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3 years ago

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A hot air balloon, when fully inflated, contains 2,800,000 Liter of air. If each mole of air occupies 22.4 L of at a temperature

of 120 Celsius, at what pressure is the air at when the balloon is fully inflated?

1 answer:

emmainna [20.7K]3 years ago

3 0

Answer:

1.44 atm

Explanation:

Step 1:

We'll begin by calculating the number of mole in 2,800,000 Liter of air.

I mole of air occupy 22.4L.

Therefore, Xmol of air will occupy 2800000L i.e

Xmol of air = 2800000/22.4

Xmol of air = 125000 moles

Step 2:

Determination of the pressure when the balloon is fully inflated .

This can be obtained as follow:

Number of mole (n) of air = 125000 moles

Volume (V) = 2800000 L

Temperature (T) = 120°C = 120°C + 273 = 393K

Gas constant (R) = 0.082atm.L/Kmol

Pressure (P) =.?

PV = nRT

Divide both side V

P= nRT/V

P= (125000x0.082x393) / 2800000

P = 1.44 atm

Therefore, the pressure of the air when the balloon is fully inflated is 1.44 atm

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Strontium, a group 2 element with an electron configuration of 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 , will form a _____

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Answer:

Strontium, a group 2 element with an electron configuration of 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² , will form a +2 ion in a chemical bond.

Explanation:

An ionic bond is produced between metallic and non-metallic atoms, where electrons are completely transferred from one atom to another. During this process, one atom loses electrons and another one gains them, forming ions. Usually, the metal gives up its electrons forming a cation to the nonmetal element, which forms an anion.

So an ion is an atom or group of atoms that have an electric charge. A cation is an ion that has a positive charge.

The alkali metals in the earth (Elements IIA) lose two electrons to form a 2+ cation. Strontium is an alkali metal that loses the two electrons belonging to 5s², forming the +2 ion.

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3 years ago

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2 years ago

Enough of a monoprotic acid is dissolved in water to produce a 1.64 M solution. The pH of the resulting solution is 2.82 . Calcu

Lady bird [3.3K]

Answer:

Ka = 1.39x10⁻⁶

Explanation:

A monoprotic acid, HX, will be in equilibrium in an aqueous medium such as:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

<em>Where Ka is:</em>

Ka = [H⁺] [X⁻] / [HX]

<em>Where [] is the molar concentration in equilibrium of each specie. </em>

The equilibrium is reached when some HX reacts producing H+ and X-, that is:

[HX] = 1.64M - X

[H⁺] = X

[X⁻] = X

As pH is 2.82 = -log [H⁺]:

[H⁺] = 1.51x10⁻³M:

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[X⁻] = 1.51x10⁻³M

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4 years ago

How many formula units make up 24.2 g of magnesium chloride (MgCl2)?<br><br> Help!!

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Answer:

Approximately $1.53 \times 10^{23}$ formula units ( $0.254\; \rm mol$ ).

Explanation:

Refer to a modern periodic table for the relative atomic mass of magnesium ( $\rm Mg$ ) and chlorine ( $\rm Cl$ ):

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In other words, the mass of $1\; \rm mol$ of $\rm Mg$ atoms would be (approximately) $24.305\; \rm g$ .

Likewise, the mass of $1\; \rm mol$ of $\rm Cl$ atoms would be approximately $35.45\; \rm g$ .

One formula unit of the ionic compound $\rm MgCl_{2}$ includes exactly as many atoms as there are in the given formula. The formula mass of a compound is the mass of $1\; \rm mol$ of the formula units of this compound.

The formula $\rm MgCl_{2}$ includes one $\rm Mg$ atom and two $\rm Cl$ atoms.

Hence, every formula unit of $\rm MgCl_{2} \!$ would include the same number of atoms: one $\rm Mg\!$ atom and two $\rm Cl\!$ atoms. There would be $1\; \rm mol$ of $\rm Mg$ atoms and $2\; \rm mol$ of $\rm Cl$ atoms in $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units.

Thus, the mass of $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units would be equal to the mass of $1\; \rm mol$ of $\rm Mg$ atoms plus the mass of $2\; \rm mol$ of $\rm Cl$ atoms. (The mass of $1\; \rm mol\!\!$ of each atom could be found from the relative atomic mass of each element.)

$\begin{aligned}& M({\rm MgCl_{2}}) \\ =\; & 24.305\; {\rm g \cdot mol^{-1}} + 2\times {\rm 35.45 \; \rm g \cdot mol^{-1}} \\ =\; & 95.205\; \rm g \cdot mol^{-1}\end{aligned}$ .

In other words, the formula mass of $\rm MgCl_{2}$ is $95.205\; \rm g \cdot mol^{-1}$ .

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$\begin{aligned}n &= \frac{m({\rm MgCl_{2}})}{M({\rm MgCl_{2}})} \\ &= \frac{24.2\; \rm g}{95.205\; \rm g\cdot mol^{-1}} \\ & \approx 0.254\; \rm mol\end{aligned}$ .

Multiple $n$ by Avogadro's Number $N_{A} \approx 6.022 \times 10^{23}\; \rm mol^{-1}$ to estimate the number of formula units in $0.254\; \rm mol$ :

$\begin{aligned}N &= n \cdot N_{A} \\ &\approx 0.254\; \rm mol \times 6.022 \times 10^{23}\; \rm mol^{-1} \\ &\approx 1.53\times 10^{23}\end{aligned}$ .

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