How many cups of beer are in a keg?

The product gas is then passed through a concentrated solution of KOH to remove the CO2. After passage through the KOH solution,

Kamila [148]

<u>Answer:</u> The mass percent of nitrogen gas in the compound is 13.3 %

<u>Explanation:</u>

Assuming the chemical equation of the compound forming product gases is:

$\text{Compound}\xrightarrow[CuO(s)]{Hot}N_2(g)+CO_2(g)+H_2O(g)$

Now, the product gases are treated with KOH to remove carbon dioxide.

We are given:

$p_{Total}=726torr\\P_{water}=23.8torr\\$

So, pressure of nitrogen gas will be = $p_{Total}-p_{water}=726-23.8=702.2torr$

To calculate the number of moles of nitrogen, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of nitrogen gas = 702.2 torr = 0.924 atm (Conversion factor: 1 atm = 760 torr)

V = Volume of nitrogen gas = 31.8 mL = 0.0318 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of nitrogen gas = $25^oC=[25+273]K=298K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of nitrogen gas = ?

Putting values in above equation, we get:

$0.924atm\times 0.0318L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\n_{mix}=\frac{0.924\times 0.0318}{0.0821\times 298}=0.0012mol$

To calculate the mass of nitrogen gas, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of nitrogen gas = 28 g/mol

Moles of nitrogen gas = 0.0012 moles

Putting values in above equation, we get:

$0.0012mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.0012mol\times 28g/mol)=0.0336g$

To calculate the mass percent of nitrogen gas in compound, we use the equation:

$\text{Mass percent of nitrogen gas}=\frac{\text{Mass of nitrogen gas}}{\text{Mass of compound}}\times 100$

Mass of compound = 0.253 g

Mass of nitrogen gas = 0.0336 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{0.0336g}{0.253g}\times 100=13.3\%$

Hence, the mass percent of nitrogen gas in the compound is 13.3 %

8 0

2 years ago

PLEASE HELP! (IM GIVING BRAINLIEST) Iron is a transition metal and, as such, can form two types of "oxide" salt compounds, FeO a

belka [17]

Answer:

See explanation

Explanation:

Iron exhibits two stable oxidation states, +2 and +3 states. The difference between FeO and Fe2O3 is that, in FeO, Fe is in the +2 oxidation state while in Fe2O3, Fe is in the +3 oxidation state

In FeO, iron looses two electrons which are gained by oxygen as follows;

Fe^2+ + O^2- -----> FeO

In Fe2O3, Fe looses six electrons while oxygen has gained six electrons as shown;

2Fe^3+ + 3O^2- ------>Fe2O3

7 0

2 years ago

A buffer is made that is .10M propanoic acid (pKa = 4.9) and .05M sodium propanoate. Calculate the pH.

inna [77]

PKa= 4.9 therefore ka= 10^-4.9= 1.259x10^-5

$ka= \frac{[H^+][CH3CH2COO^-]}{[CH3CH2COOH]}$

$[CH3CH2COO^-]
$ = 0.05
$[CH3CH2COOH]$ = 0.10

Therefore 1.259x10^-5 = $\frac{[H^+][0.05]}{[0.1]}$
Rearrange the equation to make the concentration of hydrogen the subject.
Therefore $[H^+] = \frac{(1.259*10^-5)(0.1)}{0.05}$

Therefore $[H^+]= 2.513*10^-5$

pH= -log [ $H^+$ ] = -log(2.513*10^-5)= 4.59.