Answer:
100ml of a stock 50% KNO3 solutions are needed to prepare 250ml of a 20% KNO3 solution.
Explanation:
In the given question it is mentioned that
S1=50%
V2=250ml
S2= 20%
We all know that
V1S1=V2S2
∴V1= V2×S2÷S1
∴V1= V2S2×1/S1
∴V1= 250×20÷50
∴V1= 100ml
Answer:
• The actual number of moles of each element in the smallest unit of the compound. •In water (H 2 O), ammonia (NH 3), methane (CH 4), and ionic compounds, the empirical and molecular
Explanation:
Answer:
Yes it will
An example is ice, when it melts the volume goes up which means it occupies much more space
Answer:
74.4 ml
Explanation:
C₆H₈O₇(aq) + 3NaHCO₃(s) => Na₃C₆H₅O₃(aq + 3CO₂(g) + 3H₂O(l)
Given 15g = 15g/84g/mol = 0.1786mole Sodium Bicarbonate
From equation stoichiometry 3moles NaHCO₃ is needed for each mole citric acid or, moles of citric acid needed is 1/3 of moles sodium bicarbonate used.
Therefore, for complete reaction of 0.1786 mole NaHCO₃ one would need 1/3 of 0.1786 mole citric acid or 0.0595 mole H-citrate.
The question is now what volume of 0.8M H-citrate solution would contain 0.0595mole of the H-citrate? This can be determined from the equation defining molarity. That is => Molarity = moles solute / Liters of solution
=> Volume (Liters) = moles citric acid / Molarity of citric acid solution
=> Volume needed in liters = 0.0.0595 mole/0.80M = 0.0744 Liters or 74.4 ml
The answer to the problem is 4.5 kilometers. you can solve this problem by cross multiplying
