Waves made by the breeze were very different than wave created by speedboat. describe the difference

Answer:

For A: The molarity is the solution 9.706 M

For B: The molarity of the solution is 10.91 M.

For C: The molarity of the solution is 1.78 M.

Explanation:

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L.

The formula used to calculate molarity:

[tex]\text{Molarity of solution}=\frac{\text{Number of moles of solute}}{\text{Volume of solution (L)}}[/tex]                 ......(1)

For A:

Number of moles = 3.30 moles

Volume of solution = 0.340 L

Plugging values in equation 1:

[tex]\text{Molarity of solution}=\frac{3.30 mol}{0.340 L}=9.706 mol/L[/tex]

For B:

Number of moles = 6.00 moles

Volume of solution = 550 mL = 0.550 L        (Conversion factor: 1 L = 1000 mL)

Plugging values in equation 1:

[tex]\text{Molarity of solution}=\frac{6.00 mol}{0.550 L}=10.91 mol/L[/tex]

For C:

Number of moles = 0.0160 moles

Volume of solution = 9.00 mL = 0.009 L

Plugging values in equation 1:

[tex]\text{Molarity of solution}=\frac{0.0160 mol}{0.009 L}=1.78 mol/L[/tex]

Answer:

D

Explanation:

It states that D has oxygen, and is closer to the Sun.

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Answer:

I am pretty sure it is B

Explanation:

The others would either burn u like a crispy chicken nugget or freeze you like a popsicle

I'm pretty sure ita A. acceleration

Answer:

it's a friction force

The product of the reaction between [tex]CH_{3}[/tex]-CH=[tex]CH_{2}[/tex] and HBr is 2-bromo-3-methylbutane.

The reaction proceeds through the addition of a proton from HBr to the double bond, followed by the addition of a bromide ion.

The addition of the proton is stereospecific, and the bromide ion will add to the carbon atom that is least substituted by hydrogen. In this case, the carbon atom that is least substituted by hydrogen is the carbon atom that is attached to two hydrogen atoms.

Therefore, the bromide ion will add to the carbon atom that is attached to the double bond and the methyl group. The product of the reaction is 2-bromo-3-methylbutane.

here is the predominant product of the reaction of [tex]CH_{3}CHCH=CH-CH_{3}[/tex] with HBr:

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The complete question is:

Draw the predominant product(s) of the following reactions including stereochemistry when appropriate. CH CH CH -CEC-H HBr Consider EIZ stercochemistry of alkenes. Do not show stereochemistry in other cases If no reaction occurs_ raw the organic starting material. Draw one stnicture per sketcher Add additional sketchers using the drop down menu in the bottom right corner Separate multiple products using the sign from the drop-down menu: ChemDoodle Submit Answver Retry Entire Group more cirovp attempts remaining

Answer:

Corrosion is the process of deterioration of materials as a result of chemical, electrochemical or other reactions. Rusting is a part of corrosion and is a chemical process which results in the formation of red or orange coating on the surface of metals. ... Rust or rusting can affect only iron and its alloys.

Explanation:

The two emissions scenarios that most closely represent the current trend in CO2 emissions are the Representative Concentration Pathway (RCP) 4.5 and RCP 6.0 scenarios. RCP 4.5 assumes moderate emission reduction efforts, while RCP 6.0 represents a higher emission trajectory, reflecting the current trend where emissions reductions are not keeping pace with necessary targets.

The Representative Concentration Pathways (RCPs) are scenarios used to assess the potential impacts of greenhouse gas emissions on the climate system. RCP 4.5 assumes a moderate emission reduction pathway, with emissions peaking around 2040 and declining gradually. On the other hand, RCP 6.0 represents a higher emission trajectory, with emissions peaking later and declining more slowly compared to RCP 4.5. This scenario aligns with the current trend of rising CO2 emissions, indicating that global efforts to reduce emissions have not been sufficient. The current trend is closer to RCP 6.0, highlighting the challenges of achieving widespread emission reductions in various sectors of the global economy.

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Answer:

a, they are the building blocks of proteins.

Explanation:

hope this helps!

~mina

Answer:

A. They are the building blocks of proteins.

(Photo for proof at the bottom.)

Explanation:

Glycine, leucine, and lysine are all amino acids. There are 20 different kinds of amino acids, including the 3 mentioned here. Different amino acids bond together to form different structures, depending on how many there are. Generally, 50 or more amino acids bonded together makes a protein. But there are some exceptions. Amino acids (including glycine, leucine, and lysine) are the building blocks of proteins.

