Answer:
28atm
Explanation:
Using Gay lussac's law equation as follows:
P1/T1 = P2/T2
Where;
P1 = initial pressure (atm)
T1 = initial temperature (K)
P2 = final pressure (atm)
T2 = final temperature (K)
Based on the information provided in this question;
P1 = 30.0 atm, T1 = 30.0°C, P2 = ?, T2 = 10.0°C
NOTE: Absolute temperature i.e. Kelvin is required for this law
T1 = 30°C + 273K = 303K
T2 = 10°C + 273K = 283K
Using P1/T1 = P2/T2
30/303 = P2/283
Cross multiply
P2 × 303 = 30 × 283
303P2 = 8490
P2 = 8490/303
P2 = 28.02
New pressure of the gas = 28atm
The current required to accumulate the 1.22 grams of nickel in 0.5 hours is 2.23 A.
<h3>What is current?</h3>
The current is given as the product of the charge with time. In the electrochemical analysis of the nickel, there will be a reduction of the nickel ion to nickel. The formation is given as:
There is the deposition of 1 mole of Ni with 2 electrons transfer. The transfer of charge for 1 mole that is 58.7 grams Nickel is:
The mass of Ni to be deposited is 1.22 grams. The charge required is given as:
The current required to transfer 4010.7 C of charge in 1800 seconds is given as:
Thus, the current required to accumulate the 1.22 grams of nickel in 0.5 hours is 2.23 A.
Learn more about current, here:
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Answer:
D- a weak base
Explanation:
The ph scale goes from 1-10 a solution over 7ph is classified as basic. A solution thats 8.4 is only 1.4 over 7pH, making it a weak basic solution. An example of a strong base would be a solution with a pH of 9.2 (for example).
Answer: A mechanical wave is a disturbance in matter that transfers energy through the matter. A mechanical wave starts when matter is disturbed. A source of energy is needed to disturb matter and start a mechanical wave.
Answer:
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
Explanation:
Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)
In the oxidation half reaction, the oxidation number increases:
Mn changes from 0, in the ground state to Mn²⁺.
The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.
Silver changes from Ag⁺ to Ag.
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
To balance the hole reaction, we need to multiply by 2, the second half reaction:
Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2
2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)
Now we sum, and we can cancel the electrons:
2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻
