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3 years ago

How many moles of oxygen are needed to completely react with 9.5 g of sodium

1 answer:

8 0

4 Na + O₂ = 2 Na₂O

4* 23 g Na --------> 16 g O₂
9.5 g Na ------------> ?

Mass of O₂ = 9.5 * 16 / 4 * 23

Mass = 152 / 92

Mass = 1.6521 g of O₂

Molar mass O₂ = 16.0 g/mol

1 mole O₂ ------------ 16.0 g
? mole O₂ ------------ 1.6521 g

mole O₂ = 1.6521 * 1 / 16.0

≈ 0.10325 moles of O₂

hope this helps!

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Gasoline is a mixture of hydrocarbons, a major component of which is octane, CH3CH2CH2CH2CH2CH2CH2CH3. Octane has a vapor pressu

Nitella [24]

Answer:

$\Delta \:H_{vap}=40383.88\ J/mol$

Explanation:

The expression for Clausius-Clapeyron Equation is shown below as:

$\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c$

Where,

P is the vapor pressure

ΔHvap is the Enthalpy of Vaporization

R is the gas constant (8.314×10⁻³ kJ /mol K)

c is the constant.

For two situations and phases, the equation becomes:

$\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)$

Given:

$P_1$ = 13.95 torr

$P_2$ = 144.78 torr

$T_1$ = 25°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25 + 273.15) K = 298.15 K

$T_1$ = 298.15 K

$T_2$ = 75°C = 348.15 K

So,

$\ln \:\left(\:\frac{13.95}{144.78}\right)\:=\:\frac{\Delta \:H_{vap}}{8.314}\:\left(\:\frac{1}{348.15}-\:\frac{1}{298.15}\:\right)$

$\Delta \:H_{vap}=\ln \left(\frac{13.95}{144.78}\right)\frac{8.314}{\left(\frac{1}{348.15}-\frac{1}{298.15}\right)}$

$\Delta \:H_{vap}=\frac{8.314}{\frac{1}{348.15}-\frac{1}{298.15}}\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)$

$\Delta \:H_{vap}=\left(-\frac{863000.86966\dots }{50}\right)\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)$

$\Delta \:H_{vap}=40383.88\ J/mol$

4 0

2 years ago

In a rocket motor fueled with butane, how many kg of liquid oxygen should be provided with each kilogram of butane to provide fo

Travka [436]

Answer

So this is the reaction that happens.
C4H10 + O2 = CO2 + H2O 

Balanced, it is 
2C4H10 + 8O2 = 8CO2 + 10H2O 

Given 1 kg or 1000 g of butane, use stoichiometry aka factor labeling aka conversions and mole ratios to get to grams of oxygen. 

I'll do an example. Let's form water. Hydrogen is diatomic too. 
2H2 + O2 = 2H2O 

Given 1000 g of Hydrogen, I need to know how many grams of oxygen to use. To convert grams to moles,
I know that 1 mol of H2 equals 2.02 g. Then, for every mole of O2, there are 2 moles of H2. Then converting moles of O2 to grams, I know that one mole of it equals 32 grams. 
[1000 g H2] x [1 mol H2/2.02 g H2] x [1 mol O2/2 mol H2] x [32 g O2/1 mol O2] 

My answer would be 7.9 kg

3 0

3 years ago

It is a large body of air that has the same properties as the Earth's surface over which it develops.what is it?

Debora [2.8K]

The answer is AIR MASS.

4 0

3 years ago

A generator is not required to generate electrical energy when which of the following energy sources is used?

Anon25 [30]

Answer:

think it's C but my gut is telling me A

Explanation:

5 0

2 years ago

Read 2 more answers

Select the correct value for the indicated bond angle in each of the following compounds: O-S-O angle of SO2 F-B-F angle of BF3

Stels [109]

Answer:

(A) O-S-O bond angle of SO₂ molecule = 119°

(B) F-B-F bond angle of BF₃ molecule = 120°

(D) O-C-O bond angle of CO₂ molecule = 180°

(E) F-P-F bond angle of PF₃ molecule = 96.3°

(F) H-C-H bond angle of CH₄ molecule = 109.5°

Explanation:

Bond angle refers to the angle between two adjacent chemical bonds in a molecule. The bond angle is different for different molecular geometry.

The Valence shell electron pair repulsion theory predicts the molecular geometry and shape of the given molecule by the number of lone pairs on central atom and number of atoms bonded to central atom.

(A) SO₂ molecule

The number of atoms bonded to S = 2

Number of lone pairs on S = 1

Therefore, the shape of SO₂ molecule is bent and the O-S-O bond angle is 119°.

(B) BF₃ molecule

The number of atoms bonded to B = 3

Number of lone pairs on B = 0

Therefore, the shape of BF₃ molecule is trigonal planar and the F-B-F bond angle is 120°.

(C) SCl₂ molecule

The number of atoms bonded to S = 2

Number of lone pairs on S = 2

Therefore, the shape of SCl₂ molecule is bent and the Cl-S-Cl bond angle is 103°.

(D) CO₂ molecule

The number of atoms bonded to C = 2

Number of lone pairs on C = 0

Therefore, the shape of CO₂ molecule is linear and the O-C-O bond angle is 180°.

(E) PF₃ molecule

The number of atoms bonded to P = 3

Number of lone pairs on P = 1

Therefore, the shape of PF₃ molecule is trigonal pyramidal and the F-P-F bond angle is 96.3°.

(F) CH₄ molecule

The number of atoms bonded to C = 4

Number of lone pairs on C = 0

Therefore, the shape of CH₄ molecule is tetrahedral and the H-C-H bond angle is 109.5°.

7 0

3 years ago