complete the monosaccharide units produced by hydrolysis for part a. drag the appropriate labels to their respective targets.

Answer:

What 3 factors influence the amount of thermal energy in an object, and how do they each affect it? Mass, temperature, and phase. More mass, more Thermal Energy. Higher temperature, more Thermal Energy.

ASAP=abbreviation. As soon as possible. 'fill in your form and send it to us ASAP' More example sentences.

The mass of H₂O in the oceans is much larger than the mass of H₂O originally contained in the mantle by a factor of approximately 3860

The mass of H2O in the oceans can be calculated using the following formula:mass of H2O in the oceans = volume of seawater × density of seawater × percentage of H2O in seawater where the volume of seawater is the total volume of the oceans on Earth, which is approximately 1.332 billion km³.

The density of seawater is 1.025 gm/cm³, and seawater is 96.5 percent H₂O. Therefore, the mass of H2O in the oceans is:m = 1.332 × 10⁹ km³ × (1.025 gm/cm³) × (0.965)= 1.307 × 10²¹ gmTo compare the mass of H₂O in the oceans to the mass of H₂O originally contained in the mantle, we need to first find the mass of H₂O originally contained in the mantle. The total mass of the mantle is approximately 4.5 × 10²⁴ gm, and it is estimated that the mantle contains between 50 and 100 ppm of H₂O.

Taking an average value of 75 ppm and using the mass of the mantle, we can calculate the mass of H₂O originally contained in the mantle as follows: mass of H₂O in mantle = (75 ppm) × (4.5 × 10²⁴ gm)= 3.38 × 10¹⁹ gm Therefore, the mass of H₂O in the oceans is much larger than the mass of H₂O originally contained in the mantle by a factor of approximately 3860. It is unlikely that the H₂O of the oceans came from the outgassing of the mantle alone, as the amount of H₂O in the oceans is much greater than the amount of H₂O originally contained in the mantle. Other sources of water, such as comets and asteroids, are thought to have contributed to the water content of the oceans.

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Electrolyte:

The substance that conducts electricity is known as an electrolyte.

Explanation :

For example, strong and weak acids and bases, as well as salts, are examples of electrolytes. In contrast, nonelectrolytes are substances that do not conduct electricity.

Strong electrolyte: Strong electrolytes fully dissociate in solution to produce ions.

For example, all ionic compounds are strong electrolytes, and acids such as hydrochloric acid and nitric acid are strong electrolytes.

Weak electrolyte: Weak electrolytes only partially ionize in solution, resulting in a few ions and a few molecules in solution.

For example, weak acids like acetic acid are weak electrolytes.

Nonelectrolyte: Nonelectrolytes don't conduct electricity in solution because they don't produce ions. For example, sugar and ethanol are both nonelectrolytes.

The following are the identifications of each substance as a strong electrolyte, weak electrolyte, or nonelectrolyte:

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The concentration of NO3-N in mg/L = 10 mg/L.

The given water sample contains 10 mg NO3/L.

To find the concentration of NO3 in millimoles/L:

1 mg = 1/1000 g1 mole of NO3 contains 1 + 3x16 = 61 g NO3So, 1 g NO3 = 1/61 mole NO3

Thus, 10 mg NO3 contains (10/1000) * (1/61) mole of NO3= 0.000163934426 mole of NO3

Therefore, the concentration of NO3 in millimoles/L = 0.000163934426 * 1000= 0.163934426 millimoles/L

To find the concentration of NO3-N in mg/L:1 mole of NO3 contains 1 mole of N= 14 g N

Thus, 1 g NO3-N = 1/14 mole of NO3-N

Therefore, 10 mg NO3-N contains (10/1000) * (1/14) mole of NO3-N= 0.000714285714 mol of NO3-N

Therefore, the concentration of NO3-N in mg/L = 0.000714285714 * 14000= 10 mg/L.

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The products of the reaction are Zinc chloride (ZnCl₂) and Hydrogen gas (H₂)

From the question,

We are to determine the product or products when zinc (Zn) reacts with hydrochloric acid (HCl)

To determine the product or products, we will write the chemical equation for the reaction.

