Answer:
B. a nuclear reactor core (1000°C)......
CuCl2 + 2NaNO3 ----> Cu(NO3)2 + 2NaCl
using molar masses:-
Theoretical yields:-
63.54 + 2(35.45) g of CuCl2 produces 2(22.98 + 35.45) g of NaCl
134.44 g .................................................... 116.86 g
31.0 g ....................................................31.0 * 116.86 /134.44=26.95g
So percentage yield is 21.2* 100 / 26.95 = 78.7% to nearest tenth
According to Diagram B, look at the 1600 elevation until you see the descending air line touches it. Then look down at the temperature at the bottom of the graph. It is between 0 degrees to 5 degrees.
The only number that is between that range is 2 degrees C.
Answer:
2.77 mL of boiling water is the minimum amount which will dissolve 500 mg of phthalic acid.
Explanation:
We know from the problem that 18 g of phthalic acid are dissolved in 100 mL of water at 99 °C.
Now we devise the following reasoning:
If 18 g of phthalic acid are dissolved in 100 mL of water at 99 °C
Then 0.5 g of phthalic acid are dissolved in X mL of water at 99 °C
X = (0.5 × 100) / 18 = 2.77 mL of water
Answer:
plant cells and eukroyatic algae
