HELP PLEASE, GETS BRAINLIEST

Answer:

It is photosynthesis and transpiration

hope it helps

Global warming brought on by rising carbon dioxide concentrations in the atmosphere is changing weather patterns and ocean temperatures. Changes in precipitation patterns are being brought on by the warming of the atmosphere.

Because it is one of the gases in the atmosphere that causes the greenhouse effect, which causes the Earth to warm, carbon dioxide is known as a greenhouse gas. Long-wavelength infrared radiation (heat) from the Earth is absorbed by carbon dioxide molecules in the atmosphere, and some of it is then radiated back downward.

Similarly, as air temperatures rise in response to rising carbon dioxide concentrations, more water vapor escapes into  the atmosphere—which then amplifies greenhouse heating

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Answer:

Explanation:

Simple organic molecules having only carbon and hydrogen are alkanes, alkenes and alkynes.

Ways of making these molecules are referred to as synthesis

Synthesis of

(1) alkanes can be achieved by halogenation of an alkene or alkyne as shown in the reaction below

C₂H₄ + H₂ ⇒ C₂H₆

The reaction above shows an alkene (ethene) undergoing halogenation to form an alkane (ethane). This reaction occurs in the presence of a platinum or nickel catalyst

C₂H₂ + 2H₂  ⇒ C₂H₆

The reaction above shows an alkyne (ethyne) undergoing halogenation to form an alkane (ethane). This reaction also occurs in the presence of a nickel or platinum catalyst

(2) Alkanes can be achieved by Wurtz reaction; treating alkyl halides with sodium metal as shown below

CH₃CH₂Br + 2Na + BrCH₂CH₃ ⇒ CH₃CH₂CH₂CH₃ + 2NaBr

This reaction proceed in the presence of a dry ether.

(3) alkanes from Kolbe's electrolytic method; this involves the synthesis of alkanes through the electrolysis of sodium or potassium salt of a carboxylic acid

2CH₃COO⁻Na⁺ + 2H₂O  ⇒ CH₃CH₃ + 2CO₂ + H₂ + 2NaOH

(4) alkenes from dehydration of alcohol, as shown in the reaction below

C₂H₅OH ⇒ C₂H₄    (-H₂O)

The reaction above proceeds by heating the alcohol (ethanol) at high temperature in the presence of strong acid such as sulphuric acid.

Answer:

7.77×10^4=77,700

Explanation:

10^4 is 10,000, since there are 4 zeros in 10,000, move the decimal in 7.77 4 spaces to the right.

1: 77.7

2: 777.

3: 7,770.

4 77,700.

Side note: This trick ONLY works with +/-10^x value. If the value is positive the decimal moves to the right, and if the value is negative the decimal moves to the left.

Example: 7.77×(-10)^3=.00777

(-10)×(-10)×(-10)=(-10)^3= -1000

1: .777

2: .0777

3: .00777

This technique has been used to identify the presence of gases such as oxygen, methane, and carbon dioxide in the atmospheres of exoplanets.  

i) To estimate the frequency of the violet (leftmost) emission, we can use the equation v = c/λ, where v is frequency, c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the emission in meters. The wavelength of the violet emission is 400 nm or 400 x 10^-9 m, so the frequency can be calculated as v = (3.00 x 10^8 m/s) / (400 x 10^-9 m) = 7.50 x 10^14 Hz.

ii) To estimate the energy of the violet emission, we can use the equation E = hv, where E is energy, h is Planck's constant (6.63 x 10^-34 Js), and v is frequency in Hz. Substituting the frequency calculated in part (i), we get E = (6.63 x 10^-34 Js) x (7.50 x 10^14 Hz) = 4.97 x 10^-19 J.

b. The spectral lines are produced by the electrons within the atoms of this element, which can absorb or emit specific amounts of energy to move between different energy levels. These energy transitions result in the emission or absorption of photons with specific wavelengths and frequencies, giving rise to the observed emission spectrum.

c. The violet emission line represents the photon with the most energy since it has the shortest wavelength (400 nm) and highest frequency (7.50 x 10^14 Hz) among the lines shown. This highest energy does not necessarily represent the energy of the valence electrons, but rather corresponds to the specific energy transitions occurring within the atoms of the element.

d. Emission spectra can be used to determine the gases present in the atmosphere of a far-away planet by analyzing the specific wavelengths of the emitted or absorbed light from the planet. Each gas has a unique emission or absorption spectrum, allowing scientists to identify the gases present in the planet's atmosphere.

