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Free_Kalibri [48]
3 years ago
6

1. Which of the following compounds is most likely to exist as a covalent molecule? O NaBr O KNO, O Bel, C O CH_NH2?​

Chemistry
1 answer:
DIA [1.3K]3 years ago
6 0

Answer:

nabeo

Explanation:

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Identify the Lewis acid in this balanced equation: SnCl4 + 2Cl− → SnCl62−
Grace [21]

Answer:

The answer is SnCI4.

Explanation:

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A picture hanging on a wall has<br> energy.
KonstantinChe [14]
It has potential energy.
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Decide which element probably formed a compound with hydrogen that has a chemical formula most and least similar to the chemical
ss7ja [257]
Tbh I really don’t know
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Solid Iron (III) reacts with oxygen gas to form iron (III) oxide, which is the compound we see when rust is present. Based on th
Sunny_sXe [5.5K]

Answer:

Option E is correct. none of the above is correct

Explanation:

Step 1: Data given

Solid Iron (III) = Fe^3+

iron (III) oxide = Fe2O3

Molar mass Fe = 55.845 g/mol

Molar mass Fe2O3 = 159.69 g/mol

Step 2: The balanced equation:

4Fe + 3O2 → 2Fe2O3

4 moles of iron will need 2 moles of oxygen gas to fully react

⇒ This is false 4 moles of iron will need 3 moles of oxygen gas to fully react

B.12 moles of iron, if reacted completely, can produce 8 moles of iron (III) oxide.

⇒ This is false: When 12 moles of iron completely react, we can produce 12/2 = 6 moles of Fe2O3

C.9 moles of oxygen can produce 9 moles of Iron (III) oxide

⇒ This is false; 9 moles of O2 can produce 6 moles of Fe2O3

D.6 moles of oxygen can react completely to produce 6 moles of iron (III) oxide.

⇒ This is false 6 moles of O2 will react completely to produce 4 moles of Fe2O3

E.none of the above

6 0
3 years ago
Suppose a student titrates a 10.00-ml aliquot of saturated ca(oh)2 solution to the equivalence point with 16.08 ml of 0.0199 m h
Alborosie
Following chemical reaction is involved upon titration of Ca(OH)2 with HCl,
Ca(OH)2 + 2HCl ↔ CaCL2 + 2H2O

Above is an example of acid-base titration to generate salt and water. Here, H+ ions of acid (HCl) combines with OH- (ions) of base [Ca(OH)2] to generated H2O

Given,
concentration of HCl = 0.0199 M
Total volume of HCl consumed during titration = 16.08 mL = 16.08 X 10^(-3) L

∴, number of moles of H+ consumed = Molarity X Vol. of HCl (in L)
                                                           = 0.0199 X 16.08 X 10^(-3)
                                                           = 3.1999 X 10^-4 mol
Thus, total number of moles of [OH-] ions present initial = 3.1999 X 10-4 mol
So, initial conc. [OH-] ion = 
\frac{number of moles of [OH-]}{volume of solution (L)} = \frac{3.1999 X 10^(^-^4^)}{10 X 10^(^-^3^)} = 0.03199 M
6 0
3 years ago
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