The balanced chemical equation for the reaction is

Zn + 2HCl → ZnCl₂ + H₂

Hence, the products of the reaction are Zinc chloride (ZnCl₂) and Hydrogen gas (H₂)

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Reduction of a ketone with NaBH4  involves the use of sodium borohydride, a strong reducing agent. The sodium borohydride reacts with the carbonyl group of the ketone, resulting in the formation of a primary alcohol.

Here, correct option is A.

This reaction takes place in two steps. In the first step, a hydride ion is transferred from the sodium borohydride to the carbonyl carbon atom of the ketone, resulting in the formation of an intermediate alkoxide ion. In the second step, the alkoxide ion is protonated, resulting in the formation of a primary alcohol.

The use of NaBH4 to reduce a ketone is a simple and efficient way to produce primary alcohols from ketones. It is important to note that this reaction will not work on secondary or tertiary alcohols; the reaction will not proceed in these cases because the alkoxide ion formed in the first step is not stable enough to be protonated.

Therefore, correct option is A.

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Answer:

The answer is B. A hydrogen atom forms a convalent bond.........

The physical properties of alkenes and alkynes are generally similar to those of alkanes or cycloalkanes with equal numbers of carbon atoms. Alkynes have higher boiling points than alkanes or alkenes, because the electric field of an alkyne, with its increased number of weakly held π electrons, is more easily distorted, producing stronger attractive forces between molecules.

The statement is not part of the kinetic molecular theory of gases that tells us how gases generally behave is "the attractive forces between the particles of a gas are usually very strong".

The kinetic theory of gases describes a gas as a large number particles (atoms or molecules), all of which are in constant, random motion.

The rapidly moving particles constantly collide with each other and with the walls of the container.

The following are some of the assumptions of kinetic theory of gases;

Thus, the only incorrect statement about kinetic theory of gases is "the attractive forces between the particles of a gas are usually very strong".

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Explanation:

Atoms are generally made up of three fundamental particles which are the protons, neutrons and electrons.

Protons are the positively charged particles within an atom.

Electrons carry negative charges.

Neutrons do not carry any charges.

When an atom is neutral, the number of protons and electrons within the atom is the same.

A charged atom will have a difference in the number of protons and electrons they contain.

Answer:

6.B

7.B

Explanation:

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At equilibrium, the concentrations of the reactants and products remain constant over time, although the individual molecules are continuously interconverting. The concentration of water must be 0.227 M for the mixture to be at equilibrium.

To determine whether the concentration of water (H₂O) will increase, decrease, or remain the same as equilibrium is established, we need to compare the given concentrations of reactants and products with their stoichiometric coefficients in the balanced equation.

The balanced equation for the reaction is:

CH₃CO₂H + C₂H₅OH -> CH₃CO₂C₂H₅ + H₂O

Let's analyze each mixture (a-d) and determine the change in the concentration of water:

(a) [CH₃CO₂C₂H₅] = 0.22 M, [H₂O] = 0.10 M, [CH₃CO₂H] = 0.010 M, [C₂H₅OH] = 0.010 M

In this mixture, the concentration of water remains the same because the given concentrations are in equilibrium.

(b) [CH₃CO₂C₂H₅] = 0.22 M, [H₂O] = 0.0020 M, [CH₃CO₂H] = 0.0020 M, [C₂H₅OH] = 0.10 M

The concentration of water decreases because its initial concentration is higher than the equilibrium concentration.

(c) [CH₃CO₂C₂H₅] = 0.88 M, [H₂O] = 0.12 M, [CH₃CO₂H] = 0.044 M, [C₂H₅OH] = 6.0 M

The concentration of water increases because its initial concentration is lower than the equilibrium concentration.

(d) [CH₃CO₂C₂H₅] = 4.4 M, [H₂O] = 4.4 M, [CH₃CO₂H] = 0.88 M, [C₂H₅OH] = 10.0 M

The concentration of water remains the same because the given concentrations are in equilibrium.

