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2 years ago

What would cause the percent yield of a reaction to be less than 100%?

Chemistry

1 answer:

Alexandra [31]2 years ago

8 0

The product participates in a side reaction.

Explanation:

The reason why percentage yield for a chemical reaction can be less than a 100% is due to the presence of competing side reactions. As often the case and basics for all chemical reactions, mass and matter must be conserved. But sometimes the yield can be lower due to other reactions that must be considered.

A competing side reaction that can affect the yield of products during a chemical procedure is often one in which some of the products can react with some molecules present in the environment or some of the unused reactants when available.
Sometimes, a product can be hygroscopic, deliquescent or probably effervescent. This will definitely alter the expected amount of product one desires in a reaction based on stoichiometry.

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A biology student wants to conduct a study of how the amount of food affects weight gain for mice. He sets up four cages with a

drek231 [11]

The independent variable would be the variable in the research that is being manipulated by the researcher. In this case, it would be amount of food as it is what is being manipulated and changed in the research design. The dependent variable would be the variable that is being studied so, for this case, it would be the weight gain of the mice. The constants are the factors that might affect the dependent variable but is held constant or the same by the researcher throughout the experiment. These are the size of cage, amount of water, amount of sunlight, temperature and the exercise wheel.

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2 years ago

Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O

Bond [772]

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

ΔH = -793,6 kJ

I hope it helps!

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2 years ago

An isotope undergoes radioactive decay by emitting radiation that has no mass. What other characteristic does the radiation have

Advocard [28]

Answer : Option D) No charge

Explanation : An isotope undergoes radioactive decay by emitting radiation that has no mass. The radiation will not have any charge as it does not has any mass it will not emit a radiative charge.

It is known that there are some unstable radioactive isotopes which has no mass and the radiation thus has no charge in it.

7 0

2 years ago

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Why couldn't you substitute 3m H2SO4 for concentrated HNO3 when oxidizing copper?

schepotkina [342]

This question seems to be an essay question from experiment. Different solution of oxidizing agent will have different strength. Sulfuric acid or H2SO4 is weaker oxidizing agent when compared to nitric acid (HNO3). In this case, if you subtitute the H2SO4 you wouldn't be able to get the same result for the experiment.

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2 years ago

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A bottle of concentrated aqueous sulfuric acid, labeled 98.0 wt% h2so4, has a concentration of 18.0 m. (a) how many milliliters

nadya68 [22]

n this order, Ď=1.8gmL, cm=0.5, and mole fraction = 0.9 First, let's start with wt%, which is the symbol for weight percent. 98wt% means that for every 100g of solution, 98g represent sulphuric acid, H2SO4. We know that 1dm3=1L, so H2SO4's molarity is C=nV=18.0moles1.0L=18M In order to determine sulphuric acid solution's density, we need to find its mass; H2SO4's molar mass is 98.0gmol, so 18.0moles1Lâ‹…98.0g1mole=1764g1L Since we've determined that we have 1764g of H2SO4 in 1L, we'll use the wt% to determine the mass of the solution 98.0wt%=98g.H2SO4100.0g.solution=1764gmasssolutionâ†’ masssolution=1764gâ‹…100.0g98g=1800g Therefore, 1L of 98wt% H2SO4 solution will have a density of Ď=mV=1800g1.0â‹…103mL=1.8gmL H2SO4's molality, which is defined as the number of moles of solute divided by the mass in kg of the solvent; assuming the solvent is water, this will turn out to be cm=nH2SO4masssolvent=18moles(1800â’1764)â‹…10â’3kg=0.5m Since mole fraction is defined as the number of moles of one substance divided by the total number of moles in the solution, and knowing the water's molar mass is 18gmol, we could determine that 100g.solutionâ‹…98g100gâ‹…1mole98g=1 mole H2SO4 100g.solutionâ‹…(100â’98)g100gâ‹…1mole18g=0.11 moles H2O So, H2SO4's mole fraction is molefractionH2SO4=11+0.11=0.9

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2 years ago