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Y_Kistochka [10]
3 years ago
9

Question 4 (4 points)

Chemistry
1 answer:
Fiesta28 [93]3 years ago
7 0
I don’t know what is your problem
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Give the maximum number of electrons in an atom that can have these quantum numbers:\
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Please add a picture of what numbers ; )
4 0
3 years ago
When two clear, colourless liquids are mixed together, a yellow precipitate is formed. This is an example of a
Flura [38]

Answer:

Combining the two clear colorless liquids is a chemical change.

Explanation:

Because a different solid substance is formed. Tell students that a precipitate is an insoluble solid that forms when two solutions are combined and react chemically. Insoluble means that the solid will not dissolve.

3 0
3 years ago
Using the following reaction:
Romashka-Z-Leto [24]

Answer:

Mn is the oxidizing agent.

N is the reducing agent.

Explanation:

Hello!

In this case, according to the undergoing chemical reaction, it is seen that the manganese in KMnO4 has an oxidation state of 7+, in MnSO4 of 2+ and nitrogen in KNO2 is 3+ and in KNO3 is 5+; thus we have the following half-reactions:

Mn^{7+}+5e^-\rightarrow Mn^{2+}\\\\N^{3+}\rightarrow N^{5+}+2e^-

Thus, since manganese is undergoing a decrease in the oxidation state, we infer it is the oxidizing agent whereas nitrogen, undergoing an increase in the oxidation state is the reducing agent.

Best regards!

5 0
3 years ago
Kc for the reaction N2O4 <=> 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, w
Nadya [2.5K]

Answer:

p(N2O4) = 0.318 atm

p(NO2) = 7.17 atm

Explanation:

Step 1: Data given

Kc = 0.619

Temperature = 45.0 °C

Mass of N2O4 = 50.0 grams

Volume = 2.10 L

Molar mass N2O4 = 92.01 g/mol

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Calculate moles N2O4

Moles N2O4 = 50.0 grams / 92.01 g/mol

Moles N2O4 = 0.543 moles

Step 4: The initial concentration

[N2O4] = 0.543 moles/2.10 L = 0.259 M

[NO2]= 0 M

Step 5: Calculate concentration at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

[N2O4] = (0.259 -x)M

[NO2]= 2x

Step 6: Calculate Kc

Kc = 0.619=  [NO2]² / [N2O4]

0.619 = (2x)² / (0.259-x)

0.619 = 4x² / (0.259 -x)

x = 0.1373  

Step 7: Calculate concentrations

[N2O4] = (0.259 -x)M = 0.1217 M

[NO2]= 2x = 0.2746 M

Step 8: The moles

Moles = molarity * volume

Moles N2O4 = 0.1217 M * 2.10  = 0.0256 moles

Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

Step 9: Calculate partial pressure

p*V = n*R*T

⇒ with p = the partial pressure

⇒ with V = the volume = 2.10 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 45 °C = 318 K

p = (nRT)/V

p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10

p(N2O4) = 0.318 atm

p(NO2) = (0.577 *0.08206 * 318)/ 2.10

p(NO2) = 7.17 atm

6 0
3 years ago
An ice cube is placed in 80 mL of 70oC water. When the ice melts the temperature of the water is 60oC. a. How much energy was ne
tigry1 [53]

Answer:

800 degrees C g, I found this answer by using the formula to find how much energy was need to melt the ice. First you figure out the temperature change of ur ice (starting temp and ur final temp), and then you times that with the mass of the water.

8 0
3 years ago
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