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alexira [117]
3 years ago
10

A student designed an experiment to test the effect on the magnitude of the magnetic field that is generated around the wire loo

p. The student increased the current that passes through the wire and tests the magnitude of the magnetic force that is generated. The experimental setup is shown here. Which statements accurately describe this experimental design? Select ALL that apply. A) The student's dependent variable is the size of the wire loop. B) The student's independent variable is the size of the wire loop. C) The student's dependent variable is the magnitude of the magnetic field that is generated. D) The student's independent variable is the magnitude of the magnetic field that is generated. E) The student's independent variable is the amount of current that is being passed through the wire.
Chemistry
1 answer:
White raven [17]3 years ago
5 0

Answer:

C

Explanation:

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An electrochemical cell has the following standard cell notation: Al(s) | Al3+(aq) || Mg2+(aq) | Mg(s)
Soloha48 [4]

Answer:

a. Al(s) ⇄ Al⁺³(aq) + 3e⁻ (oxidation)

Mg²⁺(aq) + 2e⁻ ⇄ Mg(s) (reduction)

b. ΔE° = + 0.715 V

c.  It's an electrolytic cell, because it's a nonspontaneous reaction.

d. 2Al(s) + 3Mg²⁺(aq) ⇄ 2Al⁺³(aq) + 3Mg(s)

Explanation:

a. By the notation given, first is represented the oxidation reaction and then the reduction reaction, so they are:

Al(s) ⇄ Al⁺³(aq) + 3e⁻ (oxidation)

Mg²⁺(aq) + 2e⁻ ⇄ Mg(s) (reduction)

b. The standard potential of the cell (ΔE°) is the reduction potential of the oxidation less the reduction potential of the reduction. The reduction potentials are:

Al(s) = -1.66 V

Mg(s) = -2.375 V

ΔE° = -1.66 - (-2.375)

ΔE° = + 0.715 V

c. It's an electrolytic cell.

A galvanic cell is spontaneous, so the cathode (reduction) has a higher E° than the cathode (oxidation). In this case, the oxidation reaction has a higher E°, so the reaction is nonspontaneous and it's necessary an external force to it happen, so it's an electrolytic cell.

d. 2Al(s) + 3Mg²⁺(aq) ⇄ 2Al⁺³(aq) + 3Mg(s)

The number of electrons must be the same, so the oxidation reaction is multiplied by 2, and the reduction reaction by 3.

5 0
4 years ago
How many moles of gas are in 4.4 mL at 1.2 atm?
kirza4 [7]
The Correct Answer Is 3.2
3 0
3 years ago
Sucrose is very soluble in water. at 25◦c, 211.4 grams of sucrose will dissolve in 100 g of water. given that the density of the
stiv31 [10]
Molarity of solution is mathematically expressed as,
M = \frac{x\text{weight of solute(g)}}{\text{Molecular weight X Volume of solution(l)}}

We know that volume = mass/density
Given: mass of solution = 100 g, Density = 1.34 g/ml
∴ volume = 100/1.34 = 88.49 ml = 0.08849 l

Also, we know that molecular weight of sucrose = 342.3 g/mol
 ∴M = \frac{x\text{211.4}}{\text{342.3 X 0.08849}}
       = 6.979 M

Thus, molarity of solution is 6.979 M
8 0
3 years ago
Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s
bazaltina [42]

<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

k=\frac{0.693}{t_{1/2}}

where,

t_{1/2} = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

k=\frac{0.693}{9}=0.077s^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}     ......(1)

where,

k = rate constant  = 0.077s^{-1}

t = time taken for decay process = 50.7 sec

[A_o] = initial amount of the reactant = ?

[A] = amount left after decay process =  0.0741 M

Putting values in equation 1, we get:

0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}

[A_o]=3.67M

Now, calculating the time taken by using equation 1:

[A]=0.0055M

k=0.077s^{-1}

[A_o]=3.67M

Putting values in equation 1, we get:

0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s

Hence, the time taken by the reaction is 84.5 seconds

6 0
4 years ago
What would be the minimum energy Emin required to excite a hydrogen atom from its lowest energy level
Wewaii [24]

The minimum energy required to excite a hydrogen atom from its lowest energy level is 10.2 eV.

<h3>What is excitation?</h3>

The term excitation has to do with the promotion of an electron from a lower to a higher energy level.

In this case, we are dealing with the hydrogen atom having only one electron. Thus, the minimum energy required to excite a hydrogen atom from its lowest energy level is 10.2 eV.

Learn more about energy level:brainly.com/question/17396431

#SPJ1

3 0
2 years ago
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