describe three characteristics that trees used to build treehouses should have. why are these characteristics essential

We have air (21% O2 and 79% N2) at 23 bar and 30 C. 4. What is the ideal molar volume (m^3/kmol)? a. b. What is the Z factor? Wh

k0ka [10]

Answer:

The ideal molar volume is $\frac{V}{n} =V_z= 0.001095 \ m^3/mol$

The Z factor is $Z = 0.09997$

The real molar volume is $\frac{V_r}{n} = V_k= 0.0001095\ \frac{m^3}{mol}$

Explanation:

From the question we are told that

The pressure is $P = 23 \ bar = 23 *10^5 Pa$

The temperature is $T = 30 ^ oC = 303 \ K$

According to the ideal gas equation we have that

$PV = nRT$

=> $\frac{V}{n}=V_z= \frac{RT}{P}$

Where $\frac{V}{n }$ is the molar volume and R is the gas constant with value

$R = 8.314 \ m^3 \cdot Pa \cdot K^{-1}\cdot mol^{-1}$

substituting values

$\frac{V}{n} =V_z= \frac{ 8.314 * 303}{23 *10^{5}}$

$\frac{V}{n} =V_z= 0.001095 \ m^3/mol$

The compressibility factor of the gas is mathematically represented as

$Z = \frac{P * V_z}{RT}$

substituting values

$Z = \frac{23 *10^{5} * 0.001095}{8.314 * 303}$

$Z = 0.09997$

Now the real molar volume is evaluated as

$\frac{V_r}{n} = V_k= \frac{Z * RT }{P}$

substituting values

$\frac{V_r}{n} = V_k= \frac{0.09997 * 8.314 * 303}{23 *10^{5}}$

$\frac{V_r}{n} = V_k= 0.0001095\ \frac{m^3}{mol}$

8 0

3 years ago

Molybdenum can form a wide series of halide compounds, including four different fluoride compounds. The percent by mass of molyb

azamat

The formula and names of the compounds are:

1. Formula of compound => MoF₃

Name of compound => Molybdenum trifluoride

2. Formula of compound => MoF₄

Name of compound => Molybdenum tetrafluoride

3. Formula of compound => MoF₅

Name of compound => Molybdenum pentafluoride

4. Formula of compound => MoF₆

Name of compound => Molybdenum hexafluoride

1. Determination of the name and formula of the molybdenum fluoride having 63.0% of molybdenum.

Molybdenum (Mo) = 63.0%

Fluorine (F) = 100 – 63 = 37%

<h3>Formula =? </h3>

Mo = 63.0%

F = 37%

Divide by their molar mass

Mo = 63.0 / 96 = 0.656

F = 37 / 19 = 1.947

Divide by the smallest

Mo = 0.656 / 0.656 = 1

F = 1.947 / 0.656 = 3

Therefore,

Formula of compound => MoF₃

Name of compound => Molybdenum trifluoride

2. Determination of the name and formula of the molybdenum fluoride having 56.0% of molybdenum.

Molybdenum (Mo) = 56.0%,

Fluorine (F) = 100 – 56 = 44%

<h3>Formula =? </h3>

Mo = 56%

F = 44%

Divide by their molar mass

Mo = 56 / 96 = 0.583

F = 44 / 19 = 2.316

Divide by the smallest

Mo = 0.583 / 0.583 = 1

F = 2.316 / 0.583 = 4

Therefore,

Formula of compound => MoF₄

Name of compound => Molybdenum tetrafluoride

3. Determination of the name and formula of the molybdenum fluoride having 50.0% of molybdenum.

Molybdenum (Mo) = 50.0%,

Fluorine (F) = 100 – 50 = 50%

<h3>Formula =? </h3>

Mo = 50%

F = 50%

Divide by their molar mass

Mo = 50 / 96 = 0.520

F = 50 / 19 = 2.632

Divide by the smallest

Mo = 0.520 / 0.520 = 1

F = 2.632 / 0.520 = 5

Therefore,

Formula of compound => MoF₅

Name of compound => Molybdenum pentafluoride

4. Determination of the name and formula of the molybdenum fluoride having 46.0% of molybdenum.

Molybdenum (Mo) = 46.0%,

Fluorine (F) = 100 – 46 = 54%

<h3>Formula =? </h3>

Mo = 46%

F = 54%

Divide by their molar mass

Mo = 46 / 96 = 0.479

F = 54 / 19 = 2.842

Divide by the smallest

Mo = 0.479 / 0.479 = 1

F = 2.842 / 0.479 = 6

Therefore,

Formula of compound => MoF₆

Name of compound => Molybdenum hexafluoride

Learn more: brainly.com/question/11185156

7 0

2 years ago

Mass of CuSO4● XH2O (g) = 1.6 grams BEFORE HEATING

Leokris [45]

0.06105 moles is the number of moles of water lost.

<h3>What is molar mass?</h3>

Molar mass is defined as the mass in grams of one mole of a substance.

Given data:

Mass of water = 1.1g

Molar mass water = 18.016 g/mol

Moles of water =?

These quantities are related by the following equation;

Moles = $\frac{Mass}{Molar \;mass}$

Substituting the values of the quantities and solving for moles, we have;

Moles = $\frac{1.1 }{18.016}$ = 0.06105 moles.

Hence, the 0.06105 moles is the number of moles of water lost.

Learn more about molar mass here:

brainly.com/question/12127540

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7 0

2 years ago

Calculate the concentration of so42− ions in a 0.010 m aqueous solution of sulfuric acid. express your answer to four decimal pl

dimulka [17.4K]

<span>Answer: 0.00649M

The question is incomplete,
</span>

<span>You are told that the first ionization of the sulfuric acid is complete and the second ionization of the sulfuric acid has a constant Ka₂ = 0.012
</span>
<span>
With that you can solve the question following these steps"
</span>

<span>1) First ionization:
</span>
<span>
H₂SO₄(aq) --> H⁺ (aq) + HSO₄⁻ (aq)

Under the fully ionization assumption the concentration of HSO4- is the same of the acid = 0.01 M

2) Second ionization
</span>

<span>HSO₄⁻ (aq) ⇄ H⁺ + SO₄²⁻ with a Ka₂ = 0.012
</span>

<span>Do the mass balance:
</span>

<span><span> HSO₄⁻ (aq) H⁺ SO₄²⁻</span>
</span>
<span /><span /><span> 0.01 M - x x x

</span><span>Ka₂ = [H⁺] [SO₄²⁻] / [HSO₄⁻]</span>
<span /><span>
=> Ka₂ = (x²) / (0.01 - x) = 0.012
</span><span />

<span>3) Solve the equation:

</span><span>x² = 0.012(0.01 - x) = 0.00012 - 0.012x</span>
<span /><span>
x² + 0.012x - 0.0012 = 0
</span><span />

<span>Using the quadratic formula: x = 0.00649
</span><span />

<span>So, the requested concentratioN is [SO₄²⁻] = 0.00649M</span>

4 0

3 years ago

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For the following electrochemical cell, determine the oxidizing agent, reducing agent, and whether the cell is a voltaic cell or

lyudmila [28]

1637947 d hdidmbdjrn

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