K   ×441 K

n = 1.58 /36.21 /mol

n = 0.044 mol

Now we will calculate the number of molecules by using Avogadro number.

1 mol = 6.022×10²³ molecules

0.044 mol × 6.022×10²³ molecules/ 1mol

0.26×10²³ molecules

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Answer: 0.3794 moles of Iodine gas are produced.

Explanation:

2KI(aq)+Cl_2(
g)\rightarrow 2KCl(aq)+I_2

Volume of iodine gas produced at STP =8.5 L

At STP, the 1 mol of gas occupies volume = 22.4 L

So, 8.5 L of volume will be occupied by:\frac{1}{22.4 L}\times 8.5=0.3794 moles

0.3794 moles of Iodine gas are produced.

5 0
3 years ago
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NaCl has a Delta. Hfus = 30. 2 kJ/mol. What is the mass of a sample of NaCl that needs 732. 6 kJ of heat to melt completely? Use
tia_tia [17]

The mass of NaCl sample has been 24.3 g. Thus, option A is correct.

The heat of fusion has been the amount of heat required to convert  1 mole of substance into solid to liquid state.

The heat required has been given as:

Q=m\Delta H

<h3>Computation for the mass of NaCl</h3>

The given solution has heat of fusion, \Delta H_{fus}=30.2\rm kJ/mol

The heat required to melt the sample has been, Q=732.6\;\rm kJ/mol

Substituting the values for the mass of NaCl

\rm 732.6\;kJ/mol=Mass\;\times\;30.2\;kJ/mol\\\\&#10;Mass=\dfrac{732.6}{30.2}\;g\\\\&#10; Mass=24.3\;g\\

The mass of NaCl sample has been 24.3 g. Thus, option A is correct.

Learn more about heat of fusion, here:

brainly.com/question/87248

5 0
3 years ago
Menthol, the substance we can smell in mentholated cough drops, is composed of c, h, and o. a 9.045×10−2 −mg sample of menthol i
Ket [755]

Answer:

            Empirical Formula  =  C₁₀H₂₀O

Solution:

Data Given:

                      Mass of Menthol  =  9.045 × 10⁻² mg  =  9.045 × 10⁻⁵ g

                      Mass of CO₂  =  0.2546 mg  =  0.0002546 g

                      Mass of H₂O  =  0.1043 mg  =  0.0001043 g

Step 1: Calculate %age of Elements as;

                      %C  =  (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

                      %C  =  (0.0002546 ÷ 9.045 × 10⁻⁵) × (12 ÷ 44) × 100

                      %C  =  (2.814) × (12 ÷ 44) × 100

                      %C  =  2.814 × 0.2727 × 100

                      %C  =  76.73 %


                      %H  =  (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

                      %H  =  (0.0001043 ÷ 9.045 × 10⁻⁵) × (2.02 ÷ 18.02) × 100

                      %H  =  (1.153) × (2.02 ÷ 18.02) × 100

                      %H  =  1.153 × 0.1120 × 100

                     %H  =  12.91 %


                      %O  =  100% - (%C + %H)

                      %O  =  100% - (76.73% + 12.91%)

                      %O  =  100% - 89.64%

                     %O  =  10.36 %

Step 2: Calculate Moles of each Element;

                      Moles of C  =  %C ÷ At.Mass of C

                      Moles of C  = 76.73 ÷ 12.01

                     Moles of C  =  6.3888 mol


                      Moles of H  =  %H ÷ At.Mass of H

                      Moles of H  = 12.91 ÷ 1.01

                      Moles of H  =  12.7821 mol


                      Moles of O  =  %O ÷ At.Mass of O

                      Moles of O  = 10.36 ÷ 16.0

                      Moles of O  =  0.6475 mol

Step 3: Find out mole ratio and simplify it;

                C                                        H                                     O

            6.3888                              12.7821                            0.6475

     6.3888/0.6475                  12.7821/0.6475                 0.6475/0.6475

               9.86                                   19.74                                   1

             ≈ 10                                      ≈ 20                                     1

Result:

         Empirical Formula  =  C₁₀H₂₀O₁

8 0
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Explanation: Solution:

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The conversion factor is 1 mole of CrSO4 is equal to its molar mass which is 132 g CrSO3

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Solve question 4a least reactive to most reactive
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Nickel (II) oxide, iron (III) oxide, chromium (III) oxide, magnesium oxide
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