Here's a photo of Edge incase you're doubtful.

Please click the heart if this helped.

a. Metals such as zinc (Zn) or aluminum (Al) could be reasonable guesses as they have higher standard reduction potentials than copper. b. Reduction half-reaction is Cu₂+(aq) + 2e⁻ → Cu(s), Oxidation half-reaction is Unknown metal (M)(s) → M₂+(aq) + 2e⁻.

a. To make a reasonable guess for the unknown metal in the electrochemical cell, we can consider the standard reduction potentials (E°) of different metals. The standard reduction potential is a measure of the tendency of a metal to undergo reduction (gain electrons) compared to a standard hydrogen electrode (SHE).

Since the free energy change under standard conditions is negative (-293.4 kJ/mol), it indicates a spontaneous redox reaction, where the copper solution undergoes reduction and the unknown metal solution undergoes oxidation.

A reasonable guess for the unknown metal would be a metal with a higher standard reduction potential than copper (Cu). This is because for the overall reaction to be spontaneous, the unknown metal should have a greater tendency to undergo oxidation (lose electrons) compared to copper. Based on this, metals such as zinc (Zn) or aluminum (Al) could be reasonable guesses as they have higher standard reduction potentials than copper.

b. The reduction and oxidation half-reactions can be written as follows

Reduction half-reaction

Cu₂+(aq) + 2e- → Cu(s)

Oxidation half-reaction

Unknown metal (M)(s) → M₂+(aq) + 2e⁻

In the reduction half-reaction, copper ions (Cu₂⁺) in solution are reduced and gain two electrons to form solid copper (Cu).

In the oxidation half-reaction, the unknown metal (M) in solid form is oxidized, losing two electrons to form metal ions (M₂⁺) in solution.

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Answer:

B. Ca

Explanation:

Let's look at the electron configurations of all four elements (I am going to write it in noble gas configuration to make it simpler):

Mg electron configuration: [Ne]3s2

Ca electron configuration: [Ar]4s2

Ar electron configuration: [Ar]

K electron configuration: [Ar] 4s1

We notice that Ca has two electrons in the 4s sublevel, which satisfies what the question is asking for.

The answer is thus B. Ca.

Answer:

E

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BRAINLIEST PLEASE

The pH after the addition of 30.0 mL of NaOH at 25 °C is 8.82.

To calculate the pH after the addition of NaOH, we need to determine the moles of CH3COOH and NaOH that react, and then use the stoichiometry to find the resulting concentration of CH3COOH and OH-.

Given:

Volume of CH3COOH = 30.0 mL = 0.0300 L

Concentration of CH3COOH = 0.50 M

Ka for CH3COOH = 1.8×10^(-5)

Volume of NaOH = 30.0 mL = 0.0300 L

Concentration of NaOH = 0.50 M

First, we calculate the moles of CH3COOH:

moles of CH3COOH = concentration × volume

moles of CH3COOH = 0.50 M × 0.0300 L

moles of CH3COOH = 0.015 mol

Since CH3COOH and NaOH react in a 1:1 ratio, the moles of NaOH are also 0.015 mol.

Next, we calculate the moles of OH- produced:

moles of OH- = moles of NaOH

moles of OH- = 0.015 mol

Now, we calculate the concentration of OH-:

concentration of OH- = moles of OH- / total volume

concentration of OH- = 0.015 mol / (0.0300 L + 0.0300 L)

concentration of OH- = 0.250 M

Using the equilibrium expression for the dissociation of water, we can calculate the concentration of H+ (or H3O+):

[H+][OH-] = Kw

[H+] = Kw / [OH-]

[H+] = 1.0 × 10^(-14) / 0.250 M

[H+] = 4.0 × 10^(-14) M

Finally, we calculate the pH:

pH = -log[H+]

pH = -log(4.0 × 10^(-14))

pH = 8.82

The pH after the addition of 30.0 mL of NaOH at 25 °C is 8.82.

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Answer:

24 hrs it takes for the earth to rotate

Answer:

c

Explanation:

Answer:

m = 204.16 grams

Explanation:

Given that,

No. of moles, n = 3.49

Molar mass of NaCl, M = 58.5 g/mol

We need to find the mass of 3.49 moles NaCl​. We know that,

[tex]n=\dfrac{m}{M}\\\\m=n\times M\\\\m=3.49 \times 58.5\\\\m=204.16\ g[/tex]

So, there are 204.16 grams in 3.49 moles NaCl​.

