Which of the following elements may occur in the greatest number of different oxidation states?(A) C
(B) F
(C) O
(D) Ca
(E) Na

1. Option C, unsaturated compounds are those which have multiple bonds (double, triple bonds between carbon atoms). Among the given, only C has double bond between Carbon atoms hence unsaturated compound.

2. Option A, there is only one pair of electrons between nitrogen atoms . One single covalent bond between two Nitrogen atoms by two electrons or one electron pair.

Answer:

2.34 %

Explanation:

Since the density of the Jello, ρ = 1.14 g/mL and ρ = m/v where m = mass of jello and v = volume of jello = 475 mL.

So, m = ρv

substituting the values of the variables into the equation, we have

m = ρv

m = 1.14 g/mL × 475 mL = 541.5 g

Since we have 13 g of sugar in the jello, the total mass present is 13 g + 541.5 g = 554.5 g

So, the percentage by mass of sugar present % m/m = mass of sugar present/total mass × 100 %

= 13 g/554.5 g × 100 %

= 0.0234 × 100 %

= 2.34 %

So, the percentage by mass of sugar present % m/m = 2.34 %

The mass of water that will be produced if 10.54 g of H₂ react with 95.10 g of O₂ is 94.86 g

Balanced equation

2H₂ + O₂ —> 2H₂O

Molar mass of H₂ = 2 × 1 = 2 g/mol

Mass of H₂ from the balanced equation = 2 × 2 = 4 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ from the balanced equation = 1 × 32 = 32 g

Molar mass of H₂O = (2×1) + 16 = 18 g/mol

Mass of H₂O from the balanced equation = 2 ×18 = 36 g

SUMMARY

From the balanced equation above,

4 g of H₂ reacted with 32 g of O₂ to produce 36 g of H₂O

How to determine the limiting reactant

From the balanced equation above,

4 g of H₂ reacted with 32 g of O₂

Therefore,

10.54 g of H₂ will react with = (10.54 × 32) / 4 = 84.32 g of O₂

From the calculation made above, we can see that only 84.32 g out of 95.10 g of O₂ given, is needed to react completely with 10.54 g of H₂.

Therefore, H₂ is the limiting reactant.

How to determine the mass of water produced

From the balanced equation above,

4 g of H₂ reacted to produce 36 g of H₂O

Therefore,

10.54 g of H₂ will react to produce = (10.54 × 36) / 4 = 94.86 g of H₂O

Thus, 94.86 g of H₂O were obtained from the reaction.

Learn more about stoichiometry:

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Answer:

3.6*105g CH4

Explanation:

The detailed solution is shown on the image. From the detailed chemical reaction, we calculate the mass of methane required in the standard combustion reaction of methane and use it to estimate the mass required to produce the quantity of heat required in the question.

Final answer:

To supply the energy needed for an average home in Colorado, approximately 3.58 x 10^8 grams of natural gas (methane) must be burned.

Explanation:

To calculate the amount of natural gas (methane) required to supply the energy needed for an average home in Colorado, we can use the balanced chemical equation for the combustion of methane:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

The molar enthalpy change for this reaction (ΔH°rxn) is -890.4 kJ/mol. First, we need to convert the energy requirement from kilojoules to joules:

2.00 x 107 kJ = 2.00 x 1010 J

Next, we can use the molar enthalpy change to calculate the number of moles of methane required:

moles of CH4 = energy requirement / mlar enthalpy change

moles of CH4 = 2.00 x 1010 J / (-890.4 kJ/mol)

Finally, we can convert the number of moles of methane to grams using the molar mass of methane:

molar mass of CH4 = 12.01 g/mol + 4(1.008 g/mol) = 16.04 g/mol

grams of CH4 = moles of CH4 * molar mass of CH4

Therefore, the number of grams of natural gas (methane) that must be burned to supply this energy is approximately:

grams of CH4 = (2.00 x 1010 J / (-890.4 kJ/mol)) * 16.04 g/mol

Simplifying the calculation:

grams of CH4 ≈ 3.58 x 108 g

Learn more about Amount of natural gas required here:

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C) Chemical reaction!

Hope this helps!

Answer : The moles of products and are, 4.50 and 9 moles.

Explanation : Given,

Mass of water = 5.2 g

Molar mass of water = 18 g/mole

Molar mass of = 32 g/mole

The balanced chemical reaction will be,

First we have to calculate the moles of KOH.

Now we have to calculate the limiting and excess reactant.

From the balanced reaction we conclude that

As, 1 mole of react with 2 mole of

So, 4.50 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of products and .

From the balanced chemical reaction, we conclude that

As, 1 moles of react to give 1 moles of

So, 4.50 moles of react to give 4.50 moles of

and,

As, 1 moles of react to give 2 moles of

So, 4.50 moles of react to give moles of

Therefore, the moles of products and are, 4.50 and 9 moles.