A scientist who studies the origin of living organisms through fossils

1. If you have a sample of gas at a pressure of 16 atm, what will the pressure be if the volume is halved?

vagabundo [1.1K]

Answer:

1. The pressure will be 32 atm, twice the initial pressure.

2. The pressure will be 1.83 atm, one third of the initial pressure.

Explanation:

Boyle's law is one of the gas laws that relates the volume and pressure of a certain quantity of gas kept at a constant temperature.

This law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or P * V = k

Ahora es posible suponer que tienes un cierto volumen de gas V1 que se encuentra a una presión P1 al comienzo del experimento. Si varias el volumen de gas hasta un nuevo valor V2, entonces la presión cambiará a P2, y se cumplirá:

P1*V1=P2*V2

1. In this case:

P1= 16 atm
V1
P2= ?
V2= V1÷2= $\frac{V1}{2}$ because the volume is halved.

So:

16 atm*V1= P2* $\frac{V1}{2}$

Solving:

$\frac{16 atm*V1*2}{V1}$ =P2

16 atm*2= P2

32 atm= P2

The pressure will be 32 atm, twice the initial pressure.

2. Now

P1= 5.5 atm
V1
P2= ?
V2= V1*3 because the volume is tripled.

So:

5.5 atm*V1= P2* V1*3

Solving:

$\frac{5.5 atm*V1}{3*V1}$ =P2

$\frac{5.5 atm}{3}$ = P2

1.83 atm= P2

The pressure will be 1.83 atm, one third of the initial pressure.

7 0

3 years ago

You need to make an aqueous solution of 0.207 M calcium acetate for an experiment in lab, using a 300 mL volumetric flask. How m

jeka94

Answer:

We have to add 9.82 grams of calcium acetate

Explanation:

Step 1: Data given

Molarity of the calcium acetate solution = 0.207 M

Volume = 300 mL = 0.300 L

Molar mass calcium acetate = 158.17 g/mol

Step 2: Calculate moles calcium acetate

Moles calcium acetate = molarity * volume

Moles calcium acetate = 0.207 M * 0.300 L

Moles calcium acetate = 0.0621 moles

Step 3: Calculate mass calcium acetate

Mass calcium acetate = moles * molar mass

Mass calcium acetate = 0.0621 moles * 158.17 g/mol

Mass calcium acetate = 9.82 grams

We have to add 9.82 grams of calcium acetate

5 0

3 years ago

The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react

murzikaleks [220]

Answer: The value of $K_{eq}$ is $4.84\times 10^{-5}$

Explanation:

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $\frac{0.0280}{1.00}=0.0280M$

Concentration of oxygen gas = $\frac{0.0120}{1.00}=0.0120M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Initial: 0.0280 0.0120

At eqllm: 0.0280-4x 0.0120-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $3.00\times 10^{-3}M=0.003$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M$

Now, equilibrium concentration of ammonia = $0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M$

Equilibrium concentration of oxygen gas = $0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M$

Equilibrium concentration of water = $6x=(6\times 0.0015)]=0.009M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}$

Hence, the value of $K_{eq}$ is $4.84\times 10^{-5}$

4 0

3 years ago

How many moles of water are formed when 22 mol of methane combusts?

icang [17]

Methane is a hydrocarbon which when burns in air (combustion) produces carbon dioxide and water. The equation for the reaction;
CH4 +2O2 = CO2 +2H2O
When one mole of methane combusts 2 moles of water are formed
Therefore; when 22 moles of methane combusts 44 moles of water are formed (22 ×2)

7 0

3 years ago

Calculate δg∘rxn and e∘cell for a redox reaction with n = 3 that has an equilibrium constant of k = 4.4×10−2. you may want to re

lapo4ka [179]

a) First, to get ΔG°rxn we have to use this formula when:

ΔG° = - RT ㏑ K

when ΔG° is Gibbs free energy

and R is the constant = 8.314 J/mol K

and T is the temperature in Kelvin = 25 °C+ 273 = 298 K

and when K = 4.4 x 10^-2

so, by substitution:

ΔG°= - 8.314 * 298 *㏑(4.4 x 10^-2)

= -7739 J = -7.7 KJ

b) then, to get E° cell for a redox reaction we have to use this formula:

ΔE° Cell = (RT / nF) ㏑K

when R is a constant = 8.314 J/molK

and T is the temperature in Kelvin = 25°C + 273 = 298 K

and n = no.of moles of e- from the balanced redox reaction= 3

and F is Faraday constant = 96485 C/mol

and K = 4.4 x 10^-2

so, by substitution:

∴ ΔE° cell = (8.314 * 298 / 3* 96485) *㏑(4.4 x 10^-2)

= - 2.7 x 10^-2 V

8 0

3 years ago