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LenKa [72]
3 years ago
15

50 POINTS!!!!!!!!!!!

Chemistry
2 answers:
Pani-rosa [81]3 years ago
6 0

Answer:

c

Explanation:

kifflom [539]3 years ago
5 0
A substitution is a mutation that exchanges one base for another. This can include a change in a single chemical letter. ex. changing a to g or in your case g to a. I believe the correct answer should be B.
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Fe2O3 + 2Al = 2Fe + Al2O3 is this a redox reaction
katovenus [111]

Answer:

<h3>2Al+ Fe2O3 gives 2Fe + Al2O3. The given reaction is a redox reaction. As oxidation and reduction are taking place simultaneously.</h3>

Explanation:

like this...Identify oxidation and reduction with their agents:

<h3>•2Al+ Fe2O3 →2Fe + Al2O3</h3>

<h3>•Fe2O3 is reduced to Fe whereas Al is oxidized to Al2O3</h3>

<h3>In the above reaction:</h3>

<h3>Oxidizing agent:Fe2O3</h3>

<h3>Reducing agent:Al</h3>

I hope it's help you (◠‿・)—☆

5 0
2 years ago
What is the pH of a 4.08*10-'M solution of H30+?
Mariana [72]

Answer:

8.4

Explanation:

-log(4.08x10^-9) = 8.4

- Hope that helped! Please let me know if you need further explanation.

4 0
2 years ago
Write the names and symbols for four elements in each of the following categories: (a) nonmetal, (b) metal, (c)metalloid.
Over [174]

Explanation:

Non-metals are the species that are electron deficient and they are able to accept one or more electrons from a donor atom in order to complete their octet.

For  example, carbon (C), nitrogen (N), chlorine, (Cl), phosphorus (P) etc are all non-metals.

Metals are the species that contain more number of electrons in their valence shell and in order to attain stability they easily lose an electron.

For example, sodium (Na), lithium (Li), Beryllium (Be), Magnesium (Mg) etc are all metals.

Metalloids are the species that show properties of both metals and non-metals.

For example, Boron (B), Antimony (Sb), Silicon (Si) and Germanium (Ge) etc are metalloids.

5 0
3 years ago
In Lab 10 you make a stock solution of salicylic acid, and then four dilutions. The stock solution is made by diluting 5.00 ml o
Zigmanuir [339]

Answer:

Stock  solution =  1.25 *10^-3 M

Dilution 1 = 5*10^-4 M

Dilution 2= 3.75 * 10^-4 M

Dilution 3 = 2.5 *10^-4 M

Dilution 4 = 1.25 *10^-4 M

Explanation:

<u>Step 1:</u> Data given

The stock solution is made by diluting 5.00 ml of 1.250 x 10-2 M salicylic acid in 50.00 mL of solution.

<u>Step 2</u>: Calculate the concentration of the stock solution:

M1*V1 = M2*V2

⇒ with M1 = the initial concentration = 1.250 *10^-2 M

⇒ with V1 = 5 mL = 5*10^-3 L

⇒ with M2 = TO BE DETERMINED

⇒ with V2 = 50 mL = 50 *10^-3 L

M2 = (M1*V1)/V2

M2 = (1.250 *10^-2 * 5*10^-3 L) / 50 *10^-3

M2 = 0.00125 M = 1.25 *10^-3 M

<u>Step 3:</u> Calculate dilution 1

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 *10^-3)/(25*10^-3L)

M2 = 0.0005 M = 5*10^-4 M

<u>Step 4</u>: Calculate dilution 2

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 * 7.5*10^-3)/(25*10^-3)

M2 = 0.000375 M = 3.75 * 10^-4 M

<u>Step 5:</u> Calculate dilution 3

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 5*10^-3) /(25*10^-3)

M2 = 0.00025 M = 2.5 *10^-4 M

<u>Step 6</u>: Calculate dilution 4

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 2.5*10^-3)/(25*10^-3)

M2 = 0.000125 M = 1.25 *10^-4 M

5 0
3 years ago
Calculate the standard entropy of vaporization of ethanol, C2H5OH, at 285.0 K, given that the molar heat capacity at constant pr
MArishka [77]

Answer: The standard entropy of vaporization of ethanol is 0.275 J/K

Explanation:

C_2H_5OH(l)\rightleftharpoons C_2H_5OH(g)

Using Gibbs Helmholtz equation:

\Delta G=\Delta H-T\Delta S

For a phase change, the reaction remains in equilibrium, thus \Delta G=0

\Delta H=T\Delta S

Given: Temperature = 285.0 K

\Delta H=78.3J/mol

Putting the values in the equation:

78.3J=285.0K\times \Delta S

\Delta S=0.275J/K

Thus  the standard entropy of vaporization of ethanol is 0.275 J/K

4 0
3 years ago
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