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.
Answer:
N with a charge of (-3) is nitride
Explanation:
<h3>Answer:</h3>
18.75 grams
<h3>Explanation:</h3>
- Half-life refers to the time taken by a radioactive material to decay by half of the original mass.
- In this case, the half-life of element X is 10 years, which means it takes 10 years for a given mass of the element to decay by half of its original mass.
- To calculate the amount that remained after decay we use;
Remaining mass = Original mass × (1/2)^n, where n is the number of half-lives
Number of half-lives = Time for the decay ÷ Half-life
= 40 years ÷ 10 years
= 4
Therefore;
Remaining mass = 300 g × (1/2)⁴
= 300 g × 1/16
= 18.75 g
Hence, a mass of 300 g of an element X decays to 18.75 g after 40 years.
Answer:
Ok then
Explanation:
Thank you for the free points! Have a good day.
Answer:
they are a lot tougher and are expecting only greatness, sort of like theater critiques.
Explanation:
