> 2,000
mL of a 5.0 × 10–5% (w/v) sucrose solution
5.0 × 10–3
g/mL * 2000 mL * (1 mol / 342.30 g) = 0.0292 mol
<span>
> 2,000 mL of a 5.0 ppm sucrose solution</span>
5 grams /
1000000 mL * 2000 mL* (1 mol / 342.30 g) = 0.0000292 mol
<span>
> 20 mL of a 5.0 M sucrose solution </span>
5.0 M *
0.020 L = 0.1 mol
Answer:
<span>2,000 mL
of a 5.0 ppm sucrose solution</span>
Place a burning splint near the opening of a test tube. If a popping noise occurs, it's probably hydrogen. Place a glowing splint in the test tube, and if it reignites, it could be oxygen. Place a burning splint into a test tube, and if it goes out, it could be carbon dioxide.
Answer:
87.15%
Explanation:
To find percent yield, we can use this simple equation

Where "Actual" is the amount in grams actually collected from the reaction, and "Theoretical" is, well, the theoretical amount that should have been produced.
They give us these values, so to find the percent yield, just plug the numbers in.

So, the percent yield is 87.15%
An easy trick to remember how to do this is just to divide the smaller number by the bigger number and move the decimal back two places. If you have a percent yield greater than 100%, something is wrong in the reaction.