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1 year ago

Question 3 options: how many moles of argon atoms are present in 11.2 l of argon gas at stp?

1 answer:

5 0

It is known that at if a gas is identified to be as ideal gas then, at STP, the volume of 1 mole of that gas is equal to 22.4 L. Using this fact and the given volume in this item, we determine the number of moles as that below.

n = (11.2 L of argon)(1 mol argon / 22.4 L of argon)

<span> n = 0.5 moles</span>

<span>Answer: 0.5 moles </span>

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En un balneario necesitan calentarse 1 millón de litros de agua anuales, subiendo la temperatura desde 15 ºC a 50 ºC y para ello

MAXImum [283]

Answer:

a) $m_{CH_4}=2630kg$

b) 1657 €

Explanation:

Hola,

a) En este problema, vamos a considerar el millón de litros de agua anuales, ya que con ellos podemos calcular el calor requerido para dicho calentamiento, sabiendo que la densidad del agua es de 1 kg/L:

$Q_{H_2O}=m_{H_2O}Cp(T_2-T_1)=1x10^6LH_2O*\frac{1kgH_2O}{1LH_2O}*4.18\frac{kJ}{kg\°C}(50-15) \°C\\Q_{H_2O}=146.3x10^6kJ$

Luego, usamos la entalpía de combustión del metano para calcular su requerimiento en kilogramos, sabiendo que la energía ganada por el agua, es perdida por el metano:

$Q_{H_2O}=-Q_{CH_4}=-146.3x10^5kJ=m_{CH_4}\Delta _cH_{CH_4}$

$m_{CH_4}= \frac{Q_{CH_4}}{\Delta _cH_{CH_4}} =\frac{-146.3x10^5kJ}{-890kJ/molCH_4} *\frac{16gCH_4}{1molCH_4} \\\\m_{CH_4}=2630112.36g=2630kg$

b) En este caso, consideramos que a condiciones normales de 1 bar y 273 K, 1 metro cúbico de metano cuesta 0,45 €, con esto, calculamos las moles de metano a dichas condiciones:

$n_{CH_4}=\frac{PV}{RT}=\frac{1atm*1000L}{0.082\frac{atm*L}{mol*K}*273K} =44.67mol$

Con ello, los kilogramos de metano que cuestan 0,45 €:

$44.67molCH_4*\frac{16gCH_4}{1molCH_4}*\frac{1kg}{1000g} =0.715kgCH_4$

Luego, aplicamos la regla de tres:

0.715 kg ⇒ 0.45 €

2630 kg ⇒ X

X = (2630 kg x 0.45 €) / 0.715 kg

X = 1657 €

Regards.

3 0

2 years ago

30. how many grams of boric acid, b(oh)3 (fm 61.83), should be used to make 2.00 l of 0.050 0 m solution? what kind of fl ask is

grigory [225]

Answer:

Mass = 6.183 g

Solution:

Step 1: Calculate number of moles of Boric acid using following formula,

Molarity = Moles ÷ Volume

Solving for Moles,

Moles = Molarity × Volume

Putting Values,

Moles = 0.05 mol.L⁻¹ × 2.0 L

Moles = 0.1 mol

Step 2: Calculate Mass of Boric Acid using following formula,

Moles = Mass ÷ M.mass

Solving for Mass,

Mass = Moles × M.mass

Putting values,

Mass = 0.1 mol × 61.83 g.mol⁻¹

Mass = 6.183 g

Flask used to prepare this solution is called as Volumetric flask. Take 2 L volumetric flask, add 6.183 g of Boric acid and fill it to the mark with distilled water.