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Answer:
Final molarity of ammonium cation in the solution = 0.16 M
Explanation:
Complete Question
Suppose 2.59 g of ammonium nitrate is dissolved in 200. mL of a 0.40M aqueous solution of sodium chromate. Calculate the final molarity of ammonium cation in the solution. You can assume the volume of the solution doesn't change when the ammonium nitrate is dissolved in it. Be sure your answer has the correct number of significant digits.
Solution
2NH₄NO₃ + Na₂CrO₄ → (NH₄)₂CrO₄ + 2NaNO₃
We first convert the given parameters to number of moles
Number of moles = (Mass/Molar mass)
Molar mass of NH₄NO₃ = 80.043 g/mol
Number of moles of NH₄NO₃ = (2.59/80.043) = 0.03224 mole
Number of moles = (Concentration in mol/L) × (Volume in L)
Number of moles of Na₂CrO₄ = 0.4 × 0.2 = 0.08 Mole
2 moles of NH₄NO₃ react with 1 mole of Na₂CrO₄
So, it it evident that NH₄NO₃ is the limiting reagent as it is in short supply in the amount needed for the reaction.
So, the number of moles of ammonium ion in the product is also 0.03224 mole.
Molarity = (Number of moles)/(Volume L)
Molarity of ammonium ion = (0.03224/0.2) = 0.1612 mol/L = 0.16 M
Hope this Helps!!!
Total number of students (male and female) in the college=2760. Total number of students including both male and female represents 100% (percentage) of students which is equal to 2760 number.
Number of male students is 65% (percentage) which is equal to 
= 1794. So, total number of female students are (100-65)%=35% (percentage) which is equal to 
=966 numbers.
Answer:
Theoretical yield of the reaction = 34 g
Excess reactant is hydrogen
Limiting reactant is nitrogen
Explanation:
Given there is 100 g of nitrogen and 100 g of hydrogen
Number of moles of nitrogen = 100 ÷ 28 = 3·57
Number of moles of hydrogen = 100 ÷ 2 = 50
Reaction between nitrogen and hydrogen yields ammonia according to the following chemical equation
N2 + 3H2 → 2NH3
From the above chemical equation for every mole of nitrogen that reacts, 3 moles of hydrogen will be required and 2 moles of ammonia will be formed
Now we have 3·57 moles of nitrogen and therefore we require 3 × 3·57 moles of hydrogen
⇒ We require 10·71 moles of hydrogen
But we have 50 moles of hydrogen
∴ Limiting reactant is nitrogen and excess reactant is hydrogen
From the balanced chemical equation the yield will be 2 × 3·57 moles of ammonia
Molecular weight of ammonia = 17 g
∴ Theoretical yield of the reaction = 2 × 3·57 × 17 = 121·38 g
