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Develop expressions for the mole fraction of reacting species functions of the reaction coordinate for: A system initially containing 2 mol NH3 and 5 mol O2 and undergoing the reaction: 4NH3 (g) + 5O2 (g) ® 4NO (g) + 6 H20 (g) A system initially containing 3 mol NO2, 4 mol NH3, and 1 mol N2 and undergoing the reaction: 6NO2 (g) + 8NH3 (g) ® 7N2 (g) +12H2O (g)

Answers

Answer 1

Answer:

Answer:

Individual mole fractions of all the species of the all reaction is as follows.

(a)

$y_{NH_(3)}=(2+(-4)\epsilon)/(7+\epsilon)$

$y_{O_(2)}=(5+(-5)\epsilon)/(7+\epsilon)$

$y_(NO)=(0+(4)\epsilon)/(7+\epsilon)=(4\epsilon)/(7+\epsilon)$

$y_{H_(2)O}=(0+6\epsilon)/(7+\epsilon)$

(b)

$y_{H_(2)S}=(3+(-2)\epsilon)/(8- \epsilon)$

$y_{O_(2)}=(5+(-3)\epsilon)/(8- \epsilon)$

$y_{H_(2)O}=(2\epsilon)/(8- \epsilon)$

$y_{SO_(2)}=(2\epsilon)/(8- \epsilon)$

(c)

$y_{NO_(2)}=(3+(-6)\epsilon)/(8+5\epsilon)$

$y_{NH_(3)}=(4+(-8)\epsilon)/(8+5\epsilon)$

$y_{N-{2}}=(1+7\epsilon)/(8+5 \epsilon)$

$y_{H_(2)O}=(12\epsilon)/(8+5\epsilon)$

Explanation:

(a)

Initial number of moles of $NH_(3)$ and $O_(2)$ are 2 mol and 5 mol respectively.

The given chemical reaction is as follows.

$4NH_(3)(g)+5O_(2)(g)\rightarrow 4NO(g)+6H_(2)O$

The stoichiometric numbers are as follows.

$v_{NH_(3)}=-4$

$v_{O_(2)}=-5$

$v_(NO)=4$

$v_{H_(2)O}=6$

The total number of moles initially present -7

$\Sigma v_(i)\epsilon = (-4-5+4+6)= 1\epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{NH_(3)}=(2+(-4)\epsilon)/(7+\epsilon)$

$y_{O_(2)}=(5+(-5)\epsilon)/(7+\epsilon)$

$y_(NO)=(0+(4)\epsilon)/(7+\epsilon)=(4\epsilon)/(7+\epsilon)$

$y_{H_(2)O}=(0+6\epsilon)/(7+\epsilon)$

(b)

Initial number of moles of $H_(2)S$ and $O_(2)$ are 3 mol and 5 mol respectively.

The given chemical reaction is as follows.

$2H_(2)S(g)+3O_(2)(g)\rightarrow 2H_(2)O(g)+2SO_(2)$

The stoichiometric numbers are as follows.

$v_{H_(2)S}=-2$

$v_{O_(2)}=-3$

$v_{H_(2)O}=2$

$v_{SO_(2)=2$

The total number of moles initially present -8

$\Sigma v_(i)\epsilon = (-2-3+2+2)= - \epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{H_(2)S}=(3+(-2)\epsilon)/(8- \epsilon)$

$y_{O_(2)}=(5+(-3)\epsilon)/(8- \epsilon)$

$y_{H_(2)O}=(2\epsilon)/(8- \epsilon)$

$y_{SO_(2)}=(2\epsilon)/(8- \epsilon)$

(c)

Initial number of moles of $NO_(2)$ , $NH_(3)$ and $N_(2)$ are 3 mol,4 mol and 1 mol respectively.

The given chemical reaction is as follows.

$6NO_(2)(g)+8NH_(3)(g)\rightarrow 7N_(2)(g)+12H_(2)O$

The stoichiometric numbers are as follows.

$v_{NO_(2)}=-6$

$v_{NH_(3)}=-8$

$v_{N_(2)}=7$

$v_{H_(2)O}=12$

The total number of moles initially present -8

$\Sigma v_(i)\epsilon = (-6-8+7+12)= 5 \epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{NO_(2)}=(3+(-6)\epsilon)/(8+5\epsilon)$

$y_{NH_(3)}=(4+(-8)\epsilon)/(8+5\epsilon)$

$y_{N-{2}}=(1+7\epsilon)/(8+5 \epsilon)$

$y_{H_(2)O}=(12\epsilon)/(8+5\epsilon)$

Answer 2

Answer:

Final answer:

Expressions for the mole fractions of reacting species are determined using stoichiometry and the initial molar amounts, taking into account the stoichiometric coefficients of the chemical reactions.

