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emmainna [20.7K]
3 years ago
11

How do you convert 3.9mL to hL

Chemistry
1 answer:
Reptile [31]3 years ago
5 0

3.9 ml * 1 hl
100000 ml = 0.000039 hl
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How many moles of copper are needed to react with sulfur to produce 2.9 moles of copper (I) sulfide
dalvyx [7]

Answer:

5.8 moles of copper are needed to react with sulfur to produce 2.9 moles of copper (I) sulfide.

Explanation:

2Cu+S\rightarrow Cu_2S

Moles of copper(I) sufide = 2.9 mol

According to reaction, 1 mole of copper(I) sulfide is obtained from 2 moles of copper, then 2.9 moles of copper(I) sulfide will be obtained from :

\frac{2}{1}\times 2.9 mol=5.8 mol copper

5.8 moles of copper are needed to react with sulfur to produce 2.9 moles of copper (I) sulfide.

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Does NH3 dissolve in acetone?
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What current (in a) is required to plate out 1. 22 g of nickel from a solution of ni2 in 0. 50 hour?
Mashcka [7]

The current required to accumulate the 1.22 grams of nickel in 0.5 hours is 2.23 A.

<h3>What is current?</h3>

The current is given as the product of the charge with time. In the electrochemical analysis of the nickel, there will be a reduction of the nickel ion to nickel. The formation is given as:

\rm Ni^{2+}+2e^-\;\rightleftharpoons Ni

There is the deposition of 1 mole of Ni with 2 electrons transfer. The transfer of charge for 1 mole that is 58.7 grams Nickel is:

\rm 58.7\;g=2\;\times\;96487\;C\\58.7\;g=192974\;C

The mass of Ni to be deposited is 1.22 grams. The charge required is given as:

\rm 58.7\;grams\;Ni=192974\;C\\\\1.22\;grams\;Ni=\dfrac{192974}{58.7}\;\times\;1.22\;C\\\\1.22\;grams\;Ni=4010.7\;C

The current required to transfer 4010.7 C of charge in 1800 seconds is given as:

\rm Charge=Current\;\times\;Time\\4010.7\;C=Current\;\times\;1800\;sec\\Current=2.23\;A

Thus, the current required to accumulate the 1.22 grams of nickel in 0.5 hours is 2.23 A.

Learn more about current, here:

brainly.com/question/23063355

#SPJ4

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