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Alex777 [14]
3 years ago
13

Type the correct answer in the box. Express your answer to three significant figures.

Chemistry
1 answer:
satela [25.4K]3 years ago
5 0

Answer:

The partial pressure of argon in the jar is 0.944 kilopascal.

Explanation:

Step 1: Data given

Volume of the jar of air = 25.0 L

Number of moles argon = 0.0104 moles

Temperature = 273 K

Step 2: Calculate the pressure of argon with the ideal gas law

p*V = nRT

p = (nRT)/V

⇒ with n = the number of moles of argon = 0.0104 moles

⇒ with R = the gas constant = 0.0821 L*atm/mol*K

⇒ with T = the temperature = 273 K

⇒ with V = the volume of the jar = 25.0 L

p = (0.0104 * 0.0821 * 273)/25.0

p = 0.00932 atm

1 atm =101.3 kPa

0.00932 atm = 101.3 * 0.00932 = 0.944 kPa

The partial pressure of argon in the jar is 0.944 kilopascal.

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In the reaction below, 3.46 atm each of H2 and Cl2 were placed into a 1.00 L flask and allowed to react: H2(g) + Cl2(g) <=&gt
Novosadov [1.4K]

Answer:

The equilibrium pressure of HCl is 5.40 atm

Explanation:

Kc = [HCl]^2/[H2][Cl2]

Initial pressure of H2 = 3.46 atm and Cl2 = 3.46 atm

Let the equilibrium pressure of HCl be y

From the equation of reaction, mole ratio of H2 to HCl is 1:2, equilibrium pressure of H2 is (3.46 - 0.5y)

Also, from the y of reaction, mole ratio of Cl2 to HCl is 1:2, equilibrium pressure of Cl2 is (3.46 - 0.5y)

Kc = 50.1

50.1 = y^2/(3.46 -y)(3.46 - y)

50.1(11.9716 - 3.46y + 0.25y^2) = y^2

12.525y^2 - 173.346y + 599.77716 = y^2

11.525y^2 - 173.346y + 599.77716 = 0

y^2 - 15.04y + 52.04 = 0

The value of y is obtained using the quadratic formula

y = [15.04 - (sqrt (-15.04^2 - 4×1×52.04))] ÷ 2(1) = (15.04 - 4.25) ÷ 2 = 10.79 ÷ 2 = 5.40 atm (to 2 decimal places)

6 0
3 years ago
10 points) Given the following two half-reactions, write the overall reaction in the direction in which it is product-favored, a
IceJOKER [234]

Answer:

The over all reaction :

2Fe^{3+}(aq)+Pb(s)\rightarrow 2Fe^{2+}(s)+Pb^{2+}(aq)  

The standard cell potential of the reaction is 0,.897 Volts.

Explanation:

Reduction at cathode :

Fe^{3+}(aq)+e^-\rightarrow Fe^{2+}(s)..[1]  

E^o_{Fe^{3+}/Fe^{2+}}=0.771 V

Reduction potential of Fe^{3+} to Fe^{2+}=0.771 V

Oxidation at anode:

Pb(s)\rightarrow Pb^{2+}(aq)+2e^-.[2]

E^o_{Pb^{2+}/Pb}=-0.126 V

Reduction potential of Pb^{3+} to Pb=-0.126 V

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}

Putting values in above equation, we get:

E^o_{cell}=E^o_{Fe^{3+}/Fe^{2+}}-E^o_{Pb^{2+}/Pb}

=0.771 V-(-0.126 )=0.897 V

The over all reaction : 2 × [1] + [2]

2Fe^{3+}(aq)+Pb(s)\rightarrow 2Fe^{2+}(s)+Pb^{2+}(aq)  

The standard cell potential of the reaction is 0,.897 Volts.

7 0
3 years ago
If the density of a set of keys is 8 g/cm', what is the volume if its mass is 54 g?
Savatey [412]

Answer:

the volume of a set of keys is 6.75ml

3 0
3 years ago
Colour of crushed chalk + Iodine solution​
Rama09 [41]
I guess if you mix them together, the chalk would get less harder.
5 0
3 years ago
1) A sample of copper has a mass of 4.75 g and a volume
lianna [129]

Answer:

<h2>8.92 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 4.75 g

volume = 0.5325 cm³

We have

density =  \frac{4.75}{0.5325}  \\  = 8.920187

We have the final answer as

<h3>8.92 g/cm³</h3>

Hope this helps you

5 0
3 years ago
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