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Answer:

0 587 mL

Explanation:

First we convert 180 grams of HCl into moles, using its molar mass:

Now we can use the number of moles and the given concentration to calculate the required volume, applying the definition of molarity:

Finally we convert liters into milliliters:

The closest answer is option C, 587 mL.

The volume needed to make an 8.40 M solution of HCl is 0.587 ml.

The correct option is C.

HCl is a strong acid. It is present in our stomach to help in digesting food.

Step1: covert 180 grams into moles

Molar mass of HCl is 36.46 g

\(\dfrac{180 g}{36.46 g/mol } = 4.94\; mol\; HCl\)

Step2: calculate the molarity

\(\rm Molality = \dfrac{mol}{L}\\\\\rm L = \dfrac{4.94 mol }{8.40 M} = 0.588 L\\\)

Covert the liters into ml

\(\dfrac{0.588 L}{1000} = 588 mL\)

Thus, the volume of the solution is 588 ml., nearest is option C, 587 ml.

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Based on the balanced chemical equation, the volume of 0.250 M  K₂S₂O₃ (aq) needed to completely react with 37.32 mL of 0.125 M KI₃ (aq) is 37.32 mL.

The reaction between K₂S₂O₃ (aq) and KI₃ (aq) is a redox reaction.

The balanced chemical equation of the reaction is given as follows:

2 S₂O₃²⁻ (aq) + I₃⁻ (aq) ---> S₄O₆²⁻ (aq) + 3 I⁻ (aq)

Based on the balanced equation of the reaction:

2 moles of K₂S₂O₃ (aq) are required to react with one mole KI₃ (aq).

Moles of KI₃ (aq) in  37.32 mL of 0.125 M solution;

Moles of  KI₃ (aq) = 37.32 * 0.125 M

Moles of  KI₃ (aq) = 4.665 mmoles

Moles of K₂S₂O₃ (aq) required = 2 *  4.665 mmoles

Moles of K₂S₂O₃ (aq) required = 9.33 mmoles

Volume of K₂S₂O₃ (aq) required =moles / molarity

The volume of K₂S₂O₃ (aq) required = 9.33 / 0.25

The volume of K₂S₂O₃ (aq) required = 37.32 mL

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Answer:

By observing the time between transits, we know the orbital period. Kepler's Third law can be used to determine the orbital radius of the planet if the mass of the orbiting star is known (R3=T2−Mstar/Msun, the radius is in AU and the period is in earth years)

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Ionic and Covalent Bonds- The materials utilized during the experiment, the steps taken, and the techniques used to analyze the results should all be described in the section titled "Materials and Methods."

A lab report is broken down into eight sections: title, abstract, introduction, techniques and materials, results, discussion, conclusion, and references. The title of the lab document must be descriptive of the scan and replicate what the scan analyzed.

Ionic bonds require an electron donor, the metal, and an electron accepter, the non-metal. Covalent bonding is the sharing of electrons between atoms. This kind of bonding happens between two of the equal element or elements shut to each other in the periodic table.

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There are approximately \(2.76 * 10^{24\) ammonia molecules in the given sample.

To calculate the number of ammonia molecules in the sample, we need to use Avogadro's number and the molar mass of ammonia.

The molar mass of ammonia \((NH_3)\) can be calculated by adding up the atomic masses of nitrogen (N) and hydrogen (H):

Molar mass of \(NH_3\) = (1 x atomic mass of N) + (3 x atomic mass of H)

              = (1 x 14.01 g/mol) + (3 x 1.01 g/mol)

              = 14.01 g/mol + 3.03 g/mol

              = 17.04 g/mol

Now, we can calculate the number of moles of ammonia in the sample using the formula:

Number of moles = Mass of the sample / Molar mass

Number of moles = 78.25 g / 17.04 g/mol

              ≈ 4.5865 mol (rounded to four decimal places)

Finally, we can use Avogadro's number, which represents the number of particles (atoms, molecules, etc.) in one mole of a substance. Avogadro's number is approximately \(6.022 * 10^{23\) particles/mol.

