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Lostsunrise [7]
2 years ago
3

At stp, which gaseous sample has the same number of molecules as 3.0 liter of N2(g)

Chemistry
2 answers:
Pani-rosa [81]2 years ago
5 0

The answer is: 3.0 L of CH4(g).

Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).

At STP one mole of gas occupies 22.4 liters of volume.

In this question all substances are gases, amount of substance depends on volume of the gas.

Same volume, same number of molecules.

So 3 liters of hydrogen gas (H₂) has the same number of molecules as 3 liters of methane gas (CH₄).


RSB [31]2 years ago
5 0

Answer:

Hydrogen gas has the same number of molecules as 3.0 liter of nitrogen gas.

Explanation:

Given volume of nitrogen gas at STP = 3 L

Moles of nitrogen gas = n

At STP, 1 mole of gas occupies 22.4 L of volume.

n\times 22.4 L =3 L

n=\frac{3 L}{22.4 L}=0.1339 mol

1 mol = 6.022\times 10^{23} atoms/ molecules

Number of nitrogen gas molecules:

0.1339\times 6.022\times 10^{23}=8.652\times 10^{22} molecules

So, from the options, the gas with volume equal to the volume of nitrogen gas and will have moles equal to number of moles of nitrogen with which it will also have same number of molecules of gas.

Volume of fluorine gas = 6.0 l

Volume of hydrogen gas = 3.0 L

Volume of nitrogen gas = 4.5 L

Volume of chlorine gas = 4.5 L

Volume of hydrogen gas = Given volume o nitrogen gas = 3.0 L

Hydrogen gas has the same number of molecules as 3.0 liter of nitrogen gas.

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2 years ago
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The molar mass of an imaginary molecule is is 93.89 g/mol. Determine its density at STP.
Igoryamba

Answer:

Density = 4.191 gm/L

Explanation:

Given:

Molar mass = 93.89 g/mol

Volume(Missing) = 22.4 L (Approx)

Find:

Density at STP

Computation:

Density = Mass/Volume  

Density = 93.89 / 22.4  

Density = 4.191 gm/L

3 0
1 year ago
Butane (c4h10) undergoes combustion in excess oxygen to generate gaseous carbon dioxide and water. given δh°f[c4h10(g)] = –124.7
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The value of Δ H butane (g) = -124.7 kJ/mol

The value of Δ H CO2 (g) = -393.5 kJ/mol

The value of Δ H H2O (g) = -241.8 kJ/mol

Mass of butane, m = 8.30 gm

Molar mass of butane is 58 gm/mol

Consider the reaction,

C₄H₁₀ + 6.5 O₂ = 4CO₂ + 5H₂O

Calculating the value of Δ H° rxn:

ΔH°rxn = ∑nH° f (products) - ∑nH° f (reactants)

Substituting the values we get,

Δ H° rxn = 4 (-393.5) + 5 (-241.8) - (-124.7)

= -1574 -1209 + 124.7

= -2783 - 124.7

= -2658.3 kJ/mol

Now, calculate the number of moles of butane in 8.30 gm.

Number of moles = mass/molar mass

= 8.30 / 58

= 0.143 moles

Thus, the total energy released in the reaction is,

Q = number of moles × ΔH° rxn

= 0.143 × (2658.3)

= 380.14 kJ

Hence, the total heat released in the reaction is 380.14 kJ.

6 0
2 years ago
. A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture
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Answer:

A) Mass flow rate of air = 22.892 kmol/hr

B)percentage by mass of oxygen in the product gas = 22.52%

Explanation:

We are given that the mixture containing 9.0 mole% methane in air flowing.

Thus, we have 0.09 mole of methane(CH4) and the remaining will be the air which is (100% - 9%) = 91% = 0.91

Molar mass of CH4 = 12 + 1(4) = 16 g/mol

We are given the average molecular weight of air = 29 g/mol

Thus;

Average molar mass of air and methane mixture is;

M_avg = (0.09 × 16) + (0.91 × 29)

M_avg = 27.83 g/mol

We are told that air flowing at a rate of 7 × 10² kg/h = 700 kg/h

Thus;

Mass flow rate of CH4 in air mixture = 700kg/h × (0.09CH4)/1 mix × (1/27.83kg/kmol) = 2.264 kmol/hr

Mass flow rate of air in mixture = 2.264kmol/h × 0.91kmol air/0.09kmolCH4 = 22.892 kmol/hr

We are told that the mixture is capable of being ignited if the mole percent of methane is between 5% and 15%.

Thus, for 5% of methane, the air required will be;

2.264kmol/h × 0.95kmol air/0.05kmol CH4 = 43.016 kmol/hr

Now, the dilution air needed will be =

43.016 - 22.892 = 20.124 kmol/hr

Total mass flow rate of mixture =

700kg/hr + (20.124kmol/hr × 29kg/mol) = 1283.596 kg/hr

We are told that air consist of 21 mole% Oxygen (O2).

Molar mass of oxygen = 32

Thus;

Mass fraction of oxygen in the product gas = 43.016kmol/h × (0.21molO2/1mol air) × (32kg oxygen/1kmol oxygen) × (1/1283.596kg/h) = 0.2252

Thus, written in percentage form, we have; 22.52%

So, percentage by mass of oxygen in the product gas = 22.52%

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