Answer:
Disagree with the suggestion based on the hygroscopic nature of anhydrous magnesium sulfate
Explanation:
Magnesium sulfate in the anhydrous form is a drying agent. A drying agent salts of inorganic compounds that has the capability of absorbing water to become hydrated, when placed in the presence of a wet surface or moist air
Anhydrous magnesium sulfate is therefore hygroscopic such that it absorbs water from the atmosphere and becomes hydrated and increases in size as its volume is increased according to the following chemical equation
MgSO₄(s) + 7H₂O(l) → MgSO₄·7H₂O(s)
The molar mass of anhydrous magnesium sulfate = 120.366 g/mol
The molar mass of the heptahydrate = 246.47 g/mol
Therefore, the mass of the magnesium sulfate doubles when it forms the heptahydrate, and the magnesium sulfate grows bigger.
Answer:
1 mole of platinum
Explanation:
To obtain the number of mole(s) of platinum present, we need to determine the empirical formula for the compound.
The empirical formula for the compound can be obtained as follow:
Platinum (Pt) = 117.4 g
Carbon (C) = 28.91 g
Nitrogen (N) = 33.71 g
Divide by their molar mass
Pt = 117.4 / 195 = 0.602
C = 28.91 / 12 = 2.409
N = 33.71 / 14 = 2.408
Divide by the smallest
Pt = 0.602 / 0.602 = 1
C = 2.409 / 0.602 = 4
N = 2.408 / 0.602 = 4
The empirical formula for the compound is PtC₄N₄ => Pt(CN)₄
From the formula of the compound (i.e Pt(CN)₄), we can see clearly that the compound contains 1 mole of platinum.
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16.4 grams is the mass of solute in a 500 mL solution of 0.200 M
.
sodium phosphate
Explanation:
Given data about sodium phosphate
atomic mass of Na3PO4 = 164 grams/mole
volume of the solution = 500 ml or 0.5 litres
molarity of sodium phosphate solution = 0.200 M
The formula for molarity will be used here to know the mass dissolved in the given volume of the solution:
The formula is
molarity =
putting the values in the equation, we get
molarity x volume = number of moles
0.200 X 0.5= number of moles
number of moles = 0.1 moles
Atomic mass x number of moles = mass
putting the values in the above equation
164 x 0.1 = 16.4 grams
16.4 grams of sodium phosphate is present in 0.5 L of the solution to make a 0.2 M solution.