In a 0.25 M aqueous solution of boric acid, the species with the highest concentration would be H_{3}BO_ 3}^{-} option B.

Boric acid (H_{3}BO_{3}) is a weak acid that can undergo multiple ionization steps. The given equilibrium constants (Ka) provide information about the ionization reactions. The first ionization reaction of boric acid is as follows:

[tex]H_{3}BO_{3}[/tex] ⇌ H^{+}+ H_{3}BO_ 3}^{-}

Ka1 = 7.3 x 10^-10

The second ionization reaction is:

[tex]H_{3}BO_ 3}^{-}[/tex] ⇌H^{+} + H_{3}BO_ 3}^{-2}

Ka2 = 1.8 x 10^-13

The third ionization reaction is:

H_{3}BO_ 3}^{-2} ⇌ H+ + [tex]BO_{3} ^{-3}[/tex]

Ka3 = 1.6 x [tex]10^{-14}[/tex]

To determine the species with the highest , we need to compare the equilibrium constants. The larger the Ka value, the more the reaction favors the dissociation of the species.

Based on the given Ka values, the second ionization reaction (H2BO3- ⇌ [tex]H^{+}[/tex] + H_{3}BO_ 3}^{-2}) has the highest Ka value (1.8 x 10^-13). This indicates that the concentration of H_{3}BO_ 3}^{-} in the solution would be higher compared to the other species. Therefore, option B, H_{3}BO_ 3}^{-}, would have the highest concentration in a 0.25 M aqueous solution of boric acid.

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The amounts of isopropyl 99% alcohol and sterile water you need to use to make the 70% isopropyl solution using the alligation method are 707.1 mL of isopropyl 99% and 292.9 mL of water. The correct answer is option C.

Alligation method is a type of mathematical process used to calculate the quantities of two or more components of a mixture to obtain a mixture with a specific concentration or quality. This method uses the concept of averaging and the weighted average to get the desired results.

The allegation formula for isopropyl alcohol solution: Required % strength —— Higher % strength | DifferenceLower % strength | Difference70% ——- 99% | 29%70-29=41% | 29%

So, the required ratio of the alcohol to be mixed with water should be 41:29. Thus, the volumes of the 99% alcohol and water needed are (41/70) L and (29/70) L, respectively. Let's substitute the values and calculate:

41/70 × 1L = 0.586 L of 99% isopropyl alcohol

29/70 × 1L = 0.414 L of sterile water

Now we have the volumes of the 99% isopropyl alcohol and sterile water required. By adding these, the total volume of the solution is 1L, as given in the question. Therefore, the answer is 707.1 mL of 99% isopropyl alcohol and 292.9 mL of sterile water (707.1 mL + 292.9 mL = 1000 mL = 1 L).

Hence, option C) 707.1 mL of isopropyl 99% and 292.9 mL of water is the correct answer.

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A phase diagram is a graphical representation or map that shows the different phases (solid, liquid, and gas) of a substance as a function of pressure and temperature. It provides valuable information about the conditions under which a substance exists in each phase or undergoes phase transitions.

In a phase diagram, pressure is typically represented on the y-axis, and temperature is represented on the x-axis. The diagram is divided into regions corresponding to different phases, and boundaries or lines separating these regions represent phase transitions.

The phase diagram allows us to determine the conditions at which a substance can exist as a solid, liquid, or gas, and it provides insights into the effects of temperature and pressure on the phase behavior of the substance.

It helps in understanding processes such as melting, boiling, and sublimation and provides a useful tool in various fields including chemistry, physics, and materials science.

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Answer:

the limiting reactant is aluminum

Explanation:

Answer:

q = mC∆T

q = heat = ?

m = mass = 6.30 g

C = specific heat = 0.385 J/g/deg

∆T = change in temperature = 32-20 = 12º

q = ((6.30 g)(0.385 J/g/deg)(12 deg) = 29.1 J (to 3 significant figures)

Explanation:

The correct answer is O(a) describes a transition-state complementarity model, whereas (b) describes a strict "lock and key" model.

