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3 years ago

Calculate the number of moles of O2 gas held in a sealed 2.00 L tank at 3.50 atm and 25 ℃.

Chemistry

1 answer:

Leviafan [203]3 years ago

6 0

Answer:

$n=0.286mol$

Explanation:

Hello,

In this case, we consider oxygen as an ideal gas, for that reason, we use yhe ideal gas equation to compute the moles based on:

$PV=nRT\\\\n=\frac{PV}{RT}$

Hence, at 3.50 atm and 25 °C for a volume of 2.00 L we compute the moles considering absolute temperature in Kelvins:

$n=\frac{3.50atm*2.00L}{0.082\frac{atm*L}{mol*K}(25+273)K} \\\\n=0.286mol$

Best regards.

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What is the effect on this???

Setler79 [48]

It being transmitted

7 0

3 years ago

When 3.0 grams of H2 is reacted with excess C at constant pressure, the reaction forms CH4 and releases 53.3 kJ of heat. C(s) +

Murrr4er [49]

Answer:

THE ENTHALPY OF REACTION IN KJ/MOL OF CH4 IS 7.07 KJ/MOL.

Explanation:

Mass of H2 = 3 g

Molar mass of H2 = 2 g/mol

Heat released = 53.3 kJ

Equation of the reaction:

C(s) + 2H2(g) -------> CH4(g)

First:

Calculate the number of moles of H2 that was used:

Number of moles = mass / molar mass

Number of moles = 3g / 2g

Number of moles = 1.5 moles

So therefore, when 53.3 kJ of heat was released from the reaction, 1.5 moles of hydrogen was used.

From the equation of the reaction, one mole of carbon reacts with two moles of hydrogen to form one mole of methane.

For 3 g of hydrogen, 1.5 mole of hydrogen is involved.

It means:

1.5 moles of hydrogen reacts with 0.75 moles of carbon and produces 0.75 moles of methane. This is so because the reaction occurs in 1: 2: 1 in respect to carbon, hydrogen and methane respectively.

So we can say that the production of 0.75 mole of methane will evolve 53.3 kJ of heat.

0.75 mole of methane releases 53.3 kJ of heat.

1 mole of methane will release ( 53.3 kJ * 1 / 0.75 )

= 71.0666 kJ of heat

In conclusion, the enthalpy of the reaction in kJ/ mole of CH4 is 71.07 kJ/mol.

7 0

3 years ago

A 25-mL solution of H2SO4 is completely neutralized by 18 mL of 1.0M NaOH. What is the concentration of the H2SO4?

Charra [1.4K]

There are several information's already given in the question. Based on the information's provided, the answer can be easily deduced.
V1 = 25 ml
= 25/1000 liter
= 0.025 liter
V2 = 18 ml
= 18/1000 liter
= 0.018 liter
M2 = 1.0 M
M1 = ?
Then
M1V1 = M2V2
M1 = M2V2/V1
= (1 * 0.018)/0.025
= 0.72 M
From the above deduction, it can be easily concluded that the correct option among all the options that are given in the question is the first option or option "A". I hope that this is the answer that has actually come to your help.

5 0

3 years ago

Read 2 more answers

A chemist studies the reaction below. 2NO(g) + Cl2(g) 2NOCl(g) He performs three experiments using different concentrations and

Dmitry_Shevchenko [17]

Answer:

1. $Rate =k [NO]^{2}[Cl_{2}]$

2. $k= 0.42 \frac{L^{2}}{mol^{2}*s}$

Explanation:

$Rate =k [NO]^{m}[Cl_{2}]^{n}$

$Rate1 = k[0.4]^{m}[0.3]^{n}=0.02\\Rate 2=k [0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate1}{Rate2}=\frac{0.02}{0.08} =\frac{k[0.4]^{m}[0.3]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{1}{4} =(\frac{1}{2} )^{m},\\m=2$

$Rate3 =k [0.8]^{m}[0.6]^{n}=0.16\\Rate 2= k[0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate3}{Rate2}=\frac{0.16}{0.08} =\frac{k[0.8]^{m}[0.6]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{2}{1} =(\frac{2}{1} )^{n},\\n=1$

$Rate =k [NO]^{2}[Cl_{2}]^{1}$

$Rate =k [NO]^{2}[Cl_{2}]^{1}\\Rate 1=k [0.4]^{2}[0.3]^{1} =0.02\\k*0.16*0.3=0.02\\k=\frac{0.02}{0.16*0.3}=\frac{1}{8*(\frac{3}{10} )}=\frac{5}{12} = 0.42 \frac{L^{2}}{mol^{2}*s}$

6 0

3 years ago

The quantum numbers for the last electron placed in three elements are listed below. Which of these is(are) NOT correct? Er (4 3

UNO [17]

Answer:

The three elements Erbium, Thallium and Osmium have incorrect quantum numbers for the last electron placed.

Explanation:

The 4 quantum numbers are (n,l,ml,ms):

n (Principal quantum number): it is the number of the shell (level) where the electron is placed.
l (Angular momentum quantum number or Secondary): it represents the sublevel where the electron is placed. There are 4 subleves: s, p d and f so secondary quantum number can take the number 0 (s), 1 (p), 2 (d) or 3 (f) depending on which sublevel the electron is placed.
ml (Magnetic quantum number): it represents the spatial orientation of the electron in respect of the sublevel the electron is placed. For example: if the electron occupies the s sublevel the magnetic number will be 0, if the electron occupies the p sublevel the magnetic number could be -1,0,1, if the electron occupies the d sublevel the magnetic number could be -2,-1,0,1,2 and if the electron occupies the f sublevel the magnetic number could be -3,-2,-1,0,1,2,3. You can see this in the attachment related to the correct sublevel for the example.

ms (Spin quantum number): this number represents the possible rotation of the electron so it could be 1/2 (which is represented by an up arrow) or -1/2 (represented by an down arrow).

Let's analyze the last electron of each element. You can see the attachment for better understanding. The last electron it is represented with orange color.

- Erbium:

This element has 68 electrons so following the Moeller's Diagram to fill the the electronic configuration, we found that the last electron of Erbium it is in the 4th level (shell), in the f sublevel. As Erbium has 12 electrons in the f sublevel, it is necessary to follow the Hund's rule (electrons must be placed singly in every sublevel before place a parallel electron) to placed correctly all of them. Finally, the last electron of Erbium stays in the middle of the sublevel and it is represented by a down arrow so the correct quantum numbers in the Erbium element are (4,3,1,-1/2).

- Thallium:

This element has 81 electrons and following the Moeller's Diagram, we found that it last electron it is in the 6th level, in the p sublevel. As Thallium has 1 electron in the p sublevel, it is placed singly in the sublevel. So the last electron of Thallium it is represented by an up arrow so the correct quantum numbers in the Thallium element are (6,1,-1,1/2).

- Osmium:

Osmium has 76 electrons and following the steps that we did with we the other elements, we noticed that its last electron it is in the 5th level, in the d sublevel. Following the Hund's rule the last electron of Osmium has a magnetic quantum number of -2 and its spin quantum number is -1/2, so the quantum numbers in the Osmium element are (5,2,-2,-1/2).

Note:

- Remember that the s sublevel has place for 2 electrons, the p sublevel has place for 6 electrons, the d sublevel has place for 10 electrons and the f sublevel has place for 14 electrons.

3 0

3 years ago