Number of ammonia molecules = Number of moles x Avogadro's number

Number of ammonia molecules ≈ 4.5865 mol x (\(6.022 * 10^{23\) molecules/mol)

                           ≈ \(2.76 * 10^{24\) molecules (rounded to two significant figures)

Therefore, the provided sample contains roughly \(2.76 * 10^{24\) ammonia molecules.

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The number of ammonia molecules in the sample is approximately 2.764 x \(10^{24}\) molecules.

To calculate the number of ammonia molecules in a given sample, we need to use Avogadro's number and the molar mass of ammonia.

The molar mass of ammonia (NH3) is calculated as follows:

Molar mass of N = 14.01 g/mol

Molar mass of H = 1.01 g/mol

Total molar mass of NH3 = 14.01 g/mol + (3 * 1.01 g/mol) = 17.03 g/mol

Now, we can calculate the number of moles of ammonia in the sample:

Number of moles = Mass of sample / Molar mass of NH3

Number of moles = 78.25 g / 17.03 g/mol = 4.594 moles

Next, we use Avogadro's number, which states that there are 6.022 x \(10^{23}\) molecules in one mole of a substance.

Number of molecules = Number of moles * Avogadro's number

Number of molecules = 4.594 moles * 6.022 x \(10^{23}\) molecules/mol = 2.764 x \(10^{24}\) molecules

Therefore, there are approximately 2.764 x \(10^{24}\) ammonia molecules in the given sample of 78.25 g.

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The asbestos exposure during the years 1982 and 1983 was 2.5 fibers per cubic centimeter and 0.8 fibers per cubic centimeter respectively. So asbestos exposure decreased by 1.7 fibers per cubic centimeter during the year 1983.

Breathlessness Persistent, dry cough Chest pain or tightness Lack of a dry, crackling sound in the lungs when you breathe in Wider and rounder fingers and toes are some of the symptoms of asbestos exposure.

The largest group of people exposed to asbestos is those working in the construction industry. Historically, asbestos was also used by pipe fitters and shipyard workers. In addition, asbestos was used by military personnel, auto mechanics, and many other occupations.

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The branched alkanes in boxes two and three all have molecular formula C6H14 and are isomers of the compound shown.

Isomers are compounds that has the same molecular formula but different structural formulas. Hence, isomers of compounds can be represented by the same molecular formula since they contain the same number of each atom.

The molecule shown has molecular formula C6H14. The branched alkanes in boxes two and three all have molecular formula C6H14 and are isomers of the compound shown.

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Draw cyclohexanol with a double bond and without the OH- group. next draw water

Two atoms sharing two sets or electrons result in a double bond. Always shared in pairs, electrons. Covalent bonds are referred to as each shared electron pair. A double bond between two atoms requires more energy to break than a single bond because four electrons are divided between the two atoms in a dual bond.

When two atoms join one couple of electrons, they form a single bond; when they share two pairs, they form a double bond (four electrons). To form triple bonds, three pairs free electrons (six atoms) share space.

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To solve this problem using the gas laws, we need to use the Ideal Gas Law. This law states that the product of the pressure and the volume of a gas is proportional to the absolute temperature.

The equation of the Ideal Gas Law is the following:

\(\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{\dfrac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} } \end{gathered}$} }\)

Where:

We clear for V₂:

\(\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{V_2=\frac{P_1V_1T_2}{P_2T_1 } } \end{gathered}$} }\)

Where:

Substituting the known values:

\(\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{V_2=\frac{760\not{kPa}\times0.050 \ L\times20.0\not{k} }{ 380\not{kPa}\times10.0\not{k} } } \end{gathered}$} }\)

\(\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{V_2=\frac{760 \ L}{3800 } } \end{gathered}$} }\)

\(\boxed{\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{V_2\approx0.2 \ Liters} \end{gathered}$} }}\)

Answer:

A change is called irreversible if it cannot be changed back again. For example you cannot change a cake back into its ingredients again. Irreversible changes are permanent.

Explanation:

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Yes, a chain reaction can only  be carried out using just one atom.

We know that a chain reaction has to do with the type of reaction in which there is a a continuous self sustaining reaction. In this kind of reaction, there is one radioactive material that is bombarded by a particle such as a neutron. The bombardment of the particle would make the a cycle of reactions to begin which is able to sustain itself.

