6. Given a mixture of gases: 1.00 g He,
14.0 g N2 and 10.0 g NO. What is the total pressure at 27.0 °C if the gases are confined in a 2.00 L container?

A) 0.310 atm
B) 1.24 atm
C) 13.3 atm
D) 21.6 atm

Answers

Answer 1

Considering the ideal gas law, the total pressure at 27.0 °C if the gases are confined in a 2.00 L container is 13.3 atm.

Definition of ideal gas law

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

Where:

P is the gas pressure.

V is the volume that occupies.

T is its temperature.

R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.

n is the number of moles of the gas.

Total pressure in this case

In this case you know:

mass of He= 1 gr

mass of N₂= 14 gr

mass of NO= 10 gr

molar mass of He= 4 gr/mole

molar mass of N₂= 28 gr/mole

molar mass of NO= 30 gr/mole

moles of He= mass of He÷ molar mass of He= 1 gr÷ 4 gr/mole= 0.25 moles

moles of N₂= mass of N₂÷ molar mass of N₂= 14 gr÷ 28 gr/mole= 0.5 moles

moles of NO= mass of NO÷ molar mass of NO= 10 gr÷ 30 gr/mole= 1/3 moles

total moles= moles of He + moles of N₂ + moles of NO= 0.25 moles + 0.5 moles + 1/3 moles= 13/12 moles

V= 2 L

R= 0.082 (atmL)/(molK)

T= 27 °C= 300 K

Replacing in the ideal gas law:

P×2 L = 13/12 moles×0.082 (atmL)/(molK)×300 K

Solving:

P= [13/12 moles×0.082 (atmL)/(molK)×300 K]÷ 2 L

P= 13.3 atm

Finally, the total pressure is 13.3 atm.

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Related Questions

What apparatus are used in crystallization?​

Answers

Answer:

Crystallization apparatus: (1) laboratory crucible furnace, (2) continuosly changeable transformer, (3) air cooler (“cold key”), (4) movable rings and (5) branched Tamman's test tube (“crystallization test comb”).

Calculate how many grams of rust (Fe2O3) should form when 10.0g of iron reacts with 20.0g of oxygen.
4Fe + 3O2 → 2Fe2O3

Answers

Taking into account the reaction stoichiometry, 14.31 grams of Fe₂O₃ are formed when 10.0g of iron reacts with 20.0g of oxygen.

Reaction stoichiometry

In first place, the balanced reaction is:

4 Fe + 3 O₂ → 2 Fe₂O₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Fe: 4 molesO₂: 3 molesFe₂O₃: 2 moles

The molar mass of the compounds is:

Fe: 55.85 g/moleO₂: 32 g/moleFe₂O₃: 159.7 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Fe: 4 moles ×55.85 g/mole= 223.4 gramsO₂: 3 moles ×32 g/mole= 96 gramsFe₂O₃: 2 moles ×159.7 g/mole= 319.7 grams

Limiting reagent

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 96 grams of O₂ reacts with 223.4 grams of Fe, 20 grams of O₂ reacts with how much mass of Fe?

mass of Fe= (20 grams of O₂ ×223.4 grams of Fe) ÷96 grams of O₂

mass of Fe= 46.54 grams

But 46.54 grams of Fe are not available, 10 grams are available. Since you have less mass than you need to react with 20 grams of O₂, Fe will be the limiting reagent.

Mass of Fe₂O₃ formed

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 223.4 grams of Fe form 319.7 grams of Fe₂O₃, 10 grams of Fe form how much mass of Fe₂O₃?

mass of Fe₂O₃= (10 grams of Fe×319.7 grams of Fe₂O₃)÷223.4 grams of Fe

mass of Fe₂O₃= 14.31 grams

Finally, 14.31 grams of Fe₂O₃ are formed.

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Select the correct answer.
Which statement about a rotation is true?
O A. Rotating a right trapezoid around its vertical axis will form a cone.
Rotating a right triangle around its vertical axis will form a cone.
Rotating a square around its vertical axis will form a sphere.
Rotating a rectangle around its vertical axis will form a sphere.
B.
C.
OD.
Reset
Next

Answers

Rotating a right triangle around its vertical axis will form a cone. The correct option is B.

The above one is basically the correct statement. When a two-dimensional shape is rotated around an axis, it creates a three-dimensional shape.

Rotating a right triangle around its vertical axis creates a cone because one of the sides of the triangle forms the curved surface of the cone, while the other side becomes the height.

