Answer:
(a) X electrode
(b) Y electrode
(c) Y electrode
(d) X electrode
(e) Y electrode
Explanation:
<em>A galvanic (voltaic) cell has the generic metals X and Y as electrodes. X is more reactive than Y, that is, X more readily reacts to form a cation than Y does.</em>
In the X electrode occurs the oxidation whereas in the Y electrode occurs the reduction.
Oxidation: X(s) → X⁺ⁿ(aq) + n e⁻
Reduction: Y⁺ˣ(aq) + x e⁻ → Y(s)
<em>Classify the descriptions by whether they apply to the X or Y electrode.
</em>
<em>(a) anode.</em> Is where the oxidation takes place (X electrode).
<em>(b) cathode.</em> Is where the reduction takes place (Y electrode).
<em>(c) electrons in the wire flow toward.</em> Electrons in the wire flow toward the cathode (Y electrode).
<em>(d) electrons in the wire flow away.</em> Electrons in the wire flow away from the anode (X electrode).
<em>(e) cations from salt bridge flow toward.</em> Cations from the salt bridge flow toward the cathode (Y electrode) to maintain the electroneutrality.
Answer: The final temperature in Kelvin is 1488
Explanation:
To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
where,
are the initial pressure and temperature of the gas.
are the final pressure and temperature of the gas.
We are given:
Putting values in above equation, we get:
Hence, the final temperature in Kelvin is 1488
Answer: B) Bones
Explanation:
Natural casts form when flowing water removes all of the original bone or tissue, leaving just an impression in sediment.
Answer:
C
Explanation:
Melting point, temperature at which the solid and liquid forms of a pure substance can exist in equilibrium. As heat is applied to a solid, its temperature will increase until the melting point is reached. More heat then will convert the solid into a liquid with no temperature change.
Answer : The voltage applied by the batteries is, 6.0 V
Solution : Given,
Resistance of flashlight = 2.4 ohm
Current in the circuit = 2.5 Ampere
Formula used :
where,
V = applied voltage
I = current in the circuit
R = resistance of light
Now put all the given values in the above formula, we get
Therefore, the voltage applied by the batteries is, 6.0 V
