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serious [3.7K]
3 years ago
7

Using the periodic table, elaborate the difference between the number of electrons in an oxygen-16 atom and an oxygen-18 atom.

Chemistry
2 answers:
tensa zangetsu [6.8K]3 years ago
6 0

Answer:

the answer is d because the The number of electrons are the same for all isotopes of an element. Hope this helped :)

Explanation:

Butoxors [25]3 years ago
5 0
The correct answer here is D
Have a nice day :)
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Determine the temperature of 1.42 mol of a gas contained in a 3.00-L container at a pressure of 123 kPa.
Schach [20]

Answer:

31.2K

Explanation:

The following data were obtained from the question:

n = 1.42 mol

V = 3L

P = 123 kPa = 123000Pa

Recall: 101325Pa = 1atm

123000Pa = 123000/101325 = 1.21atm

R = 0.082atm.L/Kmol

T =?

Using the ideal gas equation PV = nRT, we can obtain the temperature as follows:

PV = nRT

T = PV/nR

T = (1.21 x 3)/(1.42 x 0.082)

T = 31.2K

8 0
4 years ago
Calculate Delta G for each reaction using Delta Gf values: answer kJ ...thank you
Leni [432]

Answer:

a) \Delta G=2.6kJ

b) \Delta G=-979.57kJ

c) \Delta G=264.21kJ

Explanation:

Hello,

In this case, in each reaction we must subtract the Gibbs free energy of formation the reactants to the Gibbs free energy of formation of the products considering each species stoichiometric coefficients. In such a way, the Gibbs free energy of formations are:

\Delta _fG_{H_2}=\Delta _fG_{I_2}=0kJ/mol\\\Delta _fG_{HI}=1.3kJ/mol\\\Delta _fG_{CO_2}=-394.4kJ/mol\\\Delta _fG_{CO}=-137.3 kJ/mol\\\Delta _fG_{NH_3}=16.7 kJ/mol\\\Delta _fG_{HCl}=-95.3kJ/mol\\\Delta _fG_{MnO_2}=465.37kJ/mol\\\Delta _fG_{Mn}=0kJ/mol\\\Delta _fG_{NH_4Cl}=-342.81kJ/mol

So we proceed as follows:

a)

\Delta G=2\Delta _fG_{HI}-\Delta _fG_{H_2}-\Delta _fG_{I_2}\\\\\Delta G=2*1.3\\\\\Delta G=2.6kJ

b)

\Delta G=\Delta _fG_{Mn}+2*\Delta _fG_{CO_2}-\Delta _fG_{MnO_2}-2*\Delta _fG_{CO}\\\\\Delta G=0+2*-394.4-465.37-2*-137.3\\\\\Delta G=-979.57kJ

c)

\Delta G=\Delta _fG_{NH_3}+\Delta _fG_{HCl}-\Delta _fG_{NH_4Cl}\\\\\Delta G=16.7-95.3-(-342.81)\\\\\Delta G=264.21kJ

Regards.

6 0
3 years ago
Which of the following is not true? a. The freezing point of sea water is lower than the freezing point of pure water. b. The bo
Anna007 [38]

Answer:

The false statement is b.

The boiling point of a water sample from the Salt Lake is lower than the boiling point of pure water

Explanation:

This excersise refers to colligative properties.

Boiling point refers to the property of boiling point elevation. In a solution of ions, as water sea or water from lakes, the boiling point will be higher than water pure.

It is logical to say that, because these sort of water have ions. Remember that colligative properties depends on the solute particles.

T° boiling solution - T° pure solvent = Kb . m . i

In the freezing point depression, we have the oppossite of boiling point elevation. Freezing point of solution is lower than pure solvent, according to this:

T° freezing pure solvent - T° freezing solution = Kc . m . i

We have to always consider the i, which means Van't Hoff factor, number of ions dissolved in solution. As the i is higher, the freezing point of solution will be lower, and the boiling point of solution will be higher.

They are true statement a and c.

In solution of., [NaCl] = 0.1M, as it is a higher concentration, the molality is also higher than a solution of [NaCl] = 0.05.

As water sample form Salt Lake has certain ions, it is logical to say that the boiling point of this water is higher than boiling point of pure water.

5 0
3 years ago
M[oknjjom[pmnj <br> word to my mother y'all look
Lapatulllka [165]

Answer:

what?

Explanation:

8 0
3 years ago
An electrochemical cell is powered by the half reactions shown below.
andrezito [222]
Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation. 

To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign. 

Pb(s) --> Pb2+ +2e-      E0 = +0.13 V
Ag+ + e-  ---> Ag           E0 = +0.80 V

Adding up the E0's would yield an overall electric cell potential of +0.93 V.
7 0
3 years ago
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