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lara [203]
2 years ago
15

In the chemical stockroom, there is a 0.500 L bottle filled with 2.30 M CuSO4 solution. How many moles of CuSO4 are in the bottl

e?
A. 0.217 moles
B. 4.60 moles
C. 01.15 moles
D. 1.80 moles
Chemistry
1 answer:
Maru [420]2 years ago
7 0
Mfffffffffffffff the answer is b
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Need the names for these asap please ​
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Answer:

3_ethylpentane

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Calculate the mass of Iron in 40g of Fe203​
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Answer:

   ‏‏‎ ‎

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7 0
3 years ago
Suppose a gas mixture used for anesthesia contains 4.60 mol oxygen (O₂) and 6.00 mol nitrous oxide (N₂O). The total pressure of
julia-pushkina [17]

Considering the definition of mole fraction, the mole fraction of O₂ in the mixture is 0.434.

<h3>Definition of mole fraction</h3>

The molar fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution.

In other words, the mole fraction expresses the concentration of solute in a solution as the ratio of moles of substance to total moles of solution:

mole fraction=\frac{moles of substance}{moles of solution}

<h3>Mole fraction of O₂ in this mixture</h3>

In this case, you know a gas mixture used for anesthesia contains 4.60 mol oxygen (O₂) and 6.00 mol nitrous oxide (N₂O).

So, the total moles of the solution can be calculated as:

Total moles = moles of oxygen (O₂) + moles of nitrous oxide (N₂O)

Then:

Total moles= 4.60 moles + 6 moles

Total moles= 10.60 moles

Finally, the more fraction of O₂ can be calculated as follow:

Mole fraction of O_{2} =\frac{moles of O_{2}}{total moles}

Mole fraction of O_{2} =\frac{4.60 moles}{10.6o moles}

Solving:

<u><em>Mole fraction O₂ = 0.434</em></u>

Finally, the mole fraction of O₂ in the mixture is 0.434.

Learn more about mole fraction:

brainly.com/question/14434096

brainly.com/question/10095502

#SPJ1

3 0
2 years ago
You weigh out 0.1183 g of a complex salt to analyze for the percentage of cyanide ion in your complex salt. After dissolving the
Stels [109]

Answer:

25.35%

Explanation:

Again let me restate the the equation of the reaction;

H2O (ℓ) + 2 MnO4 - (aq) + 3 CN- (aq) → 2 MnO2 (s) + 3 CNO- (aq) + 2 OH- (aq)

Amount of potassium permanganate  reacted = 10.2/1000 * 0.08035 = 8.1957 * 10^-4 moles

If 2 moles of MnO4 - reacts with 3 moles of CN-

8.1957 * 10^-4 moles of MnO4 - reacts with 8.1957 * 10^-4 * 3/2

= 1.229 * 10^-3 moles of CN-

Mass of CN- reacted = 1.229 * 10^-3 moles of CN- * 26.02 g/mol

= 0.03 g

Hence, percentage of the cyanide = 0.03 g/0.1183 g * 100

= 25.35%

8 0
3 years ago
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