How are organisms classified as Eukarya different from organisms classified as Archaea and Bacteria?

A)They are all multicellular.
B)They all produce their own food.
C) They all reproduce sexually.
D) They all have membrane-bound nuclei

When, unknown acid, HA, having a percent dissociation of 2.16%. If the initial concentration of the acid is 0.084 M. Then, the pH of the solution is approximately 2.74. Option C is correct.

To determine the pH of the solution, we first need to calculate the concentration of H⁺ ions in the solution. The percent dissociation is given as 2.16%, which means that 2.16% of the initial concentration of the acid dissociates into H⁺ ions.

Percent dissociation = (concentration of dissociated acid / initial concentration of acid) × 100

2.16% = (concentration of H⁺ / 0.084 M) × 100

Let's solve for the concentration of H⁺;

2.16/100 = concentration of H⁺ / 0.084

0.0216 = concentration of H⁺ / 0.084

Concentration of H⁺ = 0.0216 × 0.084

Concentration of H⁺ = 0.0018144 M

Now that we have the concentration of H⁺ ions, we can calculate the pH using the formula;

pH = -log10(concentration of H⁺)

pH = -log10(0.0018144)

pH ≈ 2.74

Therefore, the pH of the solution is approximately 2.74.

Hence, C. is the correct option.

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The concentrations of the species in the 1.40 M NaCH₃COO solution are (Na⁺) = 1.40 M, (OH⁻) = 0 M, (H₂O⁺) = 0 M, (CH₃COOH⁻) = 0.00502 M, and (CH₃COOH) = 0.00502 M.

To calculate the concentrations of the species in a 1.40 M NaCH₃COO (sodium acetate) solution, we need to consider the dissociation of sodium acetate and the ionization of acetic acid.

The dissociation of sodium acetate (NaCH₃COO) can be represented as follows

NaCH₃COO → Na⁺ + CH₃COO⁻

Since sodium acetate fully dissociates in water, the concentration of Na⁺ will be the same as the initial concentration of sodium acetate: 1.40 M.

The ionization of acetic acid (CH₃COOH) can be represented as follows

CH₃COOH ⇌ H⁺ + CH₃COO⁻

The ionization constant for acetic acid (Ka) is given as 1.8×10⁻⁵

Let's denote the initial concentration of acetic acid as x M. At equilibrium, the concentration of H⁺ and CH₃COO⁻ will also be x M.

Using the ionization constant expression for acetic acid, we can write

Ka = [H⁺][CH₃COO⁻] / [CH₃COOH]

1.8×10⁻⁵ = x * x / (1.40 - x)

Since the value of x is expected to be small compared to 1.40, we can approximate the denominator as 1.40.

1.8×10⁻⁵ = x² / 1.40

Rearranging the equation

x² = 1.8×10⁻⁵ × 1.40

x² = 2.52 × 10⁻⁵

Taking the square root of both sides

x = √(2.52×10⁻⁵)

x ≈ 0.00502 M (rounded to four decimal places)

Now we can calculate the concentrations of the species

(Na⁺) = 1.40 M (same as the initial concentration of sodium acetate)

(OH⁻) = 0 M (since sodium acetate and acetic acid do not provide OH⁻ ions)

(H₂O⁺) = 0 M (water does not provide H⁺ ions)

(CH₃COOH⁻) = 0.00502 M

(CH₃COOH) = 0.00502 M

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Answer:

50000 dollars or 5 e5

Explanation:

Mr. Garibay has [tex]5.0 * 10^4[/tex] dollars in his bank.

Scientific notation represents the data in the format of expanded number or with e character.

For instance, 5.0 * 10^4 dollars = 50000 dollars or 5 e5

An electrochemical cell consists of two half-cells separated by a salt bridge or porous membrane that allows ions to flow freely between the two halves.

