Choose the atom or ion that is larger. Select one: a. Br-1 b. Br

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

Kamila [148]

Answer: The mass percent of nitrogen gas in the compound is 13.3 %

Explanation:

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

3 years ago

0.0278, how many significant figures

My name is Ann [436]

Answer:

There are three significant figures

Explanation:

When counting sig figs you don't count the zeros unless it is between a number greater than zero. The two zeros aren't between the greater numbers so there are only 3.

3 0

3 years ago

The Molar mass(M.w) in g/mol of 6.3 grams of an ideal gas that placed in 5.0 L tank at ST

BigorU [14]

Answer:

28 g/mol, N2

Explanation:

Given data:

Volume of gas = 5.0 L

Mass of gas = 6.3 g

Pressure = 1 atm

Temperature = 273 K

Molar mass of gas = ?

Solution:

We will calculate the density first.

d = mass/ volume

d = 6.3 g/ 5.0 L

d = 1.26 g/L

Molar mass:

d = PM/RT

M = dRT/P

M = 1.26 g/L× 0.0821 atm.L/mol.K × 273 K/ 1 atm

M = 28 g/mol

Molar mass of N₂ is 28 g/mol thus given gas is N₂.

7 0

3 years ago

What is the correct measurement for solubility?

Montano1993 [528]

Other commonly used units include g/L (grams of solute per liter of solution) and m/L (moles of solute per liter of solution). Solubility units always express the maximum amount of solute that will dissolve in either a given amount of solvent, or a given amount of solution, at a specific temperature.

3 0

3 years ago

A mixture of c2h2 and ch4 has a total mass of 230.9 g. when this mixture reacts completely with excess oxygen, the co2 and h2o p

Salsk061 [2.6K]

Interesting problem. Thanks for posting.

C2H2 + (3/2)02 ====> H2O + 2CO2
CH4 + 2O2 =====> 2H2O + CO2

The molar mass of C2H2 = 2*12 + 2*1 = 26
The molar mass of CH4 = 1*12 + 4*1 = 16

The number of moles of C2H2 = x
The number of moles of CH4 = y
26x + 16y = 230.9 grams

For water we get (from the C2H2). Water has a molar mass of 2*1 + 16 = 18

x*18 See the balanced equation to see what it is the same number of moles as C2H2
From the methane we get
y*18
2*y* 18. Again see the balanced equation to see where that 2 came from.
18x + 36y is the total amount of water.

Now for the CO2. CO2 has a molar mass of 12 + 2*16 = 44
From C2H2 we get 2*44*x = 88x grams of CO2
From CH4 we get 1*y*44 grams of CO2
88x + 44y for CO2

Now we total to get the grand total of water and CO2
18x + 44y + 88x + 44y = 972.7 grams total.
106x + 88y = 972.7

Two equations, two unknowns, we should be able to solve this problem
26x + 16y = 230.9
106x + 88y = 972.7

I'm not going to go through the math unless you request me to do so.
x = 8.03 moles
y = 1.38 moles

The initial amount of C2H2 was 8.03 * 26 = 208.78
The initial amount of CH4 was 16*1.38 = 22.08
The total (as a check is 230.86 which is pretty close to the given amount.
So Methane's mass in the initial givens was 22.08 grams.

6 0

3 years ago