Relaxers that contain two components and must be mixed immediately prior to use are _____.
Select one:
a. guanidine hydroxide relaxers
b. ammonium thioglycolate
c. potassium hydroxide relaxers
d. no-lye relaxers

Noble gases in Group VIII have electron affinities that are close to zero because each atom has a stable octet and does not readily accept an electron.

Both electron affinity and electron negativity have a downward trend over time. The electron affinity is also 0 if the electron negativity is 0. Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), Radon (Rn), and Oganesson are the elements with no affinity for electrons (Og). Affinities for electrons can be zero, negative, or positive. Noble gases in Group VIII have electron affinities that are close to zero because each atom has a stable octet and does not readily accept an electron. There is no tendency for noble or inert gases to gain an electron. Its electron affinity is therefore zero.

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The factors affecting the rate of sugar dissolution in water are temperature and surface area, with higher temperatures and smaller sugar particles leading to faster dissolution. Stirring further enhances the dissolution rate.

Based on the provided experiment, the data table is attached in the image below:

Conclusion:

Sugar dissolves faster in hot water compared to warm and cold water. This indicates that temperature impacts the rate of dissolution, with higher temperatures leading to faster dissolution.

Granular sugar dissolves faster than sugar cubes in the same temperature of water. This suggests that the surface area of the sugar particles influences the rate of dissolution, with smaller granules having a larger surface area and therefore dissolving more quickly.

Stirring the water speeds up the dissolution process for both granular sugar and sugar cubes. Stirring helps in increasing the contact between the sugar and water, facilitating the dissolution process.

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Answer:

84.313 g/mol

Explanation:

Magnesium Carbonate molecular weight. Molar mass of MgCO3 = 84.313 g/mol.

Answer:

The second option

Explanation:

The process of elimiation.

1. We know liquids dont have a definite shape, they will morph into whatever enviroment theyre in

3. Solids dont expand based on what container theyre in, they either dont fit or they do

4. We know that liquid expands, so we can also rule this one out.

Which leaves us with option 2 to be correct.

Answer:

the solution is acidic as its ph has already exceeded the neutral level

\(1.1 X 10^-9 M\) is the [H] in a solution with a pH of 9.92.

pH is the quantitative measure of the acidity or basicity of aqueous or other liquid solutions.

The pH of a solution is usually found by the following expression:

pH=-log [\(H^+\)]

Therefore, to find the concentration of \(H^+\) we can rearrange this expression and thus,

= [\(H^+\)] = \(10^-^{pH}\)

= [\(H^+\)] = \(10^-^{9.92}\)

=\(1.1 X 10^-9 M\)

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The energy of the proton produced when the electron jumps from the first excited state to the ground state of the hydrogen atom is 1.64 x 10^-18 J.

The energy of the proton produced can be found using the formula: E = hc/λ
where E is the energy, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength of the emitted photon.

The transition from the first excited state (n=2) to the ground state (n=1) of the hydrogen atom corresponds to the emission of a photon with a wavelength of 121.6 nm. Therefore, the energy of the proton produced is:

E = hc/λ = (6.626 x 10^-34 J·s) x (2.998 x 10^8 m/s) / (121.6 x 10^-9 m)

E = 1.64 x 10^-18 J

Alternatively, you can use the equation E = -RH(1/n1^2 - 1/n2^2), where RH is the Rydberg constant (2.18 x 10^-18 J), n1 is the initial energy level (2), and n2 is the final energy level (1). Plugging in these values gives:

E = -RH(1/1^2 - 1/2^2)

E = -2.18 x 10^-18 J (1 - 1/4)

E = 1.64 x 10^-18 J

Therefore, the energy of the proton produced when the electron jumps from the first excited state to the ground state of the hydrogen atom is 1.64 x 10^-18 J.

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Answer:False

Explanation:

The statement is false. Because, it is easy to find the average acceleration for a period of time by simply dividing the change in velocity by the period of time.

Acceleration is a physical quantity which describes the rate of change in velocity. Like velocity acceleration is a vector quantity having both magnitude and direction.

If both speed and direction are changing and  if we want to know the acceleration at any particular moment in time, calculating acceleration can be challenging.

However, since just speed is changing, it is quite simple to determine average acceleration over time. After then, acceleration is equal to the time change divided by the velocity change.

Therefore, the statement is false. And its is easy to calculate the average acceleration and difficult to find acceleration at a given moment.