To determine the concentration of water for a mixture to be at equilibrium with [CH₃CO₂C₂H₅] = 3.0 M, [CH₃CO₂H] = 0.10 M, and [C₂H₅OH] = 6.0 M, we need to set up an expression using the equilibrium constant (K) and the concentrations of the other species:

K = [CH₃CO₂C₂H₅] / ([CH₃CO₂H] * [C₂H₅OH] * [H₂O])

We can rearrange the equation to solve for [H₂O]:

[H₂O] = [CH₃CO₂C₂H₅] / (K * [CH₃CO₂H] * [C₂H₅OH])

Substituting the given concentrations:

[H₂O] = 3.0 M / (2.2 * 0.10 M * 6.0 M)

Simplifying the expression:

[H₂O] = 0.227 M

Therefore, the concentration of water must be 0.227 M for the mixture to be at equilibrium.

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The coefficient of OH⁻ in the balanced redox equation in basic solution is 6.

To provide a balanced redox equation in basic solution, I'll assume the reaction involves the oxidation of a species to a higher oxidation state.

Since we haven't provided specific reactants, we use a general example involving the oxidation of hypochlorite ion (ClO⁻) to chlorate ion (ClO₃⁻) using water (H₂O) as a reactant.

The balanced redox equation in basic solution is:

3ClO⁻ + 6OH⁻ → 5ClO₃⁻ + 3H₂O

In this equation, the coefficient of OH⁻ is 6.

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The thickness of the "circle" formed by rolling a 285 mg sample of gold with a radius of 0.78 cm is approximately 76.15 microns.

To find the thickness of the "circle" formed by rolling a gold sample, we can use the following steps:

Calculate the volume of the gold sample:

Volume = Mass / Density

V = 285 mg / 19.3 g/cm^3

Note: It's important to ensure consistent units.

Here, we convert milligrams (mg) to grams (g) to match the density unit.

Calculate the radius squared:

r^2 = (0.78 cm)^2

Calculate the thickness (height) of the "circle":

Height = Volume / (π * r^2)

Convert the thickness from centimeters to microns:

Thickness (in microns) = Height * 10,000

Let's calculate it:

Calculate the volume:

V = 285 mg / 19.3 g/cm^3

V = 0.01474 cm^3

Calculate the radius squared:

r^2 = (0.78 cm)^2

r^2 = 0.6084 cm^2

Calculate the height:

Height = V / (π * r^2)

Height = 0.01474 cm^3 / (π * 0.6084 cm^2)

Height ≈ 0.007615 cm

Convert the thickness to microns:

Thickness (in microns) = Height * 10,000

Thickness ≈ 76.15 microns

Therefore, the thickness of the "circle" formed by rolling a 285 mg sample of gold with a radius of 0.78 cm is approximately 76.15 microns.

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The pH of the solution after 20.0 ml of base have been added is 10.67.

Hydrochloric acid has a pH of less than 7.

It also goes by the name Muriatic acid. It is an odorless, colorless solution of the chemical hydrogen chloride in water. For human and most animal digestion, it is a part of stomach acid.

HCl(aq) + NaOH(aq) NaCl(aq) + H2O (1)

I mol HCl reacts with 1 mol NaOH

Mol HCI in 42.0mL of 0.135M Solution:

Mol HCI = 42.0mL / 1000mL/L 0.135 mol/L = 5.67*10^-3 mol

a) Mol NaOH in 0.00mL of 0.150 M solution

Mol NaOH = 0.20 ml/1000mL/L 0.150 mol/L = 3 × 10-3 mol

5.67*10^-3 mol NaOH will react with the 3.0*10^-3 mol HCI and there will be 2.67*10^-5 mol unreacted NaOH in 42 + 20 = 60 mL solution. This is 0.060 L.

Molarity of NaOH solution = (2.67*10^-5)mol/

0.060 L = 7.46*10^-4 M

[OH-] = 4.45*10^-4M

pOH = -log (4.45*10^-4)

POH = 3.35

pH = 14.00-3.35

pH = 10.67

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The true statement concerning racemic mixtures is that racemic mixtures are optically inactive (statement c). Racemic mixtures contain equal amounts of both enantiomers, resulting in no net rotation of plane-polarized light. The other statements (a, b, and d) are either false or have exceptions.

Racemic mixtures are mixtures that contain equal amounts of both enantiomers, which are mirror images of each other. Let's evaluate each statement to determine which ones are true.

a. Melting points between both enantiomers will vary:
This statement is false. Enantiomers have the same chemical composition, and their melting points are usually very similar or even identical. The only difference between enantiomers is their spatial arrangement.

b. Two enantiomers rotate plane-polarized light to an equal extent in the same directions, so the rotation doubles:
This statement is false. Enantiomers rotate plane-polarized light in equal amounts but in opposite directions. When a racemic mixture is formed, the rotations of the two enantiomers cancel each other out, resulting in no net rotation. As a result, racemic mixtures are optically inactive.

c. Racemic mixtures are optically inactive:
This statement is true. Racemic mixtures do not rotate plane-polarized light because the rotations of the two enantiomers cancel each other out. This is why racemic mixtures are considered optically inactive.

d. The boiling points between the enantiomers are identical:
This statement is generally true. Enantiomers have the same boiling points since their chemical compositions are the same. However, there can be some minor differences in boiling points due to factors like impurities or interactions with the surrounding environment.

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Answer:

first question is the third option and the second question is 849K.

Explanation:

Answer:

third option

849K

Explanation:

Answer: 41,650

Explanation: A kilopascal is one thousand pascals, so 41.65 times 1000 equals 41,650

Answer:

I think c) laying on an air mattress

Answer:

     3S₈  +  28Br₂ => 8S₃Br₇

Explanation:

Start with either sulfur (S) or bromine (Br) and balance ...

    3S₈  +  Br₂ => 8S₃Br₇    or      S₈  +  7/2Br₂ => S₃Br₇

Balance the remaining reactant ...

    3S₈  +  56/2Br₂ => 8S₃Br₇    

Remove fractions by multiplying by the fraction's denominator

    2(3S₈  +  56/2Br₂ => 8S₃Br₇)     =>     6S₈  +  56Br₂ => 16S₃Br₇      

Reduce to smallest whole number ratio => standard equation at STP ...

        3S₈  +  28Br₂ => 8S₃Br₇

Explanation:

A. Chemical potential energy.

According to the research, the correct option is C. A scatterplot identifies the problem and its variables showing quantitative data that relate two variables.

It is a control and support tool that shows a geometric design that presents the existing links and the level of correlation between variables and how one variable influences another.

In this sense, given that it is a tool or graphic representation it is widely used in statistics, that helps to identify the possible association between two related sets of data, it seeks to correlate said variables in order to better control the process and improve it.

Therefore, we can conclude that according to the research, the correct option is C. A scatterplot identifies the problem and its variables showing quantitative data that relate two variables.

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5 mols of O_2 need 4 mol NH_3

Now

Moles of Ammonia

Moles of NO=0.3mol

Mass of NO

So

Moles of Oxygen

Remaining moles

Mass

Yield

Answer:

2 only

Explanation:

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Answer:

The answer is two and three.

Explanation:

Answer:

Explanation:

this is a decomposition reaction because here a compound is broken into two two parts.

The dry gas would occupy 1.46 L at standard conditions.

When gas is collected over water, the vapor pressure of the water affects the total pressure measured. To account for this, we need to use Dalton's law of partial pressure, which states that the total pressure of a gas mixture is the sum of the partial pressures of each gas component.

First, we need to calculate the partial pressure of the collected gas. We can do this by subtracting the vapor pressure of water at 20 degrees Celsius (17.5 mm Hg) from the total pressure measured:

Partial pressure of gas = total pressure - vapor pressure of water
Partial pressure of gas = 622.0 mm Hg - 17.5 mm Hg
Partial pressure of gas = 604.5 mm Hg

Next, we can use the ideal gas law (PV = nRT) to calculate the volume of the dry gas at standard conditions (0 degrees Celsius and 1 atm):

PV = nRT
V = nRT/P

where P is the partial pressure of the gas (604.5 mm Hg converted to atm), n is the number of moles of gas (which we can calculate using the volume of the collected gas and the known molar volume of a gas at STP), R is the gas constant, and T is the temperature in Kelvin (273 K).

V = (40 L)(0.0821 L·atm/mol·K)(293 K)/(0.793 atm)
V = 1.46 L

Therefore, the dry gas would occupy 1.46 L at standard conditions.

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Answer:

Pulling back on the plunger allows you to put in more air into the pump and the more air you have inside a fixed volume, the greater the pressure becomes. (D)