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DENIUS [597]
3 years ago
11

Which of the following will reduce copper? zinc mercury fluorine chlorine

Chemistry
2 answers:
garri49 [273]3 years ago
6 0

<u>Answer:</u> The correct answer is Zinc.

<u>Explanation:</u>

Reactivity is defined as the tendency of an element to loose or gain electrons. Metals which loose electrons easily will be considered as more reactive metal.

The reactivity of elements will be judged on the basis of reactivity series. The elements which lie above in the series are more reactive than the elements which lie below in the series. The metals which lie above in the series will easily reduce the metals which lie below in the series.

Copper is the metal which lie above mercury and below zinc.

Thus, zinc is more reactive metal than copper and hence will easily reduce copper from its reaction.

Hence, the correct answer is Zinc.

laiz [17]3 years ago
4 0
Zinc because the only metals that would be able to reduce copper ions in solution would be hydrogen, lead, tin, nickel, iron, zinc, aluminum, Magnesium, sodium, calcium, potassium, and lithium. and according to your answer choices Zinc is the answer.
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What is the entropy change of the surroundings
KiRa [710]

Answer: The entropy change of the surroundings will be -17.7 J/K mol.

Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol

Amount of Acetone given = 10.8 g

Number of moles is calculated by using the formula:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of acetone = 58 g/mol

Number of moles = \frac{10.8}{58}=0.1862moles

If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then

0.1862 moles will have = \frac{32.3}{1}\times 0.1862=5.828kJ/mol

To calculate the entropy change for the system, we use the formula:

\Delta S_{sys}=\frac{\Delta H_{vap}}{T(\text{ in K)}}

Temperature = 56.2°C = (273 + 56.2)K = 329.2K

Putting values in above equation, we get

\Delta S_{sys}=\frac{5.828}{329.2}=0.0177kJ/Kmol=17.7J/Kmol   (Conversion Factor: 1 kJ = 1000J)

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\Delta S_{system}+\Delta S_{surrounding}=0

\Delta S_{surrouding}=-\Delta S_{system}=-17.7J/Kmol

5 0
3 years ago
What is the percent yield of a reaction in which 200. g of phosphorus trichloride reacts with excess water to form 91.0 g of hcl
Snowcat [4.5K]
Answer is: <span>yield of a reaction is 56,4%.
</span>Chemical reaction: PCl₃ + 3H₂O → 3HCl + H₃PO₃.
m(PCl₃) = 200 g.
m(HCl) = 91,0 g.
n(PCl₃) = m(PCl₃) ÷ M(PCl₃).
n(PCl₃) = 200 g ÷ 137,33 g/mol.
n(PCl₃) = 1,46 mol.
n(HCl) = m(HCl) ÷ M(HCl).
n(HCl) = 91 g ÷ 36,45 g/mol.
n(HCl) = 2,47 mol.
From reaction: n(PCl₃) : n(HCl) = 1 : 3.
n(HCl) = 1,46 mol · 3 = 4,38 mol.
Yield of reaction: 2,47 mol ÷ 4,38 mol · 100% = 56,4%.
4 0
3 years ago
Read 2 more answers
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