Since you didn't give the actual volume (or any of the experimental values) I can only tell you how to do it. Do the calculation using the real (determined) volume of the flask. Then, re-do the calculation with v = 125ml. Take the two values and calculate % error; m = measured vol; g = guessed vol.
<span>[mW (m) - mW (g)]/mW (m) x 100% </span>
<span>(they want % error so, if it is negative, just get rid of the sign) </span>
Answer:
Yes
Explanation:
A supercritical fluid has good properties for both liquid and as for extraction properties, the advantages then include:
- The fact that it has a lower viscosity than liquid CO2 allowing it to move through and around coffee beans more thoroughly with creating back pressure
- Its density is comparable to that of liquid CO2 meaning there is much CO2 per litre as there is liquid form making it more efficient
- It has a higher diffusivity than liquid CO2 which aids with penetration of the coffee beans on a molecular level
This experiment would not work with tea leaves because they also contain caffeine
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⁻
Answer:
8.2 x 106^-11
Explanation:
To begin this problem you must remember the basic rule of scientific notation, which is, must be between 1-10. .000000000082 is much smaller than 1. However by moving the decimal 11 spots to the right, we can make it 8.2
Continue to move the decimal to the right until the value is in the 1-10 range. Make sure to count the moves to the right.
Once the decimal is in the right spot count the spots moved.
Since the number is wayyy smaller than the answer given the number will be negative 10^-11, in order to make it what is was before.