One half-cell contains an oxidizing agent, which is responsible for accepting electrons, while the other half-cell contains a reducing agent, which is responsible for donating electrons. In an electrochemical cell, the overall reaction must be balanced so that no charge accumulates in either half-cell. The equation that represents the net cell reaction for the given electrochemical cell is as follows.Co(s) + 2Ag⁺(aq) → Co⁺²(aq) + 2Ag(s)The given electrochemical cell consists of the following half-reactions:Anode: Co(s) → Co²⁺(aq) + 2e⁻Cathode: 2Ag⁺(aq) + 2e⁻ → 2Ag(s)The Co metal oxidizes to Co²⁺ ions at the anode, producing two electrons. The Ag⁺ ions are reduced to Ag metal at the cathode, receiving two electrons. These two half-reactions combine to yield the net cell equation.

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Answer: 35.078 to 3 significant figures

Explanation:

Solving;

34.56 + 42.235 * 0.0123 - 0.001

We solve the multiplication first according to the BODMAS rule.

So,

42.235 * 0.0123 = 0.51949

Doing the addition,

34.56 + 0.51949 - 0.001

This gives;

35.079 - 0.001

And finally the subtraction gives;

= 35.078

Answer:

Option 2 and 4 are correct

Explanation:

The reactants in the attached image have more enthalpy and hence less stability as they are more reactive. Thus, Product is more stable than the reactants.

This is an addition reaction in which two reactants add up to form the product.

Very less activation energy is required as the reactants themselves are unstable, possess high energy and hence are very reactive.

Reactants have more energy than the products.  

Answer:

Gravitational potential energy depends on an object's weight and its height above the ground (GPE = weight x height).Explanation:

Answer:

The gas will fill the entire 25 mL of the container. Hope this helps

Explanation:

Answer:

1. 7 (a neutral solution)

Answer: 10-7= 0.0000001 moles per liter

2. 5.6 (unpolluted rainwater)

Answer: 10-5.6 = 0.0000025 moles per liter

3. 3.7 (first acid rain sample in North America)

Answer: 10-3.7 = 0.00020 moles per liter

The concentration of H+ in the Hubbard Brook sample is 0.00020/0.0000025, which is 80 times higher than the H+ concentration in unpolluted rainwater.

Explanation:

The reaction between 1,3-cyclohexadiene and maleic acid (cis-2-butene-dioic acid) leads to the formation of a Diels-Alder adduct.

The product formed is a bicyclic compound resulting from the cycloaddition reaction between 1,3-cyclohexadiene and maleic acid. The reaction involves the formation of a new six-membered ring fused with the cyclohexene ring of 1,3-cyclohexadiene.

The reaction proceeds via the concerted [4 + 2] cycloaddition mechanism, where the diene (1,3-cyclohexadiene) reacts with the dienophile (maleic acid) to form a new sigma bond and two new pi bonds. The resulting adduct exhibits a fused ring system with a double bond in the newly formed six-membered ring.

The specific arrangement of substituents and stereochemistry in the product can vary depending on the orientation of the reactants and the conditions of the reaction. It is important to consider the regioselectivity and stereochemistry when drawing the complete structure of the product.

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Answer:

Meaning:from the valley town; the settlement in the valley. Sounding like the proper English surname it was derived from, Dalton is all about being smart and trendy with equal parts cool and nerdy. Dalton, which comes from “dael” or “valley,” and “tun,” or “settlement,” may have peaked in the 1990s, but that's okay.

Explanation:

a unit used in expressing the molecular weight of proteins, equivalent to atomic mass unit.

Answer:

A unit used in expressing the molecular weight of proteins, equivalent to atomic mass unit.

Or in simpler terms a unit of measure.

Explanation:

Hope this helps and have a great say!!!!

