0.300 M IKI represents the
concentration which is in molarity of a potassium iodide solution. This means
that for every liter of solution there are 0.300 moles of potassium iodide. Knowing
that molarity is a ratio of solute to solution.
By using a conversion factor:
100 ml x (1L / 1000 mL) x (0.300
mol Kl / 1 L) x (166.0g / 1 mol Kl) = 4.98 g
Therefore, in the first
conversion by simply converting the unit of volume to liter, Molarity is in L
where the volume is in liters. The next step is converted in moles from volume
by using molarity as a conversion factor which is similar to how density can be
used to convert between volume and mass. After converting to moles it is simply
used as molar mass of Kl which is obtained from periodic table to convert from
mole to grams.
In order to get the grams of IKI
to create a 100 mL solution of 0.600 M IKI, use the same formula as above:
100 ml x (1L / 1000 mL) x (0.600
mol Kl / 1 L) x (166.0g / 1 mol Kl) = 9.96 g
Answer:
The anwer is not D the anwer is A
Explanation:
Answer:
The answer to your question is 0.33 moles of H₃PO₄
Explanation:
Data
moles of Ca(OH)₂ = 0.050
moles of H₃PO₄ = ?
Process
1.- Write the balanced chemical equation
3Ca(OH)₂ + 2H₃PO₄ ⇒ Ca₃(PO₄)₂ + 6H₂O
Reactants Elements Products
3 Ca 3
12 H 12
14 O 14
2.- Calculate the moles of phosphoric acid
3 moles of calcium hydroxide --------------- 2 moles of phosphoric acid
0.5 moles of calcium hydroxide ----------- x
x = (0.5 x 2)/3
x = 0.33 moles
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Answer:
0.231 mol/L
Explanation:
The first step is to write the balanced equation for this reaction:
The second step is to find the number of moles in the acid:
number of moles = volume * concentration
= 0.035 L * 0.275 mol/L
= 0.009625 mol
The third step is to use the molar ratio from the balanced chemical equation to find the number of moles of NaOH that can neutralize 0.009625 mol of sulphuric acid.
n(sulphuric acid) : n(sodium hydroxide)
1 : 2
0.009625 mol : x
x = 0.01925 mol
Fourth step is to calculate the concentration of sodium hydroxide:
