Recall that matter cannot move without energy. What is the energy source that moves the red dye? What is the energy source that moves the blue dye? 2 points

Answer:

Answer:

The letter I

Explanation:

H, I, J, K, L, M, N, O, P

Answer:

Ahhhhhhhhhhhhhhh. Me too. I really needed to hear that today.

Explanation:

thanks for that.

Answer:

1

4

5

6

Explanation:

all magnets have opposite poles (north and south)

magnetic domains are aligned

moving electrons generate a magnetic filed

movement of electrons creates a magnets field

attracted to iron, nickel and cobalt

Answer:

I think 5

Explanation:

The piece of styrofoam will float and the piece of oak and gold ring will sink. This is because the density of an object determines whether it will float or sink in a fluid. Density is the amount of mass an object has in relation to its volume. Objects that are less dense than the fluid they are in will float, while objects that are more dense will sink.

In this scenario, styrofoam is less dense than water, so it will float. Oak and gold are more dense than water, so they will sink. The gold ring, in particular, has a high density due to its composition of a heavy metal. Even though it is small in size, its density is greater than that of water, causing it to sink. Density is an important concept in understanding buoyancy and how objects behave in fluids.

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Chromium is the metal impurity that makes rubies red and emeralds green.


Impurities are the foreign substances that enter into the crystal lattice of the mineral or gemstone. The addition of impurities causes the gemstone to change its color. In the case of rubies and emeralds, the presence of a particular metal impurity is responsible for the red and green color of the gemstones, respectively.

Chromium is the metal impurity that causes the ruby to appear red. It enters into the lattice of aluminum oxide (Al2O3), which is the main constituent of rubies. The amount of chromium present in the crystal lattice determines the color of the ruby. The more the amount of chromium present in the crystal lattice, the deeper the color of the ruby.

In the case of emeralds, the metal impurity that causes the green color is chromium and vanadium. The addition of chromium and vanadium to the crystal lattice of beryl (Be3Al2(SiO3)6) gives it a green color. The color of the emerald depends on the amount of chromium and vanadium present in the crystal lattice.

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The correct choice is A) −5Δ[Br−]/Δt. The rate expressed as Δ[Br2]/Δt is proportional to -5 times the rate of change of Br−.

In the given reaction 5Br−(aq) + BrO3−(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(aq), the stoichiometric coefficients provide information about the relationship between the reactants and products. To determine the rate expressed as Δ[Br2]/Δt, we need to compare it with the rate of change of the other species.

Based on the balanced equation, for every 5 moles of Br− consumed, 3 moles of Br2 are produced. Therefore, the rate of change of Br−, Δ[Br−]/Δt, is related to the rate of change of Br2, Δ[Br2]/Δt, by a factor of -5/3.

The other choices, B) −0.6Δ[Br−]/Δt, C) 3Δ[BrO3−]/Δt, and D) −Δ[H2O]/Δt, do not correspond to the correct relationship based on the stoichiometric coefficients of the reaction. Therefore, the correct answer is A) −5Δ[Br−]/Δt.

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The base in the following chemical interaction is CH3NH2, which is methyl amine. It creates its conjugate acid, the methyl ammonium ion, when it receives a proton.

The conjugate base's formula is the acid's formula less one hydrogen. The base that reacts changes into its conjugate acid. The conjugate acid's formula is the base's formula adds one hydrogen ion.

Because it is categorized as both a Lewis acid and a Bronsted-Lowry acid, methanol. It will become a Bronsted-Lowry acid as a result of the proton donation, and the methanol will also take electrons as a result.

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Answer:

changes that dont result in the formation of new substance with new properties.

The mucosa or mucous membrane layer which is the innermost layer of the alimentary canal wall, adsorbs nutrients .

The wall of alimentary canal possesses four layers - serosa , muscularis , submucosa , and mucosa . Serosa is the outermost layer and is made up of thin mesothelium and connective tissue . Muscularis is formed by smooth muscles usually arranged into an inner circular and an outer longitudinal layer . Sub-mucosa layer is formed of loose connection tissues congaing nerves , blood and lymph vessels .

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The magnification of the magnifying glass is approximately 1.44.

To calculate the magnification of the magnifying glass, we can use the formula:

Magnification = (image distance) / (object distance)

In this case, the magnifying glass is held 10.4 cm from the mole, so the object distance (d₀) is 10.4 cm. The focal length (f) of the magnifying glass is given as 15.0 cm.

Using the formula, we can calculate the magnification (M):

M = (image distance) / (10.4 cm)

The magnification of a magnifying glass is typically assumed to be the ratio of the image distance to the least distance of distinct vision (25 cm for a relaxed eye). Therefore, we need to find the image distance (dᵢ) in terms of the least distance of distinct vision.

Using the lens formula: 1/f = 1/d₀ + 1/dᵢ

We can rearrange the formula to solve for the image distance (dᵢ):

1/dᵢ = 1/f - 1/d₀

Substituting the given values:

1/dᵢ = 1/15 cm - 1/10.4 cm

Calculating the result:

1/dᵢ ≈ 0.0667 cm⁻¹

dᵢ ≈ 15 cm

Now we can substitute the values for image distance (dᵢ) and object distance (d₀) into the magnification formula:

M = 15 cm / 10.4 cm

Simplifying the calculation:

M ≈ 1.44

Therefore, the magnification of the magnifying glass in this scenario is approximately 1.44.

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Hope this helps!

Answer:

The change from a liquid phase to a gaseous phase.

so the answer would be it changed to a gaseous phase

---------------

This is what occurs when the vapor pressure of the liquid is equal to the atmospheric pressure exerted on the liquid.

Answer:Hydrogen

Explanation:is the most abundant element in the universe (75%by weight or 88% of all of the atoms of the universe)

The correct IUPAC name for the molecule can be determined by identifying the longest carbon chain and numbering it to give the substituents the lowest possible numbers. The correct name is:

3-methyl-6,6-diiodo-8-chlorononane

Explanation:

The longest carbon chain is 9 carbons, so the parent name is nonane. Numbering the chain from the end closest to the first substituent gives the methyl group the number 3. The other substituents are then numbered accordingly:

The two iodine atoms are on carbons 6 and 6 (diiodo).

The chlorine atom is on carbon 8 (chloro).

Therefore, the correct IUPAC name for the molecule is 3-methyl-6,6-diiodo-8-chlorononane.

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Answer:

a

Explanation:

A chemical equation is said to balanced only when the number of atoms of each element is the same in both reactants and products side. Rest all options are absurd. Hence, option a is correct.

Balancing of the reaction has no relation with coefficient, molecules or charge of the elements

Answer:

different types of waves are due to how and where they are formed

Explanation:

Don't give me brainliest I only know first part.

Explanation:

Oxidation States. +2,3,4,6,7

Electrons Per Shell 2 8 15

Electron Configuration [Ar] 4s2 3d5

1s2 2s2 2p6 3s2 3p6 4s2 3d5

The lower first ionization energy of sodium is due to the relatively weak attraction between the outermost electron and the nucleus, as well as the shielding effect provided by the inner electrons.

The ionization energy refers to the amount of energy required to remove an electron from an atom or ion in its gaseous state. In the case of sodium, the first ionization energy is significantly lower than the second ionization energy. This can be explained by understanding the electron configuration and the principles of electron shielding and effective nuclear charge.

Sodium has an atomic number of 11, meaning it has 11 protons in its nucleus and 11 electrons surrounding it. These electrons are arranged in energy levels or shells, with the first shell containing 2 electrons and the second shell containing 8 electrons. The outermost electron in sodium is in the third energy level.

The first ionization energy is the energy required to remove the outermost electron from the atom. In sodium, this electron is relatively far from the nucleus and experiences less attraction to the positively charged protons.

Additionally, the outer electron in sodium experiences significant electron shielding from the inner electrons, meaning that the inner electrons partially shield the outer electron from the full attractive force of the nucleus.

As a result, it is easier to remove the outermost electron in sodium, and hence, the first ionization energy is relatively low. Once the outermost electron is removed, sodium becomes a positively charged ion (Na+).

The second ionization energy refers to the energy required to remove an electron from the Na+ ion, which now has a stronger effective nuclear charge due to the reduced electron-electron repulsion and decreased shielding effect. Consequently, it is more difficult to remove an electron from the Na+ ion, leading to a higher second ionization energy compared to the first ionization energy.

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An unstable atomic nucleus loses energy during radioactive decay and changes into a more stable state, frequently by producing radiation in the form of particles or electromagnetic waves.

a. Th-234 alpha decay:

Th-234 -> He-4 + Ra-230

In this process, Th-234 releases an alpha particle, which is a helium-4 nucleus, and transforms into Ra-230.

b. Fe-59 beta decay:

Fe-59 -> Co-59 + e- + anti-neutrino

In this process, Fe-59 releases a beta particle, which is an electron, and transforms into Co-59. At the same time, an anti-neutrino is also released.

c. Tc-99 gamma decay:

Tc-99m -> Tc-99 + gamma

In this process, Tc-99m transitions from a higher energy state to a lower energy state and releases a gamma ray.

d. C-111 electron capture:

C-111 + e- -> B-11 + gamma

In this process, C-111 captures an electron and transforms into B-11. At the same time, a gamma ray is also emitted.

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The complete question:

Write a balanced nuclear equation for each decay process indicated.

a. The isotope Th-234 decays by an alpha emission.

b. The isotope Fe-59 decays by a beta emission.

c. The isotope Tc-99 decays by a gamma emission.

d. The isotope C-1ll decays by a electron capture.

Answer:

???????????????????????

Answer:

river or lake

Explanation:

Answer:

The periodic table is a set of elements organized by their atomic number, from lowest (hydrogen 1) to highest (oganesson 118). That number is the amount of protons in the nucleus of an atom of the element. The table is split into three groups of elements (metals, nonmetals and metalloids). As you go down the table the elements are more reactive.

Explanation:

^^

A large activation energy is required to break the double bonds of unsaturated hydrocarbons.

Unsaturated hydrocarbons are defined as the hydrocarbons in which double or triple bonds are present between two adjacent carbon atoms. They are known as alkenes and alkynes respectively. The general formula for these hydrocarbons is CnH2n and CnH2n-2

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The third quantum number (m_l) of a 3s² electron in phosphorus is 0.

The third quantum number, denoted as m_l, represents the magnetic quantum number and describes the orientation of an orbital within a subshell. It can have integer values ranging from -l to +l, where l is the azimuthal quantum number.

In the electron configuration of phosphorus, we see that the 3s subshell is being filled. The azimuthal quantum number (l) for the 3s subshell is 0. Since the electron is in the 3s² subshell, there are two electrons present in the 3s orbital.

For the two electrons in the 3s orbital, they will have opposite spins due to the Pauli exclusion principle. However, the magnetic quantum number (m_l) for both electrons in the 3s orbital will be the same, which is 0.

Therefore, the third quantum number (m_l) of a 3s² electron in phosphorus is 0. This means that both electrons in the 3s orbital have the same orientation within the subshell.

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1. The volume of water in the tank at time t is given by the equation

Volume(t) = 30 + t.

2.The appropriate initial condition for the ODE is S(0) = 0, as there is no salt initially in the tank.

3. It is not separable because the variables S(t) and t are not separable on opposite sides of the equation.

4. The solution can be expressed in terms of the integral as:
\(S(t) = (70 * \int e^{(t^{2/2} + 30t)} dt) / e^{(t^{2/2} + 30t)})\)

5.  the salt concentration in the tank as t→infinity is zero.

1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being added and drained. The tank is being filled at a rate of 2 liters per minute and drained at a rate of 1 liter per minute.

Since the tank starts with an initial volume of 30 liters, the volume of water in the tank at time t can be calculated using the equation:
Volume(t) = Initial volume + (Rate of filling - Rate of draining) * t

Volume(t) = 30 + (2 - 1) * t

So, the volume of water in the tank at time t is given by the equation

Volume(t) = 30 + t.

2. Let S(t) denote the amount of salt in the fish tank at time t in grams.

To show that S(t) satisfies the ODE S'(t) = 70 - (t+30)S(t),

we need to take the derivative of S(t) with respect to t and substitute it into the given ODE.

Taking the derivative of S(t), we have:

S'(t) = 0 - (1+0)S(t) + 0

S'(t) = -S(t)

Substituting this into the given ODE, we get:

-S(t) = 70 - (t+30)S(t)

Simplifying the equation, we have:

S'(t) = 70 - (t+30)S(t)

Therefore, S(t) satisfies the ODE S'(t) = 70 - (t+30)S(t).

The appropriate initial condition for the ODE is S(0) = 0,

as there is no salt initially in the tank.

3. This ODE is a first-order linear ordinary differential equation. It is not separable because the variables S(t) and t are not separable on opposite sides of the equation.

4. To solve the initial value problem for S(t) using the method of integrating factors, we first rewrite the ODE in standard form:

S'(t) + (t+30)S(t) = 70

The integrating factor is given by:
\(\mu(t) = e^{(\int (t+30) dt)} = e^{(t^2/2 + 30t)\)

Multiplying both sides of the equation by μ(t), we have:
\(e^{(t^2/2 + 30t)} * S'(t) + e^{(t^2/2 + 30t)} * (t+30)S(t) = 70 * e^{(t^2/2 + 30t)\)

Applying the product rule to the left side of the equation, we get:
\((e^{(t^{2/2} + 30t) * S(t))' = 70 * e^{(t^{2/2} + 30t)})\)

Integrating both sides of the equation with respect to t, we have:
\(\int (e^{(t^2/2 + 30t)} * S(t))' dt = \int (70 * e^{(t^2/2 + 30t))} dt\)

Using the fundamental theorem of calculus, the left side becomes:
\(e^{(t^2/2 + 30t)} * S(t) = \int (70 * e^{(t^2/2 + 30t))} dt\)

Simplifying the right side by integrating, we get:
\(e^{(t^2/2 + 30t)} * S(t) = 70 * \int e^{(t^2/2 + 30t)} dt\)

At this point, the integration of \(e^{(t^2/2 + 30t)\) becomes difficult to express in terms of elementary functions.

Hence, the solution can be expressed in terms of the integral as:
\(S(t) = (70 * \int e^{(t^2/2 + 30t)} dt) / e^{(t^2/2 + 30t)\)

5. As t approaches infinity, the exponential term \(e^{(t^2/2 + 30t)\) becomes very large, causing the salt concentration S(t) to approach zero. Therefore, the salt concentration in the tank as t→infinity is zero.

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The salt concentration in the tank as t approaches infinity is 70/3.

1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being poured into the tank and the rate at which water is being drained from the bottom.

At a rate of 2 litres per minute, water is being poured into the tank. So after t minutes, the amount of water poured into the tank is 2t litres.

At a rate of 1 litre per minute, water is being drained from the tank. So after t minutes, the amount of water drained from the tank is t litres.

Since the tank was initially filled with 30 litres of fresh water, the volume of water in the tank at time t is given by:
Volume(t) = 30 + 2t - t
Volume(t) = 30 + t

2. Let S(t) denote the amount of salt in the fish tank at time t. To determine the ODE for S(t), we need to consider the salt being poured into the tank and the salt being drained from the tank.

The salt concentration in the water being poured into the tank is 35 grams per litre. So the amount of salt being poured into the tank per minute is 35 * 2 = 70 grams.

The amount of salt being drained from the tank per minute is S(t)/Volume(t) * 1.

Therefore, the ODE for S(t) is:
S'(t) = 70 - S(t)/Volume(t)

The initial condition for this ODE is S(0) = 0, since there was no salt in the tank initially.

3. The ODE S'(t) = 70 - S(t)/Volume(t) is a first-order linear ODE. It is not separable since the variables S(t) and Volume(t) are mixed together.

4. To solve the initial value problem for S(t), we can rewrite the ODE as:
Volume(t) * S'(t) + S(t) = 70 * Volume(t)

This is a linear ODE of the form y'(t) + p(t)y(t) = g(t), where p(t) = 1/Volume(t) and g(t) = 70 * Volume(t).

To solve this type of ODE, we can multiply both sides by an integrating factor, which is the exponential of the integral of p(t).

The integrating factor is exp(integral of 1/Volume(t) dt) = exp(ln(Volume(t))) = Volume(t).

Multiplying both sides of the ODE by the integrating factor, we get:
Volume(t) * S'(t) + S(t) = 70 * Volume(t)
Volume(t) * S'(t) + Volume(t) * S(t) = 70 * Volume(t)^2
( Volume(t) * S(t) )' = 70 * Volume(t)^2

Integrating both sides with respect to t, we get:
Volume(t) * S(t) = 70/3 * Volume(t)^3 + C

S(t) = 70/3 * Volume(t)^2 + C/Volume(t)

Using the initial condition S(0) = 0, we can solve for C:
0 = 70/3 * 30^2 + C/30
C = -70000

Therefore, the solution for S(t) is:
S(t) = 70/3 * Volume(t)^2 - 70000/Volume(t)

5. As t approaches infinity, the volume of water in the tank becomes very large. In this case, we can approximate the volume of the tank as t, since the rate at which water is being poured in is 2 litres per minute. So the salt concentration in the tank as t approaches infinity is given by:
S(t)/Volume(t) = (70/3 * t^2 - 70000/t) / t
As t approaches infinity, the second term (-70000/t) approaches 0, so the salt concentration in the tank as t approaches infinity is:
S(t)/Volume(t) = 70/3 * t^2 / t = 70/3 * t

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The 1000 kg ball would have 62500 J of kinetic energy; The 10 kg ball would have 125 J of kinetic energy; The 100 kg person would also have 1250 J of kinetic energy.

Kinetic energy is the energy an object possesses due to its motion. It is defined as the energy that an object has as a result of its motion, and is dependent on the mass and velocity of the object.

To calculate the kinetic energy of an object, we use the formula:

KE = \(1/2 * m * v^{2}\)

where m is the mass of the object and v is its velocity.

Given that a 100 kg ball is traveling at 5 m/s, we can calculate its kinetic energy:

KE = \(1/2 * 100 kg * (5 m/s)^2\) = 1250 J

Using the same formula, we can calculate the kinetic energy for the following scenarios:

A 1000 kg ball traveling at 5 m/s:

KE =\(1/2 * 1000 kg * (5 m/s)^2\)= 62500 J

The 1000 kg ball would have 62500 J of kinetic energy, which is 50 times greater than the kinetic energy of the 100 kg ball.

A 10 kg ball traveling at 5 m/s:

KE = \(1/2 * 10 kg * (5 m/s)^2\)= 125 J

The 10 kg ball would have 125 J of kinetic energy, which is 10 times smaller than the kinetic energy of the 100 kg ball.

A 100 kg person traveling at 5 m/s:

KE = \(1/2 * 100 kg * (5 m/s)^2\) = 1250 J

The 100 kg person would also have 1250 J of kinetic energy, which is the same as the kinetic energy of the 100 kg ball.

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Answer:

The earlier bog bodies that were discovered in the 1600s might have not been preserved properly due to a lack of knowledge on how to preserve them or a lack of awareness of their significance. It is also possible that they might have decayed and decomposed over time and not survived till the present day. However, the bog bodies found after the late 1800s were preserved and studied extensively due to the increasing awareness and understanding of their historical and archaeological significance.

Explanation:

Hope this helped!! Have a great day/night!!

Answer:

The ocean absorbs most of the excess heat

Explanation:

Increasing ocean temperatures affect marine species and ecosystems. Rising temperatures cause coral bleaching and the loss of breeding grounds for marine fishes and mammals.