The rotation axis goes through the apex of the triangle and creates a point at the other end of the curved surface.

Rotating a right trapezoid around its vertical axis would result in a frustum (a truncated cone). Rotating a square or rectangle around its vertical axis would form a cylinder, not a sphere.

Thus, the correct option is B.

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For electrical currents, the switch must be in the
a. Closed
b. Open
c. Reverse
d. Doesn't matter
position for the current to flow. (3.3.1)

Answers

Closed because it will flow through

26 What is the charge on each ion in these compounds?
(a) CaS
(b) MgF2
(c) Cs,O
(d) ScCl,
(e) Al,S,

Answers

The charges present on the following ionic compounds are Ca²⁺, Mg²⁺, Cs⁺, Sc⁺, Al³⁺, S²⁻, F⁻, O²⁻, Cl⁻.

Ionic compounds are held together by ionic bonds are classed as ionic compounds. Elements can gain or lose electrons in order to attain their nearest noble gas configuration. The formation of ions (either by gaining or losing electrons) for the completion of octet helps them gain stability.

In a reaction between metals and non-metals, metals generally loose electrons to complete their octet while non-metals gain electrons to complete their octet. Metals and non-metals generally react to form ionic compounds.

Ionic compounds include salts, oxides, hydroxides, sulphides, and the majority of inorganic compounds. Ionic solids are held together by the electrostatic attraction between the positive and negative ions.

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pllllllllsssssssssssssss helpp
Which of the these is a balanced chemical equation?

Question 18 options:

H2O + CHO2 → H2CO3


H2O + CO2 → H2CO3


3H2O + 2CO2 → 2H2CO3


2H2O + 2CO2 → H4CO4

Answers

Answer:

The balanced chemical equation is:

3H2O + 2CO2 → 2H2CO3

Explanation:

Given:
180.0 mL chloric acid (HCIO3)
440.0 mL of 1.75 M strontium hydroxide (Sr(OH)2)
Wanted: [HCIO3] necessary to neutralize Sr(OH)?

Answers

The molarity of the 180.0 mL chloric acid, HClO₃ solution needed to neutralize the 440.0 mL of 1.75 M strontium hydroxide, Sr(OH)₂ is 8.56 M

How do i determine the molarity of the chloric acid, HClO₃?

We'll begin by writing the balanced equation for the reaction. This is given below:

2HClO₃ + Sr(OH)₂ —> Sr(ClO₃)₂ + 2H₂O

The mole ratio of the acid, HClO₃ (nA) = 2The mole ratio of the base, Sr(OH)₂ (nB) = 1Volume of Sr(OH)₂ (Vb) = 440.0 mLMolarity of Sr(OH)₂ (Mb) = 1.75 M Volume of HClO₃ (Va) = 180.0 mLMolarity of HClO₃ (Ma) =?

The molarity of the chloric acid, HClO₃ solution necessary can be obtained as follow:

MaVa / MbVb = nA / nB

(Ma × 180) / (1.75 × 440) = 2

Cross multiply

Ma × 180  = 1.75 × 440 × 2

Divide both side by 180

Ma = (1.75 × 440 × 2) / 180

Ma = 8.56 M

Thus, we can conclude that the molarity of the chloric acid, HClO₃ solution is 8.56 M  

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Complete question:

Given that 180.0 mL chloric acid (HCIO3) reacted with 440.0 mL of 1.75 M strontium hydroxide (Sr(OH)2). What is the molarity of HCIO3 necessary to neutralize Sr(OH)?

Wanted: [HCIO3] necessary to neutralize Sr(OH)?

A gas at a pressure of 2.0 atm occupies a volume of 20 liters. If the volume is decreased to 10 liters and the pressure is increased to 4.0 atm, what will be the final volume of the gas?

Answers

The initial condition of the gas can be expressed using the combined gas law:

P1V1 / T1 = P2V2 / T2

where P1 = 2.0 atm, V1 = 20 L, and T1 is the initial temperature (which we will assume remains constant).

If the volume is decreased to 10 L and the pressure is increased to 4.0 atm, the new conditions can be expressed as:

P2 = 4.0 atm, V2 = ?, and T2 = T1

We can rearrange the combined gas law to solve for V2:

V2 = (P1V1T2) / (P2T1)

Plugging in the given values, we get:

V2 = (2.0 atm x 20 L x T1) / (4.0 atm x T1) = 10 L

Therefore, the final volume of the gas is 10 liters.
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