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steposvetlana [31]
4 years ago
8

You constructed an electrochemical cell with lead at the cathode and zinc at the anode and measured a positive cell potential of

0.63 V. When placed in the same beaker, which combination of metal and ions will react spontaneously
Chemistry
1 answer:
anastassius [24]4 years ago
8 0

Answer:

Pb2+ + Zn = Pb + Zn2+

Explanation:

During the reaction Zinc will be oxidized by loosing two electrons while Pb will be reduced by gaining two electrons

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What year did mendeleev organized his periodic table in order of increasing atomic number
ELEN [110]

Mendeleev organized his periodic table in order of increasing atomic number in 1869.

6 0
3 years ago
Which state of matter is being described below?
ankoles [38]

Answer:

C. Gas

Explanation:

I think this is the right answer because it fits the description

8 0
2 years ago
Calculate the energy in joules and calories required to heat 25.0 g of water from 12.5C to 25.7C
Alex Ar [27]
The heat (Q) required to raise the temp of a substance is:<span>Q=m∗Cp∗ΔT</span><span> where m is the mass of the object (25.0g in this case), Cp is the specific heat capacity of the substance (for water Cp = 1.00cal/gC, or 4.18J/gC,
and Dt is the change in temp.
You'll have to solve this twice, once with the Cp in calories, and once with the Cp in joules.
</span><span>1380.72 Joules</span>
8 0
3 years ago
Show me the periodic table
yawa3891 [41]
I think this is what you wanted, so good luck!

8 0
3 years ago
An unbalanced equation is shown. In this reaction, 200.0 g of FeS2 is burned in 100.0 g of oxygen, and 55.00 g of Fe2O3 is produ
Crazy boy [7]
<span>4FeS2 + 11O2 = 2Fe2O3 + 8SO2</span>

Percent yield is calculated as the actual yield divided by the theoretical yield multiplied by 100.

Actual yield = 55 g ( 1 mol / 159.69 g ) = 0.34 mol Fe2O3

To find for the theoretical yield, we first determine the limiting reactant.

100 g O2 ( 1 mol / 32 g) = 3.13 mol O2
200 g FeS2 (1 mol / 119.98g) = 1.67 mol FeS2

Therefore, the limiting reactant is O2.

Theoretical yield = 3.13 mol O2 ( 2 mol Fe2O3 / 11 mol O2 ) = 0.57 mol Fe2O3

Percent yield = (0.34 mol / 0.57 mol) x 100 = 59.74%
3 0
3 years ago
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