Explanation:

To develop expressions for the mole fraction of reacting species as functions of the reaction coordinate for the given systems, we will examine each reaction individually. For the reaction 4NH3 (g) + 5O2 (g) ® 4NO (g) + 6 H2O (g), we can use stoichiometry to correlate the molar amounts of each species with reaction progress. Given the initial amounts, we will track how the molar amount changes for each mole of NH3 reacted.

Starting with 2 mol NH3 and 5 mol O2, the mole ratio from NH3 to NO and H2O is 1:1 and 1:1.5, respectively. The mole ratio from NH3 to O2 is 4:5. If x moles of NH3 react, the mole fractions for each species at any point in the reaction can be expressed as follows:

Mole fraction of NH3: (2 - x)/(Total moles)
Mole fraction of O2: (5 - 5x/4)/(Total moles)
Mole fraction of NO: (4x)/(Total moles)
Mole fraction of H2O: (6x)/(Total moles)

Note that 'Total moles' is the sum of the ongoing moles of all species. The mole fractions must always add up to 1 at any point during the reaction.

For the second reaction 6NO2 (g) + 8NH3 (g) ® 7N2 (g) +12H2O (g), with initial amounts of 3 mol NO2, 4 mol NH3, and 1 mol N2, similar steps are taken. For every mole of NH3 reacted, the corresponding changes in molar amounts can be calculated from the stoichiometry of the balanced equation.

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Which is the weakest type of intramolecular force/bond?a. Polar covalent b. Ionic c. Metallic d. Nonpolar covalent

Answers

Answer:

Non polar covlant

Explanation:

What is the percent composition of nitrogen in sodium nitride

Answers

The percent composition of nitrogen in sodium nitride is 16.86%

What is sodium nitride?

Sodium nitride is an inorganic compound that is used in preservatives and antidote for cyanide poisoning.

The molar mass of Na3N is 83 M

The mass of Na3N is 14 g

$% \;of \;N = (mass of N)/( molar mass of Na3N ) * 100$

$\% \;of \;N = (mass\; of \;N)/(molar\; mass\; of\; Na3N) * 100\n\n\n\% \;of \;N = (14)/(83) * 100 = 16.86\%$

Thus, the percent composition of nitrogen in sodium nitride is 16.86%.

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Answer:

16.86%

Explanation:

Na3N is sodium nitride.

% of N = mass of N/ molar mass of Na3N *100

% of N = 14/83*100= 16.86%

1. The atomic number of an element isdetermined by the number of:
a. protons.
b. electrons.
C. neutrons.
d. isotopes.

Answers

The answer is: A, protons

Answer:

Atomic number is protons

Explanation:

Protons = positive charge

A 0.7 ft diameter hole forms in a tank containing butane at 19 atmg and 76 degrees Fahrenheit. Determine the maximum possible mass flow rate through this leak in lb m / min, if the external pressure is 1 atm.

Answers

Answer:

Q = 3,534.4 lbm/s = 212,062 lbm/min

Explanation:

Mass flowrate of discharge or leakage mass flowrate (Q) is given as

Q = AC₀√(2ρgP)

A = Cross sectional Area of leakage = (πD²/4) = (π×0.7²)/4

A = 0.385 ft²

C₀ = discharge coefficient = 0.98 (For maximum discharge flow rate, the flow is turbulent with discharge coefficient within 1% of 0.98)

ρ = density of butane at 76°F = 35.771 lbm/ft³

g = acceleration due to gravity = 32.2 lbm.ft/lbf.s²

P = Gauge Pressure in the tank = (absolute pressure) - (external pressure) = 19 - 1 = 18 atm = 38091.9 lbf/ft²

Q = AC₀√(2ρgP)

Q = (0.385)(0.98)√(2×35.771×32.2×38091.9)

Q = 3,534.4 lbm/s = 212,062 lbm/min

Hope this Helps!!!

A patient has been diagnosed with mitochondrial disease and is experiencing the following symptoms: fatigue (lack of energy)and muscle weakness. Which organelles (cell parts) would NOT likely be immediately responsible for these symptoms? select all that apply. Mitochondria,Chloroplast, Lysosome,and Cytoplasm.NEED HELP ASAP

Answers

Answer:

Chloroplast, Lysosome

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