Given the fact that we only need to have a single radioactive atom that is bombarded for the process of radioactivity to occur tell us that it is a process that requires just one atom.

Take for instance, all we need is one Uranium fuel that would be bombarded with neutrons in order to have a chain reaction that involves  uranium.

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Copper atoms (Cu) were created by reducing copper ions (Cu2+). An illustration of a reduction reaction is this. (Mg2+).To create Mg2+ ions, Mg atoms lost their electrons. An example of an oxidation reaction is this.

Oxidation and reduction occurred in test tube 5. Here is the equation for the reaction that took place:MgSO4 (aq) + Cu (s) CuSO4 (aq) + MgCuSO4's blue hue was neutralised to a solid copper hue. Copper atoms (Cu) were created by reducing copper ions (Cu2+). An illustration of a reduction reaction is this. Cu atoms are created when Cu2+ gains electrons. Due to the presence of magnesium ions (Mg2+), the solution was greenish-yellow.To create Mg2+ ions, Mg atoms lost their electrons. An example of an oxidation reaction is this. Two electrons were lost by the magnesium atom, leaving two electrons behind.

As a result, test tube 5 experienced both oxidation and reduction.It is possible to prove that oxidation, reduction, or both took place through the reaction's electron transfer and colour changes. A reduction reaction was seen when copper ions transformed into copper atoms. An oxidation reaction was evident when magnesium atoms transformed into magnesium ions.

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Krypton, Chromium, and Oxygen with the following symbols Kr-13, Cr-2, and O-15 respectively have 23 electrons.

The atomic number of an atom determines the number of electrons it has. When the number of protons is equivalent to the number of electrons, the atom is electrically neutral. An anion, on the other hand, is an atom with a negative charge. It has gained an electron or two, or even more. Below are the chemical symbols for three different atoms or atomic anions with 23 electrons.Krypton:Kr has an atomic number of 36, indicating that it has 36 electrons. However, if we add 13 electrons to it, the total number of electrons becomes 49. Krypton with 13 additional electrons becomes Kr-13, with a total of 49 electrons.Chromium:Cr has an atomic number of 24, indicating that it has 24 electrons. Adding two more electrons to it, the total number of electrons becomes 26. The atomic anion with 26 electrons is Cr-2.Oxygen:Oxygen has an atomic number of 8, indicating that it has 8 electrons. However, if we add 15 electrons to it, the total number of electrons becomes 23. Oxygen with 15 additional electrons becomes O-15, with a total of 23 electrons.

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Answer:

The volume of Component 1 is 36 cubic inches.

Explanation:

To calculate the volume of a composite solid, we need to determine the individual volumes of the different components and then add them together.

In this case, the composite solid consists of multiple components with the following dimensions:

Component 1:

Length: 4 inches

Width: 3 inches

Height: 3 inches

To find the volume of Component 1, we multiply the length, width, and height together:

Volume of Component 1 = Length x Width x Height = 4 in x 3 in x 3 in = 36 cubic inches

Therefore, the volume of Component 1 is 36 cubic inches.

Please provide the dimensions of the remaining components of the composite solid, and I will calculate the total volume by summing up the individual volumes.

The half reaction with a more positive standard reduction potential will proceed spontaneously in a redox reaction

The half reaction with a more positive standard reduction potential will undergo reduction when compared to the half reaction with a more negative standard reduction potential.

Oxidation-reduction reactions, often known as redox reactions, are a set of chemical reactions that involve electron transfer between reactants. In a redox reaction, one reactant is oxidized, losing electrons, while the other reactant is reduced, gaining electrons.

The oxidation half-reaction is the process of losing electrons and increasing the oxidation number, whereas the reduction half-reaction is the process of gaining electrons and decreasing the oxidation number. The total reaction is referred to as the redox reaction.

Half-reaction:Half-reaction refers to the two parts of an oxidation-reduction reaction that happen separately. A half-reaction must always be either an oxidation reaction or a reduction reaction. It also describes the movement of electrons and hydrogen ions in an equation.

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Based on the given reactions, the compounds are as follows:

A: The specific product formed from the reaction between prop-1-yne and either 2HBr or H2O2.

B: The product formed when compound A reacts with 2H2O.

