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Aliun [14]
3 years ago
11

A gas originally at 27 °c and 1.00 atm pressure in a 2.6 l flask is cooled at constant pressure until the temperature is 11 °c.

the new volume of the gas is
Chemistry
1 answer:
elena55 [62]3 years ago
8 0
Charles law gives the relationship between volume and temperature of gas at constant pressure 
it states that at constant pressure, volume of gas is directly proportional to temperature 
V/T = k
where V - volume T - temperature and k - constant 
\frac{V1}{T1} =  \frac{V2}{T2}
parameters for the first instance are on the left side of the equation and parameters for the second instance are on the right side of the equation
T1 - temperature in Kelvin - 27 °C + 273 = 300 K
T2 - 11 °C + 273 = 284 K
substituting the values in the equation 
2.6 L / 300 K = V / 284 K 
V = 2.46 L
New volume of the gas is 2.46 L
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5 0
3 years ago
Se hace reaccionar 4,00 g de aluminio y 42,00 g de bromo, según la reacción: Al(s)+Br2(l)⟶AlBr3(s) Calcular las moles de AlBr3(s
NeX [460]

Answer:

0.145 moles de AlBr3.

Explanation:

¡Hola!

En este caso, al considerar la reacción química dada:

Al(s)+Br2(l)⟶AlBr3(s)

Es claro que primero debemos balancearla como se muestra a continuación:

2Al(s)+3Br2(l)⟶2AlBr3(s)

Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

n_{AlBr_3}^{por\ Al}=4.00gAl*\frac{1molAl}{27gAl} *\frac{2molAlBr_3}{2molAl}=0.145mol AlBr_3\\\\n_{AlBr_3}^{por\ Br_2}=42.00gr*\frac{1molr}{160g Br_2} *\frac{2molAlBr_3}{3molBr_2}=0.175mol AlBr_3

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.

¡Saludos!

3 0
2 years ago
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6 0
2 years ago
Each student in a class placed a 2.00 g sample of a mixture of Cu and Al in a beaker and placed the beaker in a fume hood. The s
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Answer:

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Explanation:

Equation of the reaction producing Cu(NO₃) is given below:

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mass of copper = number of moles * molar mass

mass of copper = 0.01 mol * 64 g/mol = 0.64 g

Percentage by mass of copper in the 2.00 g sample = (0.64/2.00) * 100%

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