When a -COOH group replaces a hydrogen in a hydrocarbon the result is a/an

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

What is the energy of a photon that emits a light of frequency 6.42 x 1014 Hz?<br><br>

svetlana [45]

Answer:

Option B. 4.25×10¯¹⁹ J

Explanation:

From the question given above, the following data were obtained:

Frequency (f) = 6.42×10¹⁴ Hz

Energy (E) =?

Energy and frequency are related by the following equation:

Energy (E) = Planck's constant (h) × frequency (f)

E = hf

With the above formula, we can obtain the energy of the photon as follow:

Frequency (f) = 6.42×10¹⁴ Hz

Planck's constant (h) = 6.63×10¯³⁴ Js

Energy (E) =?

E = hf

E = 6.63×10¯³⁴ × 6.42×10¹⁴

E = 4.25×10¯¹⁹ J

Thus, the energy of the photon is 4.25×10¯¹⁹ J

3 0

3 years ago

Read 2 more answers

Need help this question QUICK

tino4ka555 [31]

The answer is number 2

4 0

4 years ago

Limestone (CaCO3) is decomposed by heating to quicklime (CaO) and carbon dioxide. Calculate how many grams of quicklime can be p

Harman [31]

Answer: 2800 g

Explanation:

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.