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Svetach [21]
3 years ago
8

If a penny (with the this volume .39 cm^3) was made out of pure copper, what should its mass be?

Chemistry
1 answer:
blondinia [14]3 years ago
5 0

Answer:

  3.49 g

Explanation:

The mass is the product of volume and density:

  (8.96 g/cm³)(0.39 cm³) ≈ 3.49 g

The mass of a pure-copper penny would be 3.49 g.

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Which aqueous solution should form a precipitate with aqueous cu(no3)2? 1. kno3 2. cuso4 3. k2so4 4. k2s?
chubhunter [2.5K]
Reaction of Cu(NO₃)₂ with each salt is as follow, 

1) with KNO₃;

                  Cu(NO₃)₂  +  KNO₃   →   Cu(NO₃)₂  +  KNO₃

Both salt products are water soluble.

2) With CuSO₄;

                  Cu(NO₃)₂  +  CuSO₄   →   CuSO₄  +  Cu(NO₃)₂
Again both Salt products are water soluble.

3) With K₂SO₄;

                  Cu(NO₃)₂  +  K₂SO₄   →  CuSO₄  +  2 KNO₃
Again both salt products are water soluble.

4) With K₂S;

                  Cu(NO₃)₂  +  K₂S   →   CuS  +  2 KNO₃
In this case CuS is water insoluble, hence precipitates out.

Result:
           Option-4 is the correct answer.
8 0
3 years ago
What do oxidation reactions produce during the electrolysis of water?
jok3333 [9.3K]
Hydrogen gas, oxygen gas and water

4H2O (l) ---> 2H2O (l) + O2 (g) + 2H2 (g)
8 0
3 years ago
Read 2 more answers
How would a solution that is labeled 5.0 M would be read as?
Ivanshal [37]
5 moles because molarity signifies number of moles dissolved in One litre of water
8 0
3 years ago
Read 2 more answers
Based on the balanced equation 2C2H2 + 5O2 → 4CO2 + 2H2O calculate the number of excess reagent units remaining when 52 C2H2 mol
erica [24]

Answer:

20 molecules of oxygen gas remains after the reaction.

Explanation:

2C_2H_2 + 5O_2\rightarrow 4CO_2 + 2H_2O

Molecules of ethyne = 52

Molecules of oxygen gas = 150

According to reaction, 2 molecules of ethyne reacts with 5 molecules of oxygen gas.

Then 52 molecules of ethyne will react with:

\frac{5}{2}\times 52=130 molecules of oxygen gas.

As we can see that we have 150 molecules of oxygen gas, but 52 molecules of ethyne will react with 130 molecules of oxygen gas. So, this means that ethyne is a limiting reagent and oxygen gas is an excessive reagent.

Remaining molecules of recessive reagent = 150 - 130 = 20

20 molecules of oxygen gas remains after the reaction.

5 0
3 years ago
You carefully weigh out 16.00 g of CaCO3 powder and add it to 64.80 g of HCl solution. You notice bubbles as a reaction takes pl
bulgar [2K]

Answer: Mass of CO_2  produced in this reaction was 6.56 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

CaCO_3(s)+2HCl(aq)\rightarrow H_2O(l)+CO_2(g)+CaCl_2(aq)

Mass or reactants =  Mass of CaCO_3+ mass of HCl = 16.00 + 64.80 = 80.80 g

Mass of products  = mass of aqueous solution + mass of CO_2 + = 74.24 + x g

Mass or reactants = Mass of products

80.80 g = 74.24 + x g

x = 6.56 g

Thus mass of CO_2  produced in this reaction was 6.56 grams

7 0
3 years ago
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