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kondor19780726 [428]
3 years ago
13

The question is below

Chemistry
1 answer:
rodikova [14]3 years ago
3 0
The anode is the electrode where the oxidation occurs.

Cathode is the electrode where the reducction occurs.

Equations:

  Mn(2+) + 2e-  --->  Mn(s)                      Eo = - 1.18 V
2Fe(3+) + 2e-  ----> 2 Fe(2+)                2Eo = + 1.54 V

The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.

Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72

Answer: 2.72 V

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Answer:

2

Explanation:

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3 years ago
An ionized helium atom has a mass of 6.6 × 10-27 kg and a speed of 5.3 × 105 m/s. It moves perpendicular to a 0.78-T magnetic fi
arsen [322]

Explanation:

In a magnetic field, the radius of the charged particle is as follows.

             r = \frac{mv}{qB}

where,   m = mass,      v = velocity

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Therefore, q will be calculated as follows.

         q = \frac{mv}{rB}

            = \frac{6.6 \times 10^{-27} \times 5.3 \times 10^{5}}{0.014 m \times 0.78 T}

            = \frac{34.98 \times 10^{-22}}{0.01092}

            = 3.2 \times 10^{-19} \times \frac{1.0 e}{1.6 \times 10^{-19}C}

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Thus, we can conclude that the charge of the ionized atom is +2e.

5 0
3 years ago
Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of
Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

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

a) moles of H_2

\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

b) moles of I_2

\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles

H_2(g)+I_2(g)\rightarrow 2HI(g)

According to stoichiometry :

1 mole of I_2 require 1 mole of H_2

Thus 0.0787 moles of l_2 require=\frac{1}{1}\times 0.0787=0.0787moles of H_2

Thus l_2 is the limiting reagent as it limits the formation of product and H_2 acts as the excess reagent. (10.0-0.0787)= 9.92 moles of H_2are left unreacted.

Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

Thus 19.9 g of H_2 remains unreacted.

5 0
3 years ago
Which activity might help to increase the validity of this experiment?
MArishka [77]
A should be the answer because the more you test an experiment the more data you have to rely on changing the experiment would cause you to have different outcomes making the results different and unreliable so B, C, and D is not going to be the answer Hope this helps
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What does the sun earth and moon have in common
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There both in our solar system
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