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Musya8 [376]
3 years ago
13

How many grams of chlorine gas react when 266.28 grams of aluminum reacts

Chemistry
1 answer:
erma4kov [3.2K]3 years ago
8 0

Answer:

524.82 g Cl2

Explanation:

Balanced Equation:

2Al(s) + 3Cl(g) ⇒ 2AlCl (s)

Al: 26.982 g/mol     Cl: 35.453 g/mol

266.28g Al x 1mol Al/ 26.982g Al x 3mol Cl2/ 2mol Al x

35.453g Cl2/ 1mol Cl2 = 524.82 g Cl2

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Which of these equations is balanced? a. H2SO4 + 2Al → Al2(SO4)3 + H2 b. 2KCl + Pb(NO3)2 → 2KNO3 + PbCl2
Lapatulllka [165]
The answer is B, you just check if it is the same on the left and right side

A:
Left side - Right side
2xH - 2xH
1xS - 3xS
4xO - 12xO
2xAl - 2xAl

Therefore A is not correct

B:
Left side - right side
2xK - 2xK
1xCl - 1xCl
1xPb - 1xPb
2xN - 2xN
6xO - 6xO

B is therefore correct as both sides add up
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3 years ago
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OleMash [197]
The answer is C . Watery feel
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Which element is the most reactive?sodiumnickelcarbonoxygen.
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The answer is sodium which is the element is the most reactive.

7 0
2 years ago
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

7 0
3 years ago
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