C: The product formed when compound B reacts with K2CO3(aq).

D: The product formed from the heat-induced reaction of compound C.

E: The product formed when compound D reacts with HBr.

Based on the given reactions, let's analyze the compounds involved:

Reaction 1: prop-1-yne + 2HBr/H2O2 = A

The reactant prop-1-yne reacts with either 2HBr or H2O2 to form compound A. The specific product formed will depend on the reaction conditions.

Reaction 2: A + 2H2O = B

Compound A reacts with 2H2O (water) to form compound B.

Reaction 3: B + K2CO3(aq) = C

Compound B reacts with K2CO3(aq) (potassium carbonate dissolved in water) to form compound C.

Reaction 4: C + heat = D

Compound C undergoes a heat-induced reaction to form compound D.

Reaction 5: D + HBr = E

Compound D reacts with HBr (hydrobromic acid) to form compound E.

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Armondo could test how curves in a river affect the speed of the river as the speed of the water outside of a bend increases as a river rounds the bend. The water's velocity, however, diminishes as it approaches the interior of the curve.

A fantastic illustration of how water can alter the contour of the land is a meandering river. A river seldom turns when it is bordered by sheer rock, but it will do so when it opens up in broad valleys. Water will flow more swiftly and destroy the ground faster outside the river. Over time, it will curve too much and slow down.

To simulate a meandering river, use a stream table. The speed of the water outside of a bend increases as a river rounds the bend. The water's velocity, however, diminishes as it approaches the interior of the curve. A bar of deposited silt, like this one, is created as a result of the reduction in velocity.

The velocity of a river is the rate at which water flows through its course. Numerous elements, such as the channel's form, the slope's grade, the amount of water carried by the river, and the amount of friction brought on by jagged edges in the riverbed, all affect a river's speed.

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A proper standard curve shows D. a linear relationship when plotting Concentration vs. Absorbance.

A standard curve is used to determine the concentration of an unknown sample by comparing its absorbance to a series of known standards with known concentrations. The most common way to plot a standard curve is to graph the absorbance on the y-axis and the concentration of the standards on the x-axis. When plotted in this way, a proper standard curve should show a linear relationship, where the absorbance is proportional to the concentration of the standard.

Therefore, option D, a linear relationship when plotting Concentration vs. Absorbance, is the correct answer.

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A  chemical change is said to occur when there is the appearance of a new substance after the reaction.

We know that a chemical change is said to occur when there is the appearance of a new substance after the reaction. This can be signaled by the appearance of a solid or gas as well as a change in color.

Now what has happened here is a chemical reaction that took place as follows;

2Mg(s) + O2(g) ------> 2MgO(s)

The burning of magnesium gives off a silvery white emission spectrum. The reaction that occurs is a combustion reaction and a new substance, MgO was formed.

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Voltaic series

Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au

The more to the left, the metal is more reactive, so the most reactive is Mg

The pOH of the 0.001 M NaOH solution is approximately 3.

To determine the pOH of a solution, we need to know the concentration of hydroxide ions (OH-) in the solution.

In the case of a 0.001 M NaOH solution, we can assume that all of the NaOH dissociates completely in water to form Na+ and OH- ions. Therefore, the concentration of hydroxide ions in the solution is also 0.001 M.

The pOH is calculated using the equation:

pOH = -log[OH-]

Substituting the concentration of hydroxide ions, we have:

pOH = -log(0.001)

Using a calculator, we can evaluate the logarithm:

pOH ≈ 3

Therefore, the pOH of the 0.001 M NaOH solution is approximately 3.

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We need to determine how is the PH in the given mixtures:

A) A weak base and a strong acid:

Acid and base react to form an acidic solution and, being acidic, this results in a solution with a pH lower than 7

B) A strong base and a strong acid:

When mixing a strong base and a strong acid, neutralization occurs, that is, a neutral pH or equal to 7 is obtained

C) A strong base and acid:

If we have a weak acid, when adding a strong base the pH approaches neutrality without sudden changes, but once the acid has been neutralized, just add a few drops of excess soda to obtain a sudden increase in pH as if there were only base. free. In this case, at the equivalence point the pH> 7

ANSWER:

A) Less than 7

B) Equal to 7

C) Greater than 7

Answer: True

Explanation: