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lesya [120]
3 years ago
5

Look at the diagram. Which shows the correct arrangement of electrons in water?

Chemistry
2 answers:
mariarad [96]3 years ago
6 0

Answer:

Option 4

Explanation:

In water molecule, oxygen has -2 charge on it while the hydrogen has +1 charge and one atom of oxygen combines with two hydrogen atoms. A hydrogen need only two electrons to complete its octet (one electron is its own the second electron is shared by oxygen) and a oxygen needs 8 electrons (It has six of its own and two are shared by two hydrogen atom).

Hence, option 4 is correct

Ad libitum [116K]3 years ago
4 0

Answer:

Option 2

It's more balanced than the others

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aksik [14]
I don’t even know I’m just answering for pints
7 0
3 years ago
Which equation is set up correctly to determine the volume of a 1.5 mole sample of oxygen gas at 22°C and 100 kPa?
LenKa [72]

The volume could be calculated by using  <span>V = n RT / P </span>

In which V = Volume
n = number of Moles
R= The Gas constant
T = Temperature (ideally this would be in Kelvin, but i don't see it in the option)
P = Pressure


I believe the answer is

 <span>V = (1.5mol) (0.08205 L*kPa/K*mol) (22Celsius)/100 kPa 


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6 0
3 years ago
What is the composition of 'Clean dry air'<br><br> - Idk what subject its just science
Luba_88 [7]
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7 0
3 years ago
An ore is to be analyzed for its iron content by an oxidation-reduction titration with permanganate ion. A 4.230 g sample of the
Leya [2.2K]

Answer:

See explanation

Explanation:

The balanced redox reaction equation is;

8H+ + MnO4^- + 5Fe2+ ---------> Mn2+ + 5Fe3+ + 4H2O

Amount of KMnO4 reacted = 31.60/1000 * 0.05120 = 1.62 * 10^-3 moles

From the reaction equation;

1 mole of MnO4^- reacted with 5 moles of Fe2+

1.62 * 10^-3 moles will react with 1.62 * 10^-3 moles * 5/1 = 8.1 * 10^-3 moles

Mass of Fe2+ reacted = 8.1 * 10^-3 moles  *  56 g/mol

Mass of Fe2+ reacted = 0.45 g

Amount of iron in the sample =  0.45 g

Percentage of iron in the sample;

0.45 g/4.230 g  * 100 = 10.6 %

8 0
3 years ago
Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ∆H° for the r
Kazeer [188]

Answer:

∆H° rxn = - 93 kJ

Explanation:

Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds  broken minus bonds formed (H according to Hess Law.

We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.

              N₂ (g)   +            3H₂ (g)   ⇒                          2NH₃ (g)

1 N≡N = 1(945 kJ/mol)     3 H-H = 3 (432 kJ/mol)       6 N-H = 6 ( 389 kJ/mol)

∆H° rxn = ∑  H bonds broken  - ∑ H bonds formed

∆H° rxn = [ 1(945 kJ)   + 3 (432 kJ) ] - [ 6 (389 k J]

∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ

be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond,  N≡N,  we have to use this one .

8 0
3 years ago
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