The two reaction coordinate diagrams given represent two different models that describe the catalysis of enzyme-catalyzed reactions. Both models have a single intermediate which is the ES complex. In each case, the contribution of binding energy is given, which is 5 in (a) and by 7 in (b). The ES complex is given by 2 in (a) and 4 in (b). The activation energy for the uncatalyzed reaction is given by 5+6 in (a) and by 7+4 in (b). The models presented in the diagrams are the transition-state complementarity model (a) and the strict "lock and key" model

(b). The two models are different from each other. They differ based on the mechanism by which the enzyme catalyzes the reaction. The lock-and-key model assumes that the active site of the enzyme is rigid and complementary to the substrate that fits into it. The substrate is said to "fit" into the active site of the enzyme as a lock fits into a key. In the transition-state complementarity model, the enzyme's active site has flexibility, which allows it to interact with the substrate in such a way that it stabilizes the transition state, reducing the energy required to reach it. The activation energy for the catalyzed reaction is 5 in (a) and is 7 in (b).

Hence, the correct answer is O(a) describes a transition-state complementarity model, whereas (b) describes a strict "lock and key" model.

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In the following reduction-oxidation reaction; ZnO(s) + C(s) → Zn(s) + CO(g), the species which functions as the oxidizing agent is option (B) C(s).

In the reaction ZnO(s) + C(s) → Zn(s) + CO(g), C(s) gains oxygen and goes from being a carbon atom to a carbon dioxide molecule which contains two oxygen atoms. Therefore, C(s) has been oxidized. Carbon is being oxidized in the above reaction because it has gained oxygen.

The term oxidation refers to the loss of electrons by an atom or molecule. Oxidation is a type of chemical reaction in which a substance loses electrons to another substance. The oxidation process usually occurs with a substance's interaction with oxygen.

Therefore, the correct answer is option B) C(s).

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Answer:

7. Ionization energy

8. Lattice energy

9. Free energy

10. Activation energy

Explanation:

I. Ionization energy: The energy required to convert a ground-state atom in the gas phase to a gaseous positive ion. Thus, it is the minimum energy required to remove or detach an electron from a neutral atom in a gaseous state. Generally, atoms with relatively large atomic radii tend to have a low ionization energy.

II. Lattice energy: The energy change that occurs in the conversion of an ionic solid to widely separated gaseous ions. This ultimately implies that, it is the energy generated during the conversion of an ionic solid into gaseous ions and as such is a measure of the cohesive force binding the ions of an ionic compound using the Born-Haber cycle.

III. Free energy: the energy in a chemical or physical change that is available to do useful work. Thus, it is a thermodynamic quantity that measures the maximum capacity of a system to do work at constant temperature and pressure.

IV. Activation energy: The energy required to form the transition state in a chemical reaction. When the activation energy of a reaction is low, the rate of the reaction would be faster. Therefore, an enzyme speeds or catalyzes the rate of a reaction by lowering its activation energy.

The balanced equation is: NaHCO₃ + HCl → NaCl + H2₃.

We can see that for every mole of NaHCO₃, one mole of HCl is required to convert it to H2CO₃. Therefore, the maximum amount of strong acid that can be added to the buffer is 0.35 mol HCl.

A buffer is a solution that resists pH change upon the addition of small amounts of acid or base. It is a solution of a weak acid and its corresponding conjugate base or a weak base and its corresponding conjugate acid. A buffer system consists of a weak acid or a weak base, and its conjugate base or conjugate acid.

When an acid or base is added to the buffer solution, the equilibrium is shifted either to the left or right, thereby minimizing the change in pH. Therefore, to determine the maximum amount of strong acid that can be added to a buffer made by the mixing of 0.35 mol sodium hydrogen carbonate with 0.50 mol sodium carbonate, we first need to identify the components of the buffer.

Sodium hydrogen carbonate (NaHCO₃) is a weak acid and sodium carbonate (Na2CO₃) is its conjugate base. The maximum amount of strong acid that can be added to the buffer is the amount that will convert all of the buffer components to their conjugate acid.

In this case, we need to convert all of the NaHCO₃ to H2CO₃.

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The rate of formation of SO₃ is 4.84 × 10⁻³ m/s. The correct option is d. 4.84 × 10⁻³ M/s.

To determine the rate of formation of SO₃, we need to consider the stoichiometry of the balanced equation. The coefficient of SO₃ is 2, which means that for every 1 mole of O₂ reacted, 2 moles of SO₃ are formed.

Given that the rate of O₂ loss is 2.42 × 10⁻³ m/s, we can use the stoichiometry to find the rate of formation of SO₃.

Rate of formation of SO₃ = (2 moles SO₃ / 1 mole O₂) × (2.42 × 10⁻³ m/s)

Rate of formation of SO₃ = 4.84 × 10⁻³ m/s

Therefore, the rate of formation of SO₃ is 4.84 × 10⁻³ m/s. The correct option is d. 4.84 × 10⁻³ M/s.

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Answer:

85.5%

Explanation:

To get the experimental value, you need to convert 15.0 grams of Na2SO4 to grams of Na3PO4. You do this with stoichiometry.

Convert grams of Na2SO4 to moles with molar mass. Then convert to moles of Na3PO4 with the mole-to-mole ratio according to the balanced chemical equation. Then convert moles of Na3PO4 to grams with the molar mass.

15.0 g Na2SO4 x (1 mol/142.04 g) x (2 Na3PO4/3Na2SO4) x (163.94 g/1 mol) = 11.7 g Na3PO4

Percent Yield = (actual value/experimental value) x 100

Actual Value = 10.0 g

Experimental Value = 11.7 g

10.0g/11.7 g = 85.5%

An expression for the solubility product constant (Ksp) of manganese(II) hydroxide is Ksp = [[tex]Mn^{2+}[/tex]][[tex]OH^-[/tex]]. b) Molar solubility (S) of Mn(OH)[tex]_2[/tex] in water is  2.3 x [tex]10^{-21}[/tex]. c) The Ksp value is 2.3 x  [tex]10^{-21}[/tex].

a) The expression for the solubility product constant (Ksp) of manganese (II) hydroxide, Mn(OH)[tex]_2[/tex] is given below:

Ksp = [[tex]Mn^{2+}[/tex]][[tex]OH^-[/tex][tex]]^{2}[/tex]

Mn(OH)[tex]_2[/tex] ⇌ [tex]Mn^{2+}[/tex] + 2[tex]OH^-[/tex]

The equation shows that the stoichiometry of the reaction is one mole of Mn(OH)[tex]_2[/tex] dissociating to give one mole of [tex]Mn^{2+}[/tex] and two moles of [tex]OH^-[/tex].

b) Given the concentration of hydroxide ([tex]OH^-[/tex]) in a saturated solution of Mn(OH)[tex]_2[/tex], which is 7.5 x [tex]10^{-8}[/tex] M.

The molar solubility (S) of Mn(OH)[tex]_2[/tex] in water is determined by using the stoichiometry of the dissociation reaction and the equilibrium expression.

Using the stoichiometry of the dissociation reaction and the equilibrium expression;[[tex]Mn^{2+}[/tex]] = S[[tex]OH^-[/tex]] = 2S

Therefore, Ksp = [[tex]Mn^{2+}[/tex]][[tex]OH^-[/tex][tex]]^{2}[/tex]

= S × [tex](2S)^2[/tex]

= 4[tex]S^3[/tex]

= 4 (7.5 x[tex]10^{-8})^3[/tex]

= 2.3 x [tex]10^{-21}[/tex]

c) The Ksp value is 2.3 x  [tex]10^{-21}[/tex] and the molar solubility (S) is 6.6 x [tex]10^{-8}[/tex] M which are calculated above.

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In the Lewis structure of HCN (hydrogen cyanide), we can determine the number of bonding electrons and lone pair electrons by considering the valence electrons of each atom involved.

Hydrogen (H) is in Group 1 of the periodic table and has one valence electron. Carbon (C) is in Group 14 and has four valence electrons, and nitrogen (N) is in Group 15 with five valence electrons. Cyanide (CN) is a polyatomic ion, and since we have one carbon and one nitrogen, we have a total of nine valence electrons (four from carbon and five from nitrogen). In the Lewis structure, we first connect the atoms using a single bond between carbon and nitrogen, and carbon forms a single bond with hydrogen. This accounts for two of the nine valence electrons. To complete the octets around each atom, we distribute the remaining seven valence electrons as lone pairs around the nitrogen and carbon atoms. The lone pairs do not participate in bonding and are considered nonbonding or lone pair electrons. Therefore, in the Lewis structure of HCN, there are two bonding electrons and seven lone pair electrons (nonbonding electrons). The bonding electrons are involved in forming the bonds between the atoms, while the lone pair electrons are localized on the nitrogen and carbon atoms, fulfilling their octets.

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