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Answer:

1. 2.63x10^6

2. 39.2

Both answers have been rounded

The new temperature of the gas is 115 °C

From the question given above, the following data were obtained:

The new temperature of gas can be obtained as follow:

\( \frac{P_1}{T_1} = \frac{P_2}{T_2} \\ \\ \frac{0.96}{298} = \frac{1.25}{T_2} \\ \\ cross \: multiply \\ \\ T_2 \times 0.96 = 298 \times 1.25 \\ \\ divide \: both \: side \: by \: 0.96 \\ \\T_2 = \frac{298 \times 1.25 }{0.96} \\ \\ T_2 = 388 \: K \\ \\ subtract \: 273 \: from \: 388 \: to \: express \: the \: answer \: im \: \degree \: C \\ \\ T_2 = 388 - 273 \\ \\ T_2 = 115 \: \degree \: C \)

Therefore, the new temperature is 115 °C

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The empirical formula of the compound is CH.

To determine the empirical formula of a compound from its composition, we need to find the simplest whole-number ratio of the atoms in the compound.

Assuming a 100 gram sample, we have:

92.3 g C

7.7 g H

To convert these masses to moles, we need to divide by their respective atomic masses:

92.3 g C / 12.01 g/mol = 7.689 mol C

7.7 g H / 1.01 g/mol = 7.623 mol H

Next, we need to find the simplest whole-number ratio of C to H. We can do this by dividing both moles by the smaller of the two:

7.689 mol C / 7.623 mol H = 1.008

This means that the ratio of C to H is approximately 1:1. Therefore, the empirical formula of the compound is CH.

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--The given question is incorrect, the correct question is

"How do you determine the empirical formula of a compound that has the following composition: 92.3 percent C and 7.7 percent H?"--

Answer:H+(protons)..

Answer:

3 mole

Explanation:

2 moles of Ag give 1 mole of Mg(NO3)2

5 moles of Ag will give 5×1÷2 of Mg(NO3)2

=2•5 =3

Answer:

there will a definite decrease in solute solution

Explanation:

acid reaction acting upon negative charge.

To convert 25 g of liquid water to vapor at 100°C, we need 67.7 kJ of energy.

To change 25 g of liquid water to steam, we need to calculate the energy required for the following two processes:

Heating the water from 100°C to its boiling point at atmospheric pressure (100°C).

Vaporizing the water at its boiling point at atmospheric pressure (100°C) to steam at the same temperature.

Let's start with the first step. The specific heat capacity of water is 4.18 J/g°C, so we need:

q1 = m * c * ΔT

where

m = mass of water = 25 g

c = specific heat capacity of water = 4.18 J/g°C

ΔT = change in temperature = (100 - 0)°C = 100°C

q1 = 25 g * 4.18 J/g°C * 100°C

q1 = 10450 J

This means that we need 10450 J of energy to heat 25 g of water from 0°C to 100°C.

Now let's move on to the second step, which is vaporizing the water. The molar heat of vaporization of water is 40.6 kJ/mol.

Since we know the mass of water (25 g), we need to convert it to moles:

n = m / M

where

m = mass of water = 25 g

M = molar mass of water = 18.015 g/mol

n = 25 g / 18.015 g/mol

n = 1.387 mol

The energy required to vaporize the water is:

q2 = n * Δ\(H_v_a_p\)

where

Δ\(H_v_a_p\) = molar heat of vaporization of water = 40.6 kJ/mol

q2 = 1.387 mol * 40.6 kJ/mol

q2 = 56.3 kJ

Therefore, the total energy required to change 25 g of liquid water to steam at 100°C is the sum of q1 and q2:

q = q1 + q2

q = 10450 J + 56.3 kJ

q = 67.7 kJ

So, we need 67.7 kJ of energy to change 25 g of liquid water to steam at 100°C.

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We must take into account the following stages in order to determine the amount of energy needed to convert 25 g of liquid water into steam at 100°C: Calculate how many moles of water (H2O) are contained in 25 g.

Number of moles of H2O = mass/molar mass = 25 g / 18.015 g/mol = 1.388 mol. The molar mass of H2O is 18.015 g/mol. Determine the amount of energy needed to evaporate one mole of water. Water has a molar heat of vaporization (Hvap) of 40.6 kJ/mol. Determine the amount of energy necessary to evaporate 1.388 moles of water. 1.388 moles of water must be vaporized in order to produce 40.6 kJ/mol of energy, which equals 56.4 kJ. Hence, 56.4 kJ of energy are needed to convert 25 g of liquid water to steam at 100°C. The molar heat of vaporization for liquid water is 40.6 kj/mole.

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Answer:

Three possible blood type alleles are Iᴬ, Iᴮ and i

Explanation:

Iᴬ, Iᴮ and i are three possible blood type alleles.

Iᴬ and Iᴮ are known as co-dominant, and The i allele is recessive.

Thus, Three possible blood type alleles are Iᴬ, Iᴮ and i

-TheUnknownScientist

The rusting of a kitchen knife is a natural process that occurs when iron or steel reacts with oxygen and moisture in the air. Rusting is a form of corrosion that gradually damages the metal surface, causing it to become brittle and lose its sharpness.

The process of rusting involves an electrochemical reaction. When the metal is exposed to air and moisture, water molecules combine with carbon dioxide in the air to form a weak acid. This acid reacts with the iron or steel, breaking it down into iron ions and releasing electrons. The iron ions combine with oxygen from the air to form iron oxide, commonly known as rust. Rusting can be accelerated by factors such as high humidity, exposure to saltwater or chemicals, and scratches or damage to the knife's protective coating. To prevent rusting, it is important to keep the knife dry, clean, and properly stored, preferably in a dry environment and away from moisture.

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Answer:

c

my teacher disscuss about that

Answer:

c

Explanation:

because there are two particles that can interact via the strong nuclear force I tried my best to get the best answer for you

Its a c,e,f

Sadly i got it wrong

Flammability, Ability to rust and Ability to tarnish (C,E,F) are chemical properties of an element.

The property of an element is the characteristic feature of that element which helps us to differentiate it from another  element.

We have

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Answer:

Temporary hardness is a type of water hardness caused by the presence of dissolved bicarbonate minerals (calcium bicarbonate and magnesium bicarbonate). ... However, unlike the permanent hardness caused by sulfate and chloride compounds, this "temporary" hardness can be reduced by boiling the water.

Answer:

Density is equal to mass divided by volume. So, in order to calculate the density of a penny, you need to first calculate its’ mass via a triple beam scale or a similar tool. Then you need to calculate its’ volume.

Explanation:

True. The statement is true as a unit cell encapsulates the essential features and symmetry of the crystalline lattice and is repeated to form the complete lattice structure.

A unit cell is defined as the smallest repeating unit of a crystal lattice that, when repeated in three-dimensional space, generates the entire lattice structure. It represents the fundamental building block of the crystal lattice and contains all the structural information needed to describe the crystal. By translating the unit cell in three dimensions, the complete crystal lattice can be constructed.

The concept of a unit cell is fundamental in crystallography and is used to study and understand the arrangement and properties of crystalline materials. Different types of unit cells, such as cubic, tetragonal, and hexagonal, exist depending on the symmetry and arrangement of the lattice.

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The pressure inside a container of gas increases if more gas is added to the container due to the increase in the number of molecules striking the walls of the container per unit time and the increase in the force of the collisions between the molecules and the walls of the container.

Pressure is defined as force per unit area and is usually measured in atmospheres (atm), millimeters of mercury (mmHg), or kilopascals (kPa).The molecules of gas in a container are in constant motion and collide with the walls of the container. When more gas is added to the container, the molecules have less space to move around and collide with the walls more frequently.

This leads to an increase in the number of collisions per unit time and therefore an increase in the force per unit area exerted on the walls of the container. This increase in force leads to an increase in pressure inside the container.In summary, the pressure inside a container of gas increases if more gas is added to the container due to an increase in the number of collisions and the force of the collisions between the molecules and the walls of the container.

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The approximate potential for points on the z-axis far from the sphere is given by: V ≈ 4π k^2 R^3 / r^3

To find the approximate potential for points on the z-axis far from the sphere, we can consider the potential due to an infinitesimally small charge element on the sphere and integrate over the entire sphere.

The potential at a point P on the z-axis due to an infinitesimally small charge element dq located at (r, θ) on the sphere is given by:

dV = k dq / |r - r'|

where r' is the position vector of the charge element dq and r is the position vector of point P on the z-axis.

In spherical coordinates, the position vector r' of the charge element dq can be expressed as:

r' = R sinθ' cosφ' i + R sinθ' sinφ' j + R cosθ' k

where θ' and φ' are the angles associated with the charge element dq.

Since we are considering points far from the sphere on the z-axis, we can approximate |r - r'| as r, as the radial distance of the charge element from the origin is much smaller than the distance of point P from the origin.

Therefore, the potential at point P on the z-axis due to the entire sphere can be approximated by integrating the potential due to each charge element over the sphere:

V ≈ ∫(k dq / r)

To find dq, we can express it in terms of the charge density rho:

dq = rho(r, θ) dV'

where dV' is an infinitesimally small volume element on the sphere.

