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geniusboy [140]

3 years ago

12

let 4 moles of methanol (liquid) combust in 3 moles of gaseous oxygen to form gaseous carbon dioxide and water vapor. Suppose th

is occurs in a chamber of fixed volume and fixed temperature. If the original pressure is 1.0 atm, what is the final pressure in the chamber. Express your answer in atm. Enter a numerical value, do not enter units. Assume liquids take up negligible volume.

1 answer:

kari74 [83]3 years ago

8 0

Answer:

The final pressure is 2.0 atm

Explanation:

<u>Step 1:</u> Data given

Number of moles methanol = 4 mol

Number of moles oxygen = 3 mol

original pressure is 1.0 atm

<u>Step 2:</u> The balanced equation

2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(l)

<u>Step 3</u>: Calculate final pressure

For 2 moles of methanol consumed, we need 3 moles O2 to produce 2 moles CO2 and 4 moles H2O

We started with 3 moles of O2 gas.

Since methanol is not a gas, it doesn't count for the pressure.

V and T are fixed

This means the final pressure can be given by:

P2/P1 = n2/n1

with n2 = number moles of products

with n1 = number of moles of reactants

P2 = (6.0 mol*1.0 atm) / (3.0 mol )

P2 = Final pressure = 2.0 atm

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ASHA 777 [7]

Answer: 83.74L

Explanation:

Temp. = 295K

P = 713torr = 0.938atm

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1 year ago

Why do you think polar aprotic solvents increase the rate of an sn2 reaction

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2 years ago

"acid is responsible for the odor in rancid butter. a solution of 0.25 m butyric acid has a ph of 2.71. what is the ka for"

Salsk061 [2.6K]

Answer:- The Ka for the acid is $1.53*10^-^5$ .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

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Where, Ka is the acid ionization constant. Let's plug in the values.

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Let's calculate the value of X first using the equation:

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on taking antilog ob above equation we get:

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$[H^+]$ = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

$Ka=\frac{(0.00195)^2}{0.25-0.00195}$

$Ka=1.53*10^-^5$

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3 years ago

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