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2 years ago

What is the temperature (in K) of 16.45 moles of methane gas in a 4.95 L container at 4.68 atm?

1 answer:

4 0

Using the ideal gas law: PV=nRT
P is pressure; V is volume; n is the amount in moles; R=0.082; T is temperature in K.

(4.68)*(4.95)=(16.45)*(0.0821)*T
Solve for T.
T=17.15

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2CH4(g)⟶C2H4(g)+2H2(g)

Rasek [7]

Answer : The enthalpy change for the reaction is, 201.9 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The balanced reaction of $CH_4$ will be,

$2CH_4(g)\rightarrow C_2H_4(g)+2H_2(g)$ $\Delta H^o=?$

The intermediate balanced chemical reaction will be,

(1) $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ $\Delta H_1=-890.3kJ$

(2) $C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)$ $\Delta H_2=-136.3kJ$

(3) $2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=-571.6kJ$

(4) $2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)$ $\Delta H_4=-3120.8kJ$

Now we will multiply the reaction 1 by 2, revere the reaction 2, reverse and half the reaction 3 and 4 then adding all the equations, we get :

(1) $2CH_4(g)+4O_2(g)\rightarrow 2CO_2(g)+4H_2O(l)$ $\Delta H_1=2\times (-890.3kJ)=-1780.6kJ$

(2) $C_2H_6(g)\rightarrow C_2H_4(g)+H_2(g)$ $\Delta H_2=-(-136.3kJ)=136.3kJ$

(3) $H_2O(l)\rightarrow H_2(g)+\frac{1}{2}O_2(g)$ $\Delta H_3=-\frac{1}{2}\times (-571.6kJ)=285.8kJ$

(4) $2CO_2(g)+3H_2O(l)\rightarrow C_2H_6(g)+\frac{7}{2}O_2(g)$ $\Delta H_4=-\frac{1}{2}\times (-3120.8kJ)=1560.4kJ$

The expression for enthalpy of the reaction will be,

$\Delta H^o=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4$

$\Delta H=(-1780.6kJ)+(136.3kJ)+(285.8kJ)+(1560.4kJ)$

$\Delta H=201.9kJ$

Therefore, the enthalpy change for the reaction is, 201.9 kJ

5 0

1 year ago

What is the mass of 6.73x10^23 atoms of carbon

kap26 [50]

Answer: 13.42g

Explanation:

1mole of carbon =12g

1mole (12g) of carbon contains 6.02x10^23 atoms.

Therefore, Xg of carbon will contain 6.73x10^23 atoms i.e

Xg of carbon = (12x6.73x10^23)/6.02x10^23 = 13.42g

6 0

1 year ago

When working with food, personal items such as a cell phone should be

ratelena [41]

Answer:

should be put away in a bag or a pocket away from the food

Explanation:

5 0

2 years ago

A 500 mL sample of a 0.100 M formate buffer, pH 3.75, is treated with 5 mL of 1.00 M KOH. What is the pH following this addition

lubasha [3.4K]

<u>Answer:</u> The pH of the solution after addition of KOH is 3.84

<u>Explanation:</u>

We are given:

pH of buffer = 3.75

$pK_a$ of formic acid = 3.75

Using Henderson-Hasselbalch equation for formate buffer:

$pH=pK_a+\log(\frac{[HCOO-]}{[HCOOH]})$

Putting values in above equation, we get:

$3.75=3.75+\log(\frac{[HCOO-]}{[HCOOH]})\\\\\log(\frac{[HCOO-]}{[HCOOH]})=0\\\\\frac{[HCOO-]}{[HCOOH]}=1$

$[HCOO-]=[HCOOH]$

We are given:

Concentration of formate buffer = 0.100 M

$[HCOO-]+[HCOOH]=0.1$

$[HCOO-]=[HCOOH]=0.05M$

As, the volume of buffer is the same. So, the concentration is taken as number of moles of formate ions as well as formic acid

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of KOH = 1.00 M

Volume of solution = 5 mL

Putting values in above equation, we get:

$1.00M=\frac{\text{Moles of KOH}\times 1000}{5mL}\\\\\text{Moles of KOH}=0.005mol$

The chemical reaction for formic acid and KOH follows the equation:

$HCOOH+KOH\rightarrow HCOO^-+H_2O$

Initial: 0.05 0.005 0.05

Final: 0.045 - 0.055

Volume of solution = 500 + 5 = 505 mL = 0.505 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[HCOO^-]}{[HCOOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of formic acid = 3.75

$[HCOO^-]=\frac{0.055}{0.505}$

$[HCOOH]=\frac{0.045}{0.505}$

pH = ?

Putting values in above equation, we get:

$pH=3.75+\log(\frac{0.055/0.505}{0.045/0.505})\\\\pH=3.84$

Hence, the pH of the solution after addition of KOH is 3.84

3 0

2 years ago

A bucket of water contains 5.45 L of water, how many mL of water is this ?

nalin [4]

Answer: It is 5450 mL

Explanation: There are 1000 mL in every L and then there is an extra 450 so just add that at the end

6 0

2 years ago