Answer:
V2 = 0.75 L
Explanation:
assumin ideal gas:
∴ T1 = 300K
∴ V1 = 0.500 L
∴ T2 = 450 K
⇒ V2 = ?
assuming P, n constant:
⇒ PV1 = RT1n
⇒ V1/T1 = Rn/P = K..........(1)
⇒ V2/T2 = RN/P = K..........(2)
(1) = (2):
⇒ V1/T1 = V2/T2
⇒ V2 = (V1/T1)T2
⇒ V2 = (0.500 L/300 K)(450 K)
⇒ V2 = 0.75 L
Since the temperature is constant, therefore, this problem can be solved based on Boyle's law.
Boyle's law states that: " At constant temperature, the pressure of a certain mass of gas is inversely proportional to its pressure".
This can be written as:
P1V1 = P2V2
where:
P1 is the initial pressure = 1 atm
V1 is the initial volume = 3.6 liters
P2 is the final pressure = 2.5 atm
V2 is the final volume that we need to calculate
Substitute with the givens in the above mentioned equation to get the final volume as follows:
P2V1 = P2V2
1(3.6) = 2.5V2
3.6 = 2.5V2
V2 = 3.6 / 2.5 = 1.44 liters
The chemical formula for the compound can be written as,
CxHyOz
where x is the number of C atoms, y is the number of H atoms, and z is the number of O atoms. The combustion reaction for this compound is,
CxHyOz + O2 --> CO2 + H2O
number of moles of C:
(0.7191 g)(1 mol CO2/44 g of CO2) = 0.0163 mol CO2
This signifies that 0.0163 mole of C and the mass of carbon in the compound,
(0.0163 mols C)(12 g C/ 1 mol C) = 0.196 g C
number of moles H:
(0.1472 g H2O)(1 mol H2O/18 g H2O) = 0.00818 mol H2O
This signifies that there are 0.01635 atoms of H in the compound.
mass of H in the compound = (0.01635 mols H)(1 g of H) = 0.01635 g H
Mass of oxygen in the compound,
0.3870 - (0.196 g C + 0.01635 g H) = 0.1746 g
Moles O in the compound = (0.1746 g O)(1 mol O/16 g O) = 0.0109 mols O
The formula of the compound is,
C0.0163H0.01635O0.0109
Dividing the numbers by the least number,
C3/2H3/2O
The empirical formula of the compound is therefore,
<em> C₃H₃O₂</em>
Answer : Both solutions contain 
molecules.
Explanation : The number of molecules of 0.5 M of sucrose is equal to the number of molecules in 0.5 M of glucose. Both solutions contain 
molecules.
Avogadro's Number is 
= 
which represents particles per mole and particles may be typically molecules, atoms, ions, electrons, etc.
Here, only molarity values are given; where molarity is a measurement of concentration in terms of moles of the solute per liter of solvent.
Since each substance has the same concentration, 0.5 M, each will have the same number of molecules present per liter of solution.
Addition of molar mass for individual substance is not needed. As if both are considered in 1 Liter they would have same moles which is 0.5.
We can calculate the number of molecules for each;
Number of molecules = 
;
∴ Number of molecules = 
which will be = 
Thus, these solutions compare to each other in that they have not only the same concentration, but they will have the same number of solvated sugar molecules. But the mass of glucose dissolved will be less than the mass of sucrose.
Answer:
= 12 mL H202
Explanation:
Given that, the concentration of H2O2 is given antiseptic = 3.0 % v/v
It implies that, 3ml H2O2 is present in 100 ml of solution.
Therefore, to calculate the amount of H202 in 400.0 mL bottle of solution;
we have;
(3.0 mL/ 100 mL) × 400 mL
= 12 mL H202
