Answer:
1.41 moles H2O2(with sig figs)
Explanation:
okay so what is the molar mass of H2O2= (1.008 g/mol)2+(16.00g/mol)2= (2.016+ 32.00) g/ mol
= 34. 02 g/mol
48.0g H2O2* 1 mol H2O2/ 34.02 g H2O2= 1.41 mol H2O2
Answer:
Because the opposite charge between them
Explanation:
Answer:30 L
Explanation:
Initial Volume
=
V
1
=
60
l
i
t
e
r
Initial Temperature
=
T
1
=
546
K
Final Temperature
=
T
2
=
273
K
Final Vloume
=
V
2
=
?
?
Sol:-
Since the pressure is constant and the question is asking about temperature and volume, i.e,
V
1
T
1
=
V
2
T
2
⇒
V
2
=
V
1
⋅
T
2
T
1
=
60
⋅
273
546
=
60
2
=
30
l
i
t
e
r
⇒
V
2
=
30
l
i
t
e
r
Hence the new volume of the gas is
30
l
i
t
e
r
Answer:
fH = - 3,255.7 kJ/mol
Explanation:
Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.
Qsystem + Qcombustion = 0
Qsystem = heat capacity*ΔT
10000*(25.000 - 20.826) + Qc = 0
Qcombustion = - 41,740 J = - 41.74 kJ
So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:
n = mass/molar mass = 1/ 78
n = 0.01282 mol
fH = -41.74/0.01282
fH = - 3,255.7 kJ/mol
<span>The box with no lightbulb is the control group.
The others are treatments. The purpose of a control group is used to increase the validity of the experiment by testing
the independent variable to check that it
does not influence the results. </span>