Answer:
1.84 L
Explanation:
Using the equation for reversible work:

Where:
W is the work done (J) = -287 J.
Since the gas did work, therefore W is negative.
P is the pressure in atm = 1.90 atm.
However, work done is in joules and pressure is in atm. We can use the values of universal gas constant as a convenient conversion unit. R = 8.314 J/(mol*K); R = 0.0821 (L*atm)/(mol*K)
Therefore, the conversion unit is 0.0821/8.314 = 0.00987 (L*atm)/J
is the initial volume = 0.350 L
is the final volume = ?
Thus:
(-287 J)*0.00987 (L*atm)/J = -1.9 atm*(
- 0.350) L
= [(287*0.00987)+(1.9*0.350)]/1.9 = (2.833+0.665)/1.9 =1.84 L
The rate law for this reaction is [A]².
Balanced chemical reaction used in this experiment: A + B → P
The reaction rate is the speed at which reactants are converted into products.
Comparing first and second experiment, there is no change in initial rate. The concentration of reactant B is increased by double. Initial rate does not depands on concentration of reactant B.
Comparing first and third experiment, initial rate is nine times greater, while concentration of reactant A is three times greater. Conclusion is that concentration of reactant A is squared and the rate is [A]².
More info about rate law: brainly.com/question/16981791
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Answer:
PART A: The LDF occurs between all molecules. Dispersion forces result from shifting electron clouds, which cause weak, temporary dipole.
PART B: Dipole dipole operates only between polar molecules. This is when two polar molecules get near each other and the positively charged portion of the molecule is attracted to the negatively charged portion of another molecule.
PART C: Dipole dipole and in some cases hydrogen bonding operate between the hydrogen atom of a polar bond and a nearby small electronegative atom. Only if the atom bonded to it were F, O or N it would be hydrogen bonding. Otherwise it is dipole dipole.
Answer:
50 g Sucrose
Explanation:
Step 1: Given data
- Concentration of the solution: 2.5%
Step 2: Calculate the mass of sucrose needed to prepare the solution
The concentration of the solution is 2.5%, that is, there are 2.5 g of sucrose (solute) every 100 g of solution. The mass of sucrose needed to prepare 2000 g of solution is:
2000 g Solution × 2.5 g Sucrose/100 g Solution = 50 g Sucrose