Answer:
Mass of benzene is: 149.3 g
Explanation:
Let's use density to calculate mass.
Density = Mass / Volume
Mass = Density . volume
Be careful, because density is in g/mL and the volume is in L. So let's convert the L to mL: 0.170 L . 1000 mL / 1L = 170 mL
0.8787 g/mL . 170 mL = Mass of benzene
Mass of benzene is: 149.3 g
Answer:
ΔH = ΔH₁ + ΔH₂ - ΔH₃
Explanation:
Given that:
1. A → 2B
2. B → C + D
3. E → 2D
Assuming from the corresponding ΔH for process 1, 2 and 3 are ΔH₁, ΔH₂, ΔH₃ respectively.
To estimate the ΔH for the process A → 2C + E
We multiply 2 with equation 2 where (B → C + D)
2B → 2C + 2D ⇒ 2ΔH₂
Also, let's switch equation (3), such that we have,
2D → E -ΔH₃
The summation of all the equation result into :
A → 2C + E
where; ΔH = ΔH₁ + ΔH₂ - ΔH₃
The piece of unknown metal is in thermal equilibrium with water such that Q of metal is equal to Q of the water. We write this equality as follows:
-Qm = Qw
Mass of metal (Cm)(ΔT) = Mass of water (Cw) (ΔT)
where C is the specific heat capacities of the materials.
We calculate as follows:
-(Mass of metal (Cm)(ΔT)) = Mass of water (Cw) (ΔT)
-68.6 (Cm)(52.1 - 100) = 42 (4.184) (52.1 - 20)
Cm = 1.717 -----> OPTION C
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.