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Triss
27 days ago
10

How many molecules of ethane (C2H6) are present in 0.334 g of C2H6?

Chemistry
1 answer:
castortr0y [2.8K]27 days ago
4 0

Response:

6.6253*10²¹ molecules of ethane (C₂H₆) can be found in 0.334 g of C₂H₆.

Explanation:

Avogadro's Number signifies the quantity of particles within a certain amount of a substance, typically represented as one mole. It is defined as 6.023 * 10²³ particles per mole and applies universally across substances.

In this case:

  • C: 12 g/mole
  • H: 1 g/mole

Thus, the molar mass for ethane C₂H₆ is:

C₂H₆: 2*12 g/mole + 6*1 g/mole= 30 g/mole

Next, utilizing a proportional calculation: if 30 grams of C₂H₆ represent 1 mole, we can determine the moles in 0.334 grams of C₂H₆.

moles of C_{2} H_{6} =\frac{0.334 grams*1 mole}{30 grams}

moles of C₂H₆=0.011

Finally, observing Avogadro's number: if one mole holds 6.023 * 10²³ molecules of C₂H₆, how many are present in 0.011 moles?

molecules of C_{2} H_{6} =\frac{0.011 moles*6.023*10^{23}molecules }{1 mole}

molecules of C₂H₆= 6.6253*10²¹

Thus, 6.6253*10²¹ molecules of ethane (C₂H₆) are found in 0.334 g of C₂H₆.

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Answer:

D

Reasoning:

This is due to the belief that at one time, the galaxy was condensed into a minute spot and the Big Bang event initiated the universe's expansion, propelling galaxies outward into space, distancing them from one another. Dark energy is theorized to exert the force from the Big Bang that drives this expansion.

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17 days ago
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Why are salt and sugar both able to dissolve in water, even though the solutes have different types of chemical bonding?
castortr0y [2808]

Response: Water is a polar substance, facilitating the dissolution of ionic compounds due to the principle that similar types mix.

Ionic interactions occur between salt and water

Sugar contains hydroxyl groups that can form hydrogen bonds with water molecules.

[Hydrogen bond: this refers to the attraction between a hydrogen atom bonded to a highly electronegative atom (like F, O, or N) and another highly electronegative atom (F, O, or N)]

Thus, due to the presence of hydrogen bonds, sugar dissolves in water.

Clarification: Water molecules are polar, exhibiting partial positive charges on the hydrogen atoms and a partial negative charge on the oxygen atom. This allows them to interact with ionic compounds such as salt (NaCl). These interactions occur through the partial charges on water, which attract opposite charges. When dissolved in water, NaCl dissociates into sodium and chloride ions; sodium ions are surrounded by negatively charged oxygen from water, while chloride ions are surrounded by positively charged hydrogens from water. As a result, salt dissolves in water.

Sugar, being a covalent compound, has bonds where electrons are shared unevenly, creating slight positive and negative charges. This characteristic allows sugar to interact with the polar ends of water, facilitating its dissolution. Therefore, it can be stated that sugar dissolves in water due to both substances being polar.

In summary, water is capable of dissolving most polar or ionic substances, as seen with sugar and salt.

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1 month ago
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The final overall chemical equation is Upper Ca upper O (s) plus upper C upper O subscript 2 (g) right arrow upper C a upper C u
lions [2696]

Answer:

The enthalpy of the second intermediate equation is altered by halving its value and changing the sign.

Explanation:

Let's examine both the first and second intermediate reactions alongside the overall equation concerning the examined process;

First reaction;

Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ

Second reaction;

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

Thus, the overall reaction becomes;

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH =?

According to Hess's law, which states that the total heat change in a reaction is equal to the sum of the heat changes for each step, we cannot simply sum the enthalpies for this overall reaction. Instead, we obtain the overall enthalpy by halving the second intermediate reaction's enthalpy and changing its sign before adding, as illustrated below;

Enthalpy of Intermediate reaction 1 + ½(-Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction

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1 month ago
The concentration of Si in an Fe-Si alloy is 0.25 wt%. What is the concentration in kilograms of Si per cubic meter of alloy?
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Answer: The mass of Si in kilograms is, 19.55kg/m^3

Explanation:

Given that the Si concentration in an Fe-Si alloy is 0.25 weight percent, this translates to:

Mass of Si = 0.25 g = 0.00025 kg

Mass of Fe = 100 - 0.25 = 99.75 g = 0.09975 kg

Density of Si = 2.32g/cm^3=2.32\times 10^6g/m^3

Density of Fe = 7.87g/cm^3=7.87\times 10^6g/m^3

Next, we need to find the quantity of Si in kilograms per cubic meter of alloy.

Si concentration in kilograms = \frac{\text{Weight of Si in 100 g of alloy}}{\text{Volume of 100 g of alloy}}

Si concentration in kilograms = \frac{\text{Weight of Si in 100 g of alloy}}{\frac{\text{Wight of Fe}}{\text{Density of Fe}}+\frac{\text{Wight of Si}}{\text{Density of Si}}}

By substituting all the provided values into this formula, we arrive at:

Si concentration in kilograms = \frac{0.00025kg}{\frac{99.75g}{7.87\times 10^6g/m^3}+\frac{0.25g}{2.23\times 10^6g/m^3}}

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Calculate the molarity of 48.0 mL of 6.00 M H2SO4 diluted to 0.250 L
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Answer:

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Molality is calculated by dividing the number of moles of solute by the kilograms of solvent, which in this case is water.

Calculate moles of H₂SO₄ from molarity:

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Therefore, molality is:

m = moles/solvent mass = 0.288 moles / 0.250 kg = 1.15 m

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