In order to make 0.500 L of a 0.475 M solution, you will need to weigh out how many grams of sucrose (MM = 342 g/mol)?
1 answer:
The first step to solve this problem is to multiply the volume of solution times its concentration to find the number of moles needed, remember that M=mol/L:
Now, use the molar mass of sucrose to find the number of grams needed to make the solution. This is, multiply the number of moles needed times the molar mass:
It means that to make 0.500L of a 0.475M you will have to weigh 81.225g of sucrose.
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Answer:
Final pressure = 362.7 Pa
Explanation:
Given that,
Initial volume, V₁ = 930 ml
Initial pressure P₁ = 156 Pa
Final volume, V₂ = 400 mL
We need to find the final pressure. We know that the relation between volume and pressure is inverse i.e.
So, the final pressure is equal to 362.7 Pa.
The question is incomplete, here is the complete question:
Phosgene, , gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide with chlorine:
The value of for this reaction is 5.79 at 570 K. What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which
and
?
<u>Answer:</u> The equilibrium partial pressure of CO, is 0.257 atm, 0.257 atm and 0.008 atm respectively.
<u>Explanation:</u>
The relation of is given by:
= Equilibrium constant in terms of partial pressure
= Equilibrium constant in terms of concentration = 5.79
= Difference between gaseous moles on product side and reactant side =
R = Gas constant =
T = Temperature = 570 K
Putting values in above equation, we get:
We are given:
Initial partial pressure of CO = 0.265 atm
Initial partial pressure of chlorine gas = 0.265 atm
Initial partial pressure of phosgene = 0.00 atm
The given chemical equation follows:
<u>Initial:</u> 0.265 0.265
<u>At eqllm:</u> 0.265-x 0.265-x x
The expression of for above equation follows:
Putting values in above equation, we get:
Neglecting the value of x = 8.59 because equilibrium partial pressure cannot be greater than initial pressure
So, the equilibrium partial pressure of CO =
The equilibrium partial pressure of
The equilibrium partial pressure of
Hence, the equilibrium partial pressure of CO, is 0.257 atm, 0.257 atm and 0.008 atm respectively.