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grigory [225]
3 years ago
6

One gram of salt in 100 liters of water could be considered a _______________________ solution. A) concentrated B) dilute C) sat

urated D) supersaturated
Chemistry
2 answers:
-Dominant- [34]3 years ago
6 0

The correct answer is B) Dilute

Explanation:

In a solution, the proportions of the solute vs the solvent determine the type of solution that includes dilute/concentrated or unsaturated, saturated and supersaturated. In the case of a dilute solution, this occurs if there is a small amount of the solute in comparison to the amount of solvent, which makes it possible to add more solute as this can be dissolved. This type of solution applies to the case presented because one gram of salt (solute or substance dissolved) is a small quantity of solute in comparison to the amount of solvent (the substance that dissolves) that in this case is 100 liters of water. Thus, in this case, there is a dilute solution.

vitfil [10]3 years ago
3 0

Answer:I believe your answer would be option B) dilute. Hope this helps.

Explanation:

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Consider the reaction 2NO(g) 1 O2(g) ¡ 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reac
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<h2>a) The rate at which NO_2 is formed is 0.066 M/s</h2><h2>b) The rate at which molecular oxygen O_2 is reacting is 0.033 M/s</h2>

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of NO = -\frac{1d[NO]}{2dt} = 0.066 M/s

Rate in terms of disappearance of O_2 = -\frac{1d[O_2]}{dt}

Rate in terms of appearance of NO_2= \frac{1d[NO_2]}{2dt}

1. The rate of formation of NO_2

-\frac{d[NO_2]}{2dt}=\frac{1d[NO]}{2dt}

\frac{1d[NO_2]}{dt}=\frac{2}{2}\times 0.066M/s=0.066M/s

2. The rate of disappearance of O_2

-\frac{1d[O_2]}{dt}=\frac{d[NO]}{2dt}

-\frac{1d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s

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3 years ago
How much heat energy is required to raise the temperature of 0.358 kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of co
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In order to calculate how much heat is needed to raise the temperature you need to use the formula q =mass x specific heat x (final temperature- initial temperature) where q represents heat being absorbed or released. Before you begin you would convert kg to g because the specific heat is measure in g. So you would set up the equation as q = 358 g x .092 x (60-23 degrees Celsius) which would give you 1218.6
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3 years ago
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if 14.0 g of aluminium reacts with excess sulfuric acid to produce 75.26 g of aluminium sulfate, what is the percent yield?
vlada-n [284]

Taking into account definition of percent yield, the percent yield for the reaction is 84.88%.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 Al + 3 H₂SO₄ → Al₂(SO₄)₃ + 3 H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Al: 2 moles
  • H₂SO₄: 3 moles
  • Al₂(SO₄)₃. 1 mole
  • H₂: 3 moles

The molar mass of the compounds is:

  • Al: 27 g/mole
  • H₂SO₄: 98 g/mole
  • Al₂(SO₄)₃: 342 g/mole
  • H₂: 2 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Al: 2 moles ×27 g/mole= 54 grams
  • H₂SO₄: 3 moles ×98 g/mole= 294 grams
  • Al₂(SO₄)₃: 1 mole ×342 g/mole= 342 grams
  • H₂: 3 moles ×2 g/mole= 6 grams

<h3>Mass of aluminium sulfate formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 54 grams of aluminium form 342 grams of aluminium sulfate, 14 grams of aluminium form how much mass of aluminium sulfate?

mass of aluminium sulfate=\frac{14 grams of aluminium x342 grams of aluminium sulfate}{54 grams of aluminium}

<u><em>mass of aluminium sulfate= 88.67 grams</em></u>

Then, 88.67 grams of aluminium sulfate can be produced if 14.0 g of aluminium reacts with excess sulfuric acid.

<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

percent yield=\frac{actual yield}{theorical yield} x100

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Percent yield for the reaction in this case</h3>

In this case, you know:

  • actual yield= 75.26 grams
  • theorical yield= 88.67 grams

Replacing in the definition of percent yields:

percent yield=\frac{75.26 grams}{88.67 grams} x100

Solving:

<u><em>percent yield= 84.88%</em></u>

Finally, the percent yield for the reaction is 84.88%.

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The answer is electron
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The correct answer would be <span>0.22 N.

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