Chemistry, 28.02.2020 03:16 makayla7635

A 3.0 L flask containing helium at 145 mmHg is connected by a closed valve to a 2.0 L flask containing argon at 355 mmHg. When the valve is opened and the gases are allowed to mix equally in the two flasks, what is the total pressure (in mmHg) in the two connected flasks after mixing ?

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Assuming that both helium and argon act like ideal gases, the total pressure after mixing would be approximately $229\; \rm mmHg$ .

Explanation:

By the ideal gas equation, $P\cdot V = n \cdot R \cdot T$ , where

is the pressure of the sample.

is the volume of the container.

is the number of moles of gas particles in the sample.

is the ideal gas constant.

is the temperature of the sample.

Rewrite to obtain:

$\displaystyle n = \frac{P \cdot V}{R\cdot T}$ , and $\displaystyle P = \frac{n \cdot R \cdot T}{V}$ .

Assume that the two samples have the same temperature, . Also, assume that mixing the two gases did not affect the temperature.

Apply the equation $\displaystyle n = \frac{P \cdot V}{R\cdot T}$ to find the number of moles of gas particles in each container:

In the helium container, $V = 3.0\; \rm L$ and $P = \rm 145\; mmHg$ . Hence, $\displaystyle n_1 = \frac{P\cdot V}{R \cdot T} = \frac{(3.0\; \text{L}) \cdot (145\; \text{mmHg})}{R\cdot T}$ .In the argon container, $V = 2.0\; \rm L$ and $P = 355\; \rm mmHg$ . Hence, $\displaystyle n_2 = \frac{P\cdot V}{R \cdot T} = \frac{(2.0\; \text{L}) \cdot (355\; \text{mmHg})}{R\cdot T}$ .

After mixing, $V = 2.0 + 3.0 = 5.0\; \rm L$ . Assuming that temperature stays the same.

$\displaystyle n_1 + n_2 = \frac{(3.0\; \text{L}) \cdot (145\; \text{mmHg})}{R\cdot T} + \frac{(2.0\; \text{L}) \cdot (355\; \text{mmHg})}{R\cdot T}$ .

Apply the equation $\displaystyle P = \frac{n \cdot R \cdot T}{V}$ to find the pressure after mixing.

$\begin{aligned}P &= \displaystyle \frac{\displaystyle \displaystyle \left(\frac{(3.0\; \text{L}) \cdot (145\; \text{mmHg})}{R\cdot T} + \frac{(2.0\; \text{L}) \cdot (355\; \text{mmHg})}{R\cdot T}\right) \cdot R \cdot T}{5.0\; \rm L} \\ &= \frac{3.0\; \rm L \times 145\; \rm mmHg + 2.0\; \rm L \times 355\; \rm mmHg}{5.0\; \rm L} \\ &\approx 229\; \rm mmHg\end{aligned}$ .

Answer from: spoo262005

The total pressure is 229 atm

Explanation:

Step 1: Data given

Volume of helium flask = 3.0 L

Pressure helium flask = 145 mm Hg

Volume of argon flask = 2.0 L

Pressure argon flask = 355 mmHg

total volume = 5.0 L

Step 2: Partial pressure helium

pHe = 145 *(3/5) = 87.0 atm

Step 3: Calculate pressure argon

pAr = 355*(2/5) = 142.0 atm

Step 4: Calculate total pressure

Total pressure = 87.0 + 142.0 atm

Total pressure = 229 atm

The total pressure is 229 atm

Answer from: Quest

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.

Answer from: Quest

the link still does not work even when adding it to the url bar

explanation:

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