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3 years ago

HELP HELP URGENT STRONG ACIDS AND BASES

Chemistry

1 answer:

leonid [27]3 years ago

3 0

Answer:

Solution A is correct.

Explanation:

Strong acids or bases are assumed to dissociate completely when in a aqueous solution.

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A weather balloon is filled with helium that occupies a volume of 5.57 104 L at 0.995 atm and 32.0°C. After it is released, it r

Alchen [17]

6.52 × 10⁴ L. (3 sig. fig.)

<h3>Explanation</h3>

Helium is a noble gas. The interaction between two helium molecules is rather weak, which makes the gas rather "ideal."

Consider the ideal gas law:

$P\cdot V = n\cdot R\cdot T$ ,

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of gas particles in the gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas in degrees Kelvins.

The question is asking for the final volume $V$ of the gas. Rearrange the ideal gas equation for volume:

$V = \dfrac{n \cdot R \cdot T}{P}$ .

Both the temperature of the gas, $T$ , and the pressure on the gas changed in this process. To find the new volume of the gas, change one variable at a time.

Start with the absolute temperature of the gas:

$T_0 = (32.0 + 273.15) \;\text{K} = 305.15\;\text{K}$ ,
$T_1 = (-14.5 + 273.15) \;\text{K} = 258.65\;\text{K}$ .

The volume of the gas is proportional to its temperature if both $n$ and $P$ stay constant.

$n$ won't change unless the balloon leaks, and
consider $P$ to be constant, for calculations that include $T$ .

$V_1 = V_0 \cdot \dfrac{T_1}{T_2} = 5.57\times 10^{4}\;\text{L}\times \dfrac{258.65\;\textbf{K}}{305.15\;\textbf{K}} = 4.72122\times 10^{4}\;\text{L}$ .

Now, keep the temperature at $T_1 =258.65\;\text{K}$ and change the pressure on the gas:

$P_1 = 0.995\;\text{atm}$ ,
$P_2 = 0.720\;\text{atm}$ .

The volume of the gas is proportional to the reciprocal of its absolute temperature $\dfrac{1}{T}$ if both $n$ and $T$ stays constant. In other words,

$V_2 = V_1 \cdot\dfrac{\dfrac{1}{P_2}}{\dfrac{1}{P_1}} = V_1\cdot\dfrac{P_1}{P_2} = 4.72122\times 10^{4}\;\text{L}\times\dfrac{0.995\;\text{atm}}{0.720\;\text{atm}}=6.52\times 10^{4}\;\text{L}$

(3 sig. fig. as in the question.).

See if you get the same result if you hold $T$ constant, change $P$ , and then move on to change $T$ .

6 0

3 years ago

3,621.471 in scientific notation?

tiny-mole [99]

That would be 3.621471•10^3

3 0

3 years ago

A student analyzed an antacid tablet from a bottle containing 120 tablets purchased for $2.79. The mass of the tablet was 1.247

BaLLatris [955]

The cost of one antacid is 2.325 cents per tablet.

<u>Explanation:</u>

As per the question based on the student analysis we know that,

Total antacid tablets in a bottle = 120

Purchase Price of a bottle = $ 2.79

Cost of 1 antacid tablet $=\frac{2.79}{120}$

As we know $1 = 100 cent

The cost of 1 antacid tablet = $\frac{2.79}{120}$ × 100 cents = 2.325 cents/tablet .

Thus we came to know that it costs 2.325 cents/tablet .

5 0

3 years ago

The number of kilograms of water in a human body varies directly as the mass of the body. upper a 93-kg person contains 62 kg o

Bad White [126]

Answer;

56 kg of water

Solution;

Weight = 62 kg of water in 93 kg of a person

W=km

62=93k;

k=2/3

W=(2/3)84

=56 kg

Therefore; a 84-kg person will have 56 kg of water.

4 0

3 years ago

Which type of river is similar to a mature river, but flows more slowly so has less power to change the landscape? Select one: a

Ede4ka [16]

Answer:

The correct option is;

C. Old Age River

Explanation:

Among the three stages of the development of a river, which are the youthful, mature, and old age stages, the old age river is least dynamic

The water is very slow moving with a low gradient and lesser erosive power to alter the landscape which results in the appearance of flood planes

Examples of old age rivers include, lower Ganges, lower Nile, Indus, and Yellow rivers

Old age rivers are characterized by a broad shape, with a wide flood plane, a very gentle gradient and the water current is low.

5 0

4 years ago