The molecular formula of the compound, given that the compound made up of N and O contains 30.4% N is N₂O₄

First, we shall obtain the molar mass of the compound. Details below:

The mole of the gas is obtained as follow:

PV = nRT

1 × 0.974 = n × 0.0821 × 273

Divide both sides by 24.0553

n = 0.974 / (0.0821 × 273)

n = 0.043 mole

Thus, the molar mass is obtained as:

Molar mass = mass / mole

Molar mass = 4 / 0.043

Molar mass = 93 g/mol

Next, we shall obtain the empirical formula of the compound. details below:

Divide by their molar mass

N = 30.4 / 14 = 2.171

O = 69.6 / 16 = 4.35

Divide by the smallest

N = 2.171 / 2.171 = 1

O = 4.35 / 2.171 = 2

Thus, the empirical formula is NO₂

Finally, we shall obtain the molecular formula of the compound. This is shown below:

Molecular formula = empirical × n = mass number

[NO₂]n = 150

[14 + (2 × 16)]n = 150

46n = 93

Divide both sides by 46

n = 93 / 46

n = 2

Molecular formula = [NO₂]n

Molecular formula = [NO₂]₂

Molecular formula = N₂O₄

Thus, the molecular formula of the compound is N₂O₄

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The more soluble compounds in water are: (a) Nickel(II) hydroxide over magnesium hydroxide, b) Copper(II) sulfide over lead(II) sulfide, (c) Magnesium fluoride over silver sulfate.

Solubility is the ability of a substance to dissolve in a solvent, such as water. It depends on several factors, including the nature of the solute and the solvent, as well as their respective chemical properties.

(a) Magnesium hydroxide (Mg(OH)2) and nickel(II) hydroxide (Ni(OH)2) are both metal hydroxides. However, nickel(II) hydroxide is more soluble in water than magnesium hydroxide. This is because nickel(II) hydroxide forms a more stable complex with water molecules, resulting in better solvation and higher solubility.

(b) Lead(II) sulfide (PbS) and copper(II) sulfide (CuS) are both metal sulfides. Copper(II) sulfide is more soluble in water than lead(II) sulfide. Copper(II) sulfide has a smaller lattice energy and forms a more stable complex with water, leading to higher solubility compared to lead(II) sulfide.

(c) Silver sulfate (Ag2SO4) and magnesium fluoride (MgF2) are both ionic compounds. However, magnesium fluoride is more soluble in water than silver sulfate. This is due to the higher lattice energy of silver sulfate and the stronger ion-dipole interactions between magnesium fluoride and water molecules, resulting in greater solubility for magnesium fluoride.

The more soluble compounds in water are:

(a) Nickel(II) hydroxide over magnesium hydroxide

(b) Copper(II) sulfide over lead(II) sulfide

(c) Magnesium fluoride over silver sulfate.

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Answer: Ik its too late but the answer is yeast.

Explanation: Sugar is yeasts food. If you mix sugar with yeast, and put it in something such as bread, a bottle with a balloon over it, or etc. It will cause it to rise.

Answer:

Scientists study Earth's interior by studying

seismic waves. Seismic waves are waves of energy that travel

through Earth.

Explanation:

Answer:

D [polymers]

Explanation:

The joining of monomers (small molecules) is polymerization.

Among the salts listed, ZnCO3 (zinc carbonate) and Ba3(PO4)2 (barium phosphate) are the salts that will be substantially more soluble in acidic solution than in pure water.

In the case of ZnCO3, it is an insoluble salt in pure water due to the carbonate ion's basic nature. Carbonate ions (CO3^2-) react with water molecules to form bicarbonate ions (HCO3-) and hydroxide ions (OH-) in an equilibrium reaction. The presence of an acidic solution would shift this equilibrium toward the reactant side, favoring the formation of CO3^2- ions and increasing the solubility of ZnCO3. Similarly, Ba3(PO4)2, which is barium phosphate, is insoluble in pure water. Phosphate ions (PO4^3-) have a basic nature and tend to form insoluble salts with many cations. In an acidic solution, the excess of hydrogen ions (H+) would react with phosphate ions, forming dihydrogen phosphate ions (H2PO4-) or monohