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3 years ago

La masa de una olla es de 300g y contiene 90% de aluminio. Hallar el número de moles de aluminio de la olla. P.A.(Al= 27)

Chemistry

1 answer:

Neko [114]3 years ago

6 0

Explanation:

The mass of a pot is 300g and contains 90% aluminum. Find the number of moles of aluminum in the pot. P.A. (Al = 27)

The mass of aluminum present in the pot is:

$300 g * 90/100\\=270 g$

Hence, in the given pot 270g Al is present.

$Number of moles of Al=\frac{given mass ofAl}{its molar mass}$

The gram atomic mass of Al -27 g/mol

Given the mass of Al is 270 g

Substitute these values in the above formula:

$Number of moles of Al=\frac{given mass ofAl}{its molar mass}\\=\frac{270 g}{27 g} \\=10.0 mol$

Answer is 10.0 mol of Al is present.

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If the solubility of a gas is 10.5 g/L at 525 kPa pressure, what is the solubility of the gas when the pressure is 225 kPa? Show

Talja [164]

Answer:

4.5 g/L.

Explanation:

To solve this problem, we must mention Henry's law.

Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.

It can be expressed as: P = KS,

P is the partial pressure of the gas above the solution.

K is the Henry's law constant,

S is the solubility of the gas.

At two different pressures, we have two different solubilities of the gas.

<em>∴ P₁S₂ = P₂S₁.</em>

P₁ = 525.0 kPa & S₁ = 10.5 g/L.

P₂ = 225.0 kPa & S₂ = ??? g/L.

∴ S₂ = P₂S₁/P₁ = (225.0 kPa)(10.5 g/L) / (525.0 kPa) = 4.5 g/L.

8 0

3 years ago

How many moles of mercury, Hg, are there in 1.30 x 10^7 atoms of murcury? PLEASE HELP, IT'S URGENT! Thank you! (:

vodka [1.7K]

1 mole Hg ---------------- 6.02x10²³ atoms
?? ------------------------- 1.30 x10⁷ atoms

1.30x10⁷ x 1 / 6.02x10²³ =

= 1.30x10⁷ / 6.02x10²³ => 2.159x10⁻¹⁷ moles

hope this helps!

7 0

3 years ago

A sample of propane (C3H8) has a mass of 0. 47 g. The sample is burned in a bomb calorimeter that has a mass of 1. 350 kg and a

Nady [450]

The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.

The specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius.

The specific heat has been expressed as:

$q=mc\Delta T$

<h3 /><h3>Computation for the heat absorbed</h3>

The iron and calorimeter are in side the closed system. Thus, the energy released by the sample, has been equivalent to the energy absorbed by the calorimeter.

$q_{released}=q_{absorbed}\\
q_{released}=m_{calorimeter}\;c_{calorimeter}\;\Delta T$

The given mass of calorimeter has been, $m_{calorimeter}=1350\;\rm g$

The specific heat of the calorimeter has been, $c_{calorimeter}=5.82\;\rm J/g^\circ C$

The change in temperature of the calorimeter has been, $\Delta T=2.87^\circ \rm C$

Substituting the values for heat released:

$q_{released}= 1350\;\text g\;\times\;5.82\;\text J/\text g^\circ \text C\;\times\;2.87^\circ \text C\\
q_{released}=22,549.5\;\text J\\
q_{released}}=22.54\;\rm kJ$