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:
The sun
Explanation:
The sun is the source of energy that provides the boost for electrons during photosynthesis
To balance this equation, first we should consider balancing C because it only presents in one reactant and one product. Assuming the coefficient of C6H6 is 1, there are 6 C's in the reactant, so it generates 6CO2. Then consider balancing H for the same reason. If the coefficient of C6H6 is 1, there are 6 H's in the reactant, so it generates 3H2O.
Now that the coefficient of the products are determined, we can balance O. There are 6*2=12 O's in CO2 and 3*1=3 O's in H2O. So the total number of O in the products is 12+3 = 15. O2 is the only reactant that contains O, so to balance the equation, the coefficient of O2 should be 15/2.
Now the equation looks like:
C6H6 + 15/2O2 ⇒ 6CO2 + 3H2O.
Times both sides of the equation by 2 results the final answer:
2C6H6 + 15O2 ⇒ 12CO2 + 6H2O
Answer:
94.325 g
Explanation:
We'll begin by converting 350 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
350 mL = 350 mL × 1 L /1000 mL
350 mL = 0.35 L
Next, we shall determine the number of mole of KC₂H₃O₂ in the solution. This can be obtained as follow:
Volume = 0.35 L
Molarity of KC₂H₃O₂ = 2.75 M
Mole of KC₂H₃O₂ =?
Molarity = mole /Volume
2.75 = Mole of KC₂H₃O₂ / 0.35
Cross multiply
Mole of KC₂H₃O₂ = 2.75 × 0.35
Mole of KC₂H₃O₂ = 0.9625 mole
Finally, we shall determine the mass of KC₂H₃O₂ needed to prepare the solution. This can be obtained as illustrated below:
Mole of KC₂H₃O₂ = 0.9625 mole
Molar mass of KC₂H₃O₂ = 39 + (12×2) +(3×1) + (16×2)
= 39 + 24 + 3 + 32
= 98 g/mol
Mass of KC₂H₃O₂ =?
Mass = mole × molar mass
Mass of KC₂H₃O₂ = 0.9625 × 98
Mass of KC₂H₃O₂ = 94.325 g
Thus, the mass of KC₂H₃O₂ needed to prepare the solution is 94.325 g
