Answer:
The electrode that removes ions from solution
Explanation:
Each electrochemical cell consists of an anode and a cathode. Oxidation occurs at the anode and reduction occurs at the cathode.
At the anode, ions move from the electrode into the solution while at the cathode ions move from the solution to the electrode.
At the cathode, metal ions accept electron(s) and become deposited on the electrode hence this electrode removes ions from solution. This is reduction.
Which two solutions, when mixed together, will undergo a double replacement reaction and form a white, solid substance?
1. NaCl(aq) and LiNO3(aq)
2. KCl(aq) and AgNO3(aq) answer
3. KCl(aq) and LiCL(aq)
4. NaNO3(aq) and AgNO3(aq)
2 is the answer because AgCl is formed and that is a white ppt.
Answer : The volume of oxygen at STP is 112.0665 L
Solution : Given,
The number of moles of
= 5 moles
At STP, the temperature is 273 K and pressure is 1 atm.
Using ideal gas law equation :

where,
P = pressure of gas
V = volume of gas
n = the number of moles
T = temperature of gas
R = gas constant = 0.0821 L atm/mole K (Given)
By rearranging the above ideal gas law equation, we get

Now put all the given values in this expression, we get the value of volume.

Therefore, the volume of oxygen at STP is 112.0665 L
Physical. When you simply desolve something with water, to don't actually change it. This action can be undone.
An example of a chemical reaction is when the cells within completely change. You you burn wood, you can't go back in time and un-burn it. Does that make sense?
Answer:
8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days
Explanation:
The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.
We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.
Initial amount of Thorium-234 available as per the question is 150 grams
So now we start with 150 grams of Thorium-234





So after 120.5 days the amount of sample that remains is 8.625g
In simpler way , we can use the below formula to find the sample left

Where
is the initial sample amount
n = the number of half-lives that pass in a given period of time.