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steposvetlana [31]
3 years ago
7

In an electrolytic cell, to which electrode will a positive ion migrate and undergo reduction? the anode, which is negatively ch

arged the anode, which is positively charged the cathode, which is negatively charged the cathode, which is positively charged
Physics
2 answers:
ahrayia [7]3 years ago
4 0
The reaction of reduction always undergoes with the cathode, the positive ion will migrate towards the cathode with the negative charge whilst the anode always has oxidation reaction. These two types of reaction does not change.
sveticcg [70]3 years ago
3 0

Answer:

the cathode, which is negatively charged

Explanation:

In an electrolytic cell, there are two electrodes; the anode and the cathode. The anode is positively charged while the cathode is negatively charged.

Oxidation occurs at the anode as the positive ions loose electrons. The electrons lost at the anode travels via a conductor to the cathode. Positive ions migrate to the cathode, accept these electrons and become reduced.

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Answer:

The amount of energy needed to raise the temperature of the cylinder by 25 °C is 23.3 KJ of heat.

Explanation:

The step by step calculation can be found in the attachment below. Thank you.

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3 years ago
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4 years ago
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Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiem
statuscvo [17]

Answer:

t = 0.33h = 1200s

x = 18.33 km

Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

x=x_o+v_1t    (1)

xo = 10 km

v1 = 55km/h

car 2:

x'=v_2t    (2)

v2 = 85km/h

For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}

t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s

The position in which both cars coincides is:

x=(55km/h)(0.33h)=18.33km

6 0
3 years ago
Determine the magnitude of the current flowing through a 4.7 kilo ohms resistor if the voltage across it is (a) 1mV (b) 10 V (c)
mariarad [96]

Answer:

213 nA

2.13 mA

851e^-t μA

Explanation:

We have a pretty straightforward question here.

Ohms Law states that the current in an electric circuit is directly proportional to the voltage and inversely proportional to the resistance in the circuit. It is mathematically written as

V = IR, since we need I, we can write that

I = V/R

a) at V = 1 mV

I = (1 * 10^-3) / 4.7 * 10^3

I = 2.13 * 10^-7 A or 213 nA

b) at V = 10 V

I = 10 / 4.7 * 10^3

I = 0.00213 A or 2.13 mA

c) at V = 4e^-t

I = 4e^-t / 4.7 * 10^3

I = 0.000851e^-t A or 851e^-t μA

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