hat is the pH of a solution with a hydroxide ion concentration of 10–6M? A. pH = 12 B. pH = 8 C. pH = 6 D. pH = 7

DONT SKIP OR ELSE what solids, liquids, and gases does the human body produce

Marysya12 [62]

farting pooping peeing :3

6 0

3 years ago

In which step of a four-stroke engine cycle does the car release CO2, H2O, and CO?

In-s [12.5K]

Answer:

$\boxed{\text{exhaust stroke}}$

Explanation:

The four strokes are

Intake: The mixture of air and fuel enters the cylinder.
Compression: The mixture is compressed by a factor of about 11.
Ignition: A spark ignites the fuel, which burns and produces CO₂, CO, and H₂O.
Exhaust: The combustion products are expelled from the cylinder.

$\text{The cylinder releases CO$_{2}$, CO, and H$_{2}$O during the }\boxed{\textbf{exhaust stroke}}$

7 0

3 years ago

Aluminum is manufactured using electrolysis. Carbon electrodes are used. Describe the nature of the electrolyte.

Aleksandr [31]

Answer:

The traditional electrolyte for aluminium electrolysis is based on molten cryolite (Na3AlF6), acting as solvent for the raw material, alumina (Al2O3).Metals are found in ores combined with other elements. Electrolysis can be used to extract a more reactive metal from the ore.

Aluminum can and is used as both anodes and cathodes in electrochemical cells, but there are some peculiarities to using it as an anode in aqueous solutions. As you note, aluminum forms a passivating oxide layer quite readily, even by exposure to atmosphere. In an aqueous solution, if the potential is high enough, OH− and O2− are generated at the anode, which can then react with the aluminum to produce aluminum oxide. Al^3+ can also be generated directly. The electric field will draw the anions through the growing aluminum oxide layer towards the aluminum surface and the Al^3+ towards the solution, making the oxide layer grow both away from the electrode surface and into the surface of the electrode. In this way, coatings thicker than the normal passivation in air can be produced. However, aluminum oxide is a good electrical insulator, thus if a dense non-porous layer is grown, it will become impossible to pass current through it and growth will stop, leaving a relatively thin oxide layer (this is how the dielectric layers in electrolytic capacitors are made). This is the normal behaviour in aqueous solutions at near-neutral pH (5–7).

However, if a thick aluminum oxide layer is desired (e.g. to produce coatings on aluminum parts for dying or durability), maintaining porosity is necessary to avoid completely blocking access to the surface. One technique that is commonly used is using a low pH solution, which tends to redissolve some of the oxide and neutralize some of the formed OH−, leaving pores in the oxide layer through which the ions can travel and continue to react. These pores also give a good structure to retain dyes or lubricants, but generally need to be sealed after to protect against corrosion.

3 0

2 years ago

In the animation, you can see that the electrons occupy different orbitals according to the energy level of each orbital. A sing

lara31 [8.8K]

B) The He atom has two electrons that have parallel spin in its 1s orbital.

C) Electrons generally occupy the lowest energy orbital first.

D) The arrangement of the orbitals is the same in a multielectron atom and a single-electron atom

E) The C atom has two unpaired electrons.

Explanation:

In writing sublevel notations certain rules are followed:

The maximum number of electrons in the orbitals of sublevels are:

two for s-sublevel with one orbital

six for p-sublevel with three orbitals

ten for d-sublevel with five orbitals

fourteen for f-sublevel with seven orbitals

Aufbau's principle states that sublevels with lower energies are filled before those with higher energies.

1s 2s 2p 3s 3p 4s 3d 4p 5s increasing order of filling

Pauli exclusion principle states that no two electrons can have the same set of the four quantum numbers i.e electrons in the same orbital cannot spin in the same direction.
Hund's rule states that electrons go into degenerate orbitals singly first before pairing up.

Based on these principles and the animation, the following options are correct:

B) The He atom has two electrons that have parallel spin in its 1s orbital.

Helium has two electrons in the 1s orbital. From Pauli's exclusion principle, we see that the two electrons cannot spin in the same way. Therefore, they have parallel spin

C) Electrons generally occupy the lowest energy orbital first.

From Aufbau's principle, electrons go into orbitals with lower energies first before those with higher energies.

D) The arrangement of the orbitals is the same in a multi-electron atom and a single-electron atom

1s 2s 2p 3s 3p 4s 3d 4p 5s

Orbitals are arranged in the order of their energies from the lowest to the highest.

E) The C atom has two unpaired electrons.

Configuration for carbon atom is shown below:

1s² 2s² 2p²

Learn more:

Orbitals brainly.com/question/1832385

#learnwithBrainly

8 0

3 years ago

Read 2 more answers

An air stream enters a variable area channel at a velocity of 30 m/s with a pressure of 120 kPa and a temperature of 10°C. At a

nekit [7.7K]

Answer:

P₂= 74 kPa under constant density and ρ₂ = 1.06 kg/m³ (-38.6% of error compared with incompresible assumption) . Thus Bernoulli’s equation should not be applied

Explanation:

Assuming ideal gas behaviour of air , then

P*V= n*R*T = m / M * R *T

since

ρ= m/V = P*M /( R *T)

where

n= moles , V= volume , m= mass

ρ= density

P= pressure = 120 kPa= 120000 Pa

M= molecular weight of air = 0.21*32+0.79*28= 28.24 gr/mol = 0.02824 kg/mol

T= absolute temperature = 10°C + 273 = 283 K

R= ideal gas constant = 8.314 J/mol K

solving for ρ

ρ= P*M /( R *T) = 120000 Pa*0.02824 kg/mol/(8.314 J/mol K*283 K) = 1.47 kg/m³

then from Bernoulli's equation

P₁ + ρ*v₁²/2 = P₂ + ρ*v₂²/2

where 1 denotes inlet and 2 denotes other point , p = pressure and v= velocity . Then solving for p₂

P₁ + ρ*v₁²/2 = P₂ +ρ*P₂²/2

P₂= P₁ +ρ*v₁²/2 - ρ*v₂²/2 = P₁ +ρ/2*(v₁² - v₂²)

replacing values

P₂= P₁ +ρ/2*(v₁² - v₂²) = 120000 Pa + 1.47 kg/m³/2*[(30 m/s)²-(250 m/s)²] = 74724 Pa = 74 kPa

P₂= 74 kPa

then if the temperature remains constant

ρ₁= P₁*M /( R *T) and ρ₂= P₂*M /( R *T)

dividing both equations

ρ₂/ρ₁ = P₂/ P₁

ρ₂ = (P₂/ P₁)*ρ₁

then from Bernoulli's equation

P₁ + ρ₁*v₁²/2 = P₂ + ρ₂*v₂²/2

P₂ = P₁ + ρ₁*v₁²/2 - ρ₂*v₂²/2

therefore

ρ₂ = (P₂/ P₁)*ρ₁ = (P₁ + ρ₁*v₁²/2 - ρ₂*v₂²/2 ) /P₁ *ρ₁

P₁ * ρ₂ = P₁ *ρ₁ + ρ₁²*v₁²/2 - ρ₂*ρ₁ * v₂²/2

P₁ * ρ₂ + ρ₂*ρ₁ * v₂²/2 = P₁ *ρ₁ + ρ₁²*v₁²/2

ρ₂* (P₁ + ρ₁ * v₂²/2) = P₁ *ρ₁ + ρ₁²*v₁²/2

ρ₂ = (P₁ *ρ₁ + ρ₁²*v₁²/2)/(P₁ + ρ₁ * v₂²/2) = (P₁ + ρ₁*v₁²/2)/(P₁/ρ₁ + v₂²/2)

replacing values

ρ₂ = ( 120000 Pa + 1.47 kg/m³/2*(30 m/s)²)/(120000 Pa/1.47 kg/m³+1/2*(250 m/s)²)

ρ₂ = 1.06 kg/m³

the error of assuming constant ρ would be

e = (ρ₂ - ρ)/ρ₂= 1- ρ/ρ₂= 1- 1.47 kg/m³/1.06 kg/m³ = -0.386 (-38.6%)

thus Bernoulli’s equation should not be applied

8 0

3 years ago