I honestly don’t know sorry
Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
The answer is 1.75 x 10¹⁰ m.
Scientific notation<span> refers to a mathematical expression that is used to represent a decimal number between 1 and 10 multiplied by ten, or we can say that to write a large numbers using less digits.
Diameter of chlorine atom = </span><span>175 picometers (pm)
</span>175 pm = <span>0.000000000175 m
in scientific notation we will write </span>0.000000000175 m as <span>1.75 x 10</span>¹⁰ m
Answer:
24.5
Explanation:
15/6.3 => 2.38 half lives passed.
.5^2.38 => 0.19198 decimal representation of the percentage that is left over after 2.38 half lives have passed.
0.19198 *128 = 24.5 mg of the material remaining.