The infinitesimal volume element dV' can be expressed in spherical coordinates as:

dV' = R^2 sinθ' dθ' dφ'

Substituting dq and dV' into the integral, we have:

V ≈ ∫(k rho(r, θ) dV' / r)

V ≈ k / r ∫(rho(r, θ) R^2 sinθ' dθ' dφ')

The integration is performed over the entire sphere, so the limits of integration for θ' are 0 to π and for φ' are 0 to 2π.

V ≈ k / r ∫(rho(r, θ) R^2 sinθ' dθ' dφ') (limits: φ'=0 to 2π, θ'=0 to π)

Substituting the expression for rho(r, θ) = k R / r^2 sinθ into the integral:

V ≈ k / r ∫((k R / r^2 sinθ) R^2 sinθ' dθ' dφ') (limits: φ'=0 to 2π, θ'=0 to π)

Simplifying the expression:

V ≈ k^2 R^3 / r^3 ∫(sinθ' dθ' dφ') (limits: φ'=0 to 2π, θ'=0 to π)

The integral of sinθ' over the range 0 to π is 2.

V ≈ 2 k^2 R^3 / r^3 ∫dφ' (limits: φ'=0 to 2π)

The integral of dφ' over the range 0 to 2π is 2π.

V ≈ 2π(2 k^2 R^3 / r^3)

V ≈ 4π k^2 R^3 / r^3

Therefore, The approximate potential for points on the z-axis far from the sphere is given by:

V ≈ 4π k^2 R^3 / r^3

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Answer:

397,570 g/m^3

Explanation:

The volume of the cylinder can be calculated using its height and diameter.

Mass of the soil sample (m) = 30 g

Height of the cylinder (h) = 6 cm

Diameter of the cylinder (d) = 4 cm

First, we need to calculate the radius (r) of the cylinder

Radius (r) = diameter / 2 = 4 cm / 2 = 2 cm = 0.02 m

Now, we can calculate the volume (V) of the cylinder

V = π * r^2 * h

V = 3.14159 * (0.02 m)^2 * 0.06 m

V = 7.5398 E-5 m^3

Calculate the bulk density (ρ) using this formula

ρ = m / V

ρ = 30 g / 7.5398 E-5 m^3

ρ = 397,887 g/m^3

Answer:

The correct answer is coats the pathogen exterior so that it is recognized by the host's phagocytes. :)

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Part of ecosystem | Contains energy storage molecules? (yes or no) | Energy storage molecules flowing in? (yes or no) | Energy storage molecules flowing out? (yes or no)

Producers | Yes | No | Yes

Consumers | Yes | Yes | Yes

Decomposers | Yes | Yes | Yes

Dead matter | Yes | Yes | Yes

Abiotic matter | No | No | No

In an ecosystem, different components play different roles in terms of energy storage and flow.

Producers, such as plants, have the ability to produce energy-rich molecules through photosynthesis, storing energy in the form of carbohydrates. They do not rely on external sources of energy storage molecules, but they release energy storage molecules into the ecosystem when consumed or when they undergo decomposition.

Consumers, including herbivores, carnivores, and omnivores, obtain energy storage molecules by consuming producers or other consumers. They receive energy-rich molecules flowing into their systems through their diet and release energy storage molecules when they respire or excrete waste.

Decomposers break down organic matter, including dead plants and animals, into simpler substances and release energy storage molecules in the process. They receive energy storage molecules flowing into their systems from the breakdown of organic matter and release energy storage molecules back into the ecosystem.

Dead matter, which refers to organic material that is no longer living, contains energy storage molecules. When dead matter decomposes, the stored energy is released into the ecosystem.

Abiotic matter, which includes non-living components like minerals and gases, does not contain energy storage molecules and does not participate in the flow of energy storage molecules within the ecosystem.

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Answer:butterflies

Explanation:

Nah man you are on your own.

1. Igneous intrusion Y formed.

2. The shale layer was deposited.

3. Igneous intrusion X formed.

4. The limestone layer was deposited.

5. The mudstone layer was deposited.

B 1. The mudstone layer was deposited.

2. The limestone layer was deposited.

3. The shale layer was deposited.

4. Igneous intrusion Y was forced onto the mudstone, limestone, and shale.

5. Igneous intrusion X was forced onto the mudstone, limestone, and shale.

C 1. Igneous intrusion X formed.

2. Igneous intrusion Y formed.

3. The shale layer was deposited.  

4. The limestone layer was deposited.

5. The mudstone layer was deposited.

D 1. The layer of mudstone was deposited.

2. The layer of limestone was deposited.

3. Igneous intrusion Y was forced onto the mudstone and limestone.

4. The shale layer was deposited.

5. Igneous intrusion X was forced over the mudstone, limestone, and shale