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Arte-miy333 [17]
1 year ago
15

A voltaic cell using Cu/Cu²⁺ and Sn/Sn²⁺ half-cells is set up at standard conditions, and each compartment has a volume of 345 m

L. The cell delivers 0.17 A for 48.0 h.(a) How many grams of Cu(s) are deposited?
Chemistry
1 answer:
Nadusha1986 [10]1 year ago
4 0

The mass of copper deposited is 9.589 g .

Given ,

A volatile cell using Cu/Cu2+ and Sn/Sn2+ half cells is set up at standard conditions ,and each compartment has a volume of 345 ml .

current = 0.17 A

time = 48 hrs =48 (60) (60) =172800 sec

we know , I = Q/t

thus , Q ,charge= It = 0.17 (17280) =29376 C

Then ,atomic mass of copper is 63 g

And valency of Cu is 2.

Thus , the equivalent mass of copper is 63/2 = 31.5

We know , 96500 coulombs of electricity produce copper = 31.5 g

29376 C of electricity produce copper =9.589 g

Hence , 9.589 g of Copper is deposited .

Learn more about charge here:

brainly.com/question/18102056

#SPJ4

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Answer:

First confirm the reaction is balanced:

C3H8 + 5O2 --> 3CO2 + 4H20 (3 cabon - check; 8 hydrogen - check; 10 oxygen - check).

a) In the equation there is a 5:1 ratio between propane and oxygen.  We also know that number of mole is proportional to pressure and volume.  Since pressure is constant (STP) then the volume of O2 is 7.2 * 5 = 36 litres.

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There is a 1:3 ratio between propane and CO2.  Therefore 3 * 0.29 = 0.87 mole of CO2 is produced.

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c) We know we need more oxygen than propane (due to the 1:5 ratio) so oxygen is the limiting reagent.  Again Volume is proportional to number of mole and we see there is a 5:4 ratio between oxygen and water.  Therefore the volume of water vapour produced will be (4/5) * 15 = 12 litres.

The other questions use the same technique and will give you some much needed practice.

Explanation:

7 0
3 years ago
If 125 g of CaCO3 is mixed and reacted with 125 g of HCl, which reactant is limiting and how many grams of CO2 can be made?
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Answer:

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55 g of CO2 is made

Explanation:

First we must put down the reaction equation;

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From the reaction equation;

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Hence 1.25 moles of CaCO3 yields 1.25 moles of CO2

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3.42 moles of HCl yields 3.42 * 1/2 = 1.71 moles of CO2

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