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Bas_tet [7]
3 years ago
15

The size (radius) of an oxygen molecule is about 2.0 ×10−10m. Make a rough estimate of the pressure at which the finite volume o

f the molecules should cause noticeable deviations from ideal-gas behavior at ordinary temperatures (T= 300K ). Assume that deviatons would be noticeable when volume of the gas per molecule equals the volume of the molecule itself.
Chemistry
1 answer:
belka [17]3 years ago
8 0

Answer:

Explanation:

We can calculate the volume  of the oxygen molecule as the radius of oxygen molecule is given as 2×10⁻¹⁰m.

We know that volume=4/3×πr³

volume =4/3×π(2.0×10⁻¹⁰m)³

volume=33.40×10⁻³⁰m³

Volume of oxygen molecule=33.40×10⁻³⁰m³

we know the ideal gas equation as:

PV=nRT

k=R/Na

R=k×Na

PV=n×k×Na×T

n×Na=N

PV=Nkt

p is pressure of gas

v is volume  of gas

T is temperature of gas

N is numbetr of molecules

Na is avagadros number

k is boltzmann constant =1.38×10⁻²³J/K

R is real gas constant

So to calculate pressure using the  formula;

PV=NkT

P=NkT/V

Since there is only one molecule of oxygen so N=1

P=[1×1.38×10⁻²³J/K×300]/[33.40×10⁻³⁰m³

p=12.39×10⁷Pascal

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2.93 mol of MgF2 to grams
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The mass of 1.72 mol of magnesium fluoride is 107 grams.

To determine the mass of 1.72 mol of magnesium fluoride, we first need the chemical formula of magnesium fluoride. Magnesium forms a +2 ion (Mg+2) and fluoride forms a -1 ion (F-1). Since all compounds formed from ions have to be electrically neutral, we need 2 fluoride ions and 1 magnesium ion. Therefore, the formula for magnesium fluoride is MgF2.

Now we need to determine the molar mass of the compound from the molar mass values from the periodic table. Let's use a table to calculate this molar mass.

Molar mass of MgF2

Element Molar Mass (g/mol) Quantity Total (g/mol)

Mg 24.31 1 24.31

F 19.00 2 38.00

Total molar mass of MgF2 = 24.31 g/mol + 38.00 g/mol = 62.31 g/mol

This is the mass of one mole of the substance. If we have 1.72 mols of it, we multiply 1.72 by 62.31.

1.72 mol (62.31 g/mol) = 107 grams

We rounded to 107 to keep the correct number of significant digits in our answer.

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To test the purity of sodium bicarbonate, you dissolve a 3.50g sample in water and add sulfuric acid. if 1.04g of carbon dioxide
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Answer is: the percent purity of the sodium bicarbonate is 56.83 %.

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2. m(NaHCO₃) = 3.50 g

n(NaHCO₃) = m(NaHCO₃) ÷ M(NaHCO₃).

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3. From chemical reaction: n(NaHCO₃) : n(CO₂) = 1 : 1.

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4. the percent purity = 1.04 g/1.83 g  ·100%.

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