Answer:
You first need to construct a balanced chemical equation to describe the reaction:
KOH + HNO3 ---------> KNO3 + H2O
Work out the no. moles of HNO3 being neutralized:
Moles = Volume x Concentration = (25/1000) x 0.0150 = 0.000375 moles
From the balanced equation the molar ratio of KOH to HNO3 is 1:1 so you also need 0.000375 moles of KOH to neutralise the nitric acid
Now you can work out the volume of KOH required:
Volume = Moles/Concentration = (0.000375)/0.05 = 0.0075 dm^3 = 7.5 cm^3
Answer:
18.9 x 10¹³ grams of Bauxite Ore
Explanation:
Al₂O₃ = 50% of Bauxite Ore
Al₂O₃ = 0.5 (Bauxite Ore)--------------------------------------- (1)
Overall reaction:
2Al₂O₃ + 3C → 4Al + 3CO₂--------------------------------------- (2)
[ Al= 27 , O=16, C=12]
From (2), 2 moles of Aluminium oxide (Al₂O₃) gives 4 moles of Aluminium
In terms of grams, we can say:
Al₂O₃ = [2(27) +3(16)]
= 54 +48
=102grams
2 moles of Al₂O₃ = 2 x102grams
=204grams
4 moles of Al = 4 x 27
=108 grams
So from (2):
204 grams of Al₂O₃ = 108 grams of Aluminium
x grams of Al₂O₃ = 5.0 x 10¹³grams of Aluminium
Calculating for x:
x = (204 x 5.0 x 10¹³)/ 108
= 9.44 x 10¹³ grams
So 9.44 x 10¹³ grams of pure bauxite (Bauxite) is required.
However the to calculate the quantity of raw bauxite, we use (1):
Bauxite ore = Pure Bauxite/0.5
= 9.44 x 10¹³ grams/0.5
= 18.88 x 10¹³ grams
≈ 18.9 x 10¹³ grams
Gas and radio waves I’m pretty sure is correct
Answer:
0 M is the silver ion concentration in a solution prepared mixing both the solutions.
Explanation:
Moles of silver nitrate = n
Volume of the solution = 425 mL = 0.425 L (1 mL = 0.001 L)
Molarity of the silver nitrate solution = 0.397 M
Moles of sodium phosphate = n'
Volume of the sodium phosphate solution = 427 mL = 0.427 L (1 mL = 0.001 L)
Molarity of the sodium phosphate solution = 0.459 M
According to reaction, 3 moles of silver nitrate reacts with 1 mole of sodium phosphate, then 0.1687 moles of silver nitrate will recat with :
of sodium phosphate
This means that only 0.05623 moles of sodium phosphate will react with all the 0.1687 moles of silver nitrate , making silver nitrate limiting reagent and sodium phosphate as an excessive reagent.
So, zero moles of silver nitrate will be left in the solution after mixing of the both solutions and hence zero moles of silver ions will left in the resulting solution.
0 M is the silver ion concentration in a solution prepared mixing both the solutions.
your answer is <u>D</u><u>.</u><u> </u>physical change, because a new substance is not formed.
