Review
6 Answers
1. B Remember any burning is always
exothermic. But just cooking is endothermic.
2. If the reaction
takes twice as long, (140 /70), then the rate is ½ of the original:
R_{2}/R_{1}
=_{}
2x^{3}
(0.25) = 1
x^{3}
= 2.
x = _{}
[W] should be increased by a factor of 1.26
3. a. We need to calculate the density of air
and of helium:
air:
There
are two unknowns, n and V. There is no air in the balloon. The air in the
formula is that of air surrounding the balloon.
rearranging PV = nRT, we obtain:
n/V = P/RT. But we want g/L(density)instead of
moles/L . No problem!
Since (n/V)*molar mass = density, by
substituting:
P/RT* avg.molar
mass of air = density of air = 101.3/(8.31* [20.0
+ 273])
* 28.9 = 1.20g/L.
Helium:
Use
the same technique:
P/RT* molar mass of He= density of He = 101.3/(8.31* [20 +
273]) * 4.0 = 0.17 g/L.
_{},
= 10.0 (1.20 –0.17) = 10.3 g.
b. Rearranging PV = nRT, we obtain:
n/V = P/RT. But we want g/L
instead of moles/L
. Since (n/V)*molar mass = density, which is the same as (P/RT)*molar mass =
density
we can substitute for density in the m_{load}
formula and obtain:
m_{load} = V(r_{air }-r_{He})
m_{load} = V_{},
where M = molar mass
or by factoring:
m_{load} =_{}
c. Attach a bunch of grapes to several
balloons. Then start eating one grape at a time until the balloon neither rises
nor sinks to the floor.
d. (BONUS-type question) According to the principle of
buoyancy, the buoyant force is equal to the weight of the displaced fluid or
gas ( For instance if a cork is forced under water, the buoyant force trying to
propel it upward will be equal to the equivalent weight of water represented by
the volume of the cork.). For something not to sink, the buoyant force has to
overcome gravity. In our case, where the balloon is perfectly suspended, those
forces are equal:
F_{buoyant}_{ }= F_{gravity}
Weight
of air displaced by helium = weight of string and plastic + weight of He
m_{air}g = m_{load}g + m_{He}g where g = gravitational
acceleration, but it cancels:
m_{air}= m_{load}_{ }+ m_{He}
If
we express mass as density*volume, we obtain:
Vr_{air} = m_{load}_{
}+ Vr_{He}_{
} where V = volume of
balloon
By
isolating m and factoring we get:
m_{load} = Vr_{air}_{ }- Vr_{He}_{}
_{ }_{}
m_{load} = V(r_{air }-r_{He})
4. H_{2} + I_{2}
à 2HI
Time (s) |
Remaining Amount of Iodine (moles) |
Amount of iodine that reacted(moles) |
Amount of HI produced in moles ( see
ratio) |
0 |
2.6 |
0 |
0 |
10 |
1.9 |
2.6 - 1.9 =
0.7 |
0.7*2 = 1.4 |
20 |
1.4 |
2.6 - 1.4 =
1.2 |
1.2*2 = 2.4 |
30 |
1.2 |
2.6 - 1.2 = 1.4 |
1.4*2 = 2.8 |
Avg
rate for the last 20 s = _{}
5. O_{2(}_{g) }+ H_{2(g) }à 2 OH_{(g)} DH = 77.9 kJ (1)
O_{2(g) }à 2 O_{(g) } DH = 495 kJ (2)
H_{2(g)
}à 2 H_{(g) } DH = 435.9 kJ (3)
target
: O_{(g)
}+ H_{(g) }à OH_{(g)}
divide eq(1)
by 2:
0.5 O_{2(}_{g)
}+ 0.5 H_{2(g) }à OH_{(g)} DH = 77.9 kJ/2 = 38.95 kJ
reverse eq(2)
and divide by 2:
O_{(}_{g)}à 0.5 O_{2(g) }DH = - 495
kJ/2 = - 247.5 kJ
reverse eq(3) and divide by 2:
H_{(}_{g)
}à 0.5
H_{2(g)} DH = - 435.9 kJ/2 = -217.95 kJ
Add them up: O_{(}_{g)
}+ H_{(g) }à OH_{(g) }DH = -426.5 kJ = -427 kJ
6. Using Boyle’s Law,
we obtain V_{2} = 3V_{1.}
7.
a.
adding the preservative calcium propionate to bread to
slow the growth of mold__yes____
b.
treating a cut with iodine to inhibit the function of
bacterial proteins._ yes ____
c.
destroying enzymes by adding mercury and silver___ yes ____
d.
producing chlorophyll in early spring__no___
e. adding lactase to
milk to break down lactose____no___