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butalik [34]
3 years ago
8

Which best represents the law of conservation of mass?

Chemistry
2 answers:
disa [49]3 years ago
7 0

Answer:

<h2>D. mass of reactants → mass of products </h2>

Explanation:

The only difference between the options are the sign. The sign between the two is an arrow to represent the change.

(I know this because we're learning about it in class and it's an arrow not an equal, greater, or less than sign)

makkiz [27]3 years ago
6 0
Mass of reactants &gt mass of products
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One liter of N (g) at 2.1 bar and two liters of Ar(g) at 3.4 bar are mixed in a 4.0-L 2 flask to form an ideal-gas mixture. Calc
Vika [28.1K]

Answer:

Explanation:

From the information given:

Step 1:

Determine the partial pressure of each gas at total Volume (V) = 4.0 L

So, using:

\text{The new partial pressure for }N_2 \ gas}

P_1V_1=P_2V_2

P_2=\dfrac{P_1V_1}{V_2} \\ \\  P_2=\dfrac{2.1 \ bar \times 1\ L}{4.0 \ L} \\ \\ P_2 = 0.525 \ bar

\text{The new partial pressure for }Ar \ gas}

P_2=\dfrac{P_1V_1}{V_2} \\ \\  P_2=\dfrac{3.4 \ bar \times 2 \ L}{4.0 \ L} \\ \\ P_2 = 1.7 \ bar

Total pressure= P [N_2] + P[Ar] \ \\ \\ . \ \  \ \  \ \  \ \ \ \   \ \ \ \ \ \ \ \ = (0.525 + 1.7)Bar \\ \\ . \ \  \ \  \ \  \ \ \ \   \ \ \ \ \ \ \ \ = 2.225 \ Bar

Now, to determine the final pressure using different temperature; to also achieve this, we need to determine the initial moles of each gas.

According to Ideal gas Law.

2.1  \ bar = 2.07  \ atm \\ \\3.4 \  bar = 3.36 \  atm

For moles N₂:

PV = nRT \\ \\  n = \dfrac{PV}{RT}

n = \dfrac{2.07 \ atm \times 1 \ L }{0.08206 \ L .atm. per. mol. K \times 304 \ K}

n = 0.08297 \ mol  \ N_2

For moles of Ar:

PV = nRT \\ \\  n = \dfrac{PV}{RT}

n = \dfrac{3.36 \ atm \times 1 \ 2L }{0.08206 \ L .atm. per. mol. K \times 377 \ K}

n = 0.2172 \ mol  \ Ar

\mathtt{total \  moles = moles \ of \  N_2 + moles  \ of \ Ar}

=0.08297 mol + 0.2037 mol \\                   = 0.2867 mol gases

Finally;

The final pressure of the mixture is:

PV = nRT \\ \\ P = \dfrac{nRT}{V} \\ \\ P = \dfrac{0.2867 \ mol \times 0.08206 \ L .atm/mol .K\times 377 K}{4.0 \ L}

P = 2.217 atm

P ≅ 2.24 bar

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