0.32 L is equal to how many mL
1 answer:
You might be interested in
Hello!
The initial mass of Magnesium Sulfate Heptahydrate (MgSO₄·7H₂O) is 23,08 g
The chemical reaction for the dehydrating of Magnesium Sulfate Heptahydrate (MgSO₄·7H₂O) is the following:
MgSO₄·7H₂O(s) + Δ → MgSO₄(s) + 7H₂O(g)
We know that the sample loses 11,80 g upon heating. That mass is the mass of Water that is released as vapor. Knowing that piece of information, we can apply the following conversion factor to go from the mass of water to the moles of water and back to the mass of the original compound (mi).
Have a nice day!
The initial mass of Magnesium Sulfate Heptahydrate (MgSO₄·7H₂O) is 23,08 g
The chemical reaction for the dehydrating of Magnesium Sulfate Heptahydrate (MgSO₄·7H₂O) is the following:
MgSO₄·7H₂O(s) + Δ → MgSO₄(s) + 7H₂O(g)
We know that the sample loses 11,80 g upon heating. That mass is the mass of Water that is released as vapor. Knowing that piece of information, we can apply the following conversion factor to go from the mass of water to the moles of water and back to the mass of the original compound (mi).
Have a nice day!
<u>Answer:</u> The value of for the given reaction is 1.435
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:
Given mass of = 9.2 g
Molar mass of = 92 g/mol
Volume of solution = 0.50 L
Putting values in above equation, we get:
For the given chemical equation:
<u>Initial:</u> 0.20
<u>At eqllm:</u> 0.20-x 2x
We are given:
Equilibrium concentration of = 0.057
Evaluating the value of 'x'
The expression of for above equation follows:
Putting values in above expression, we get:
Hence, the value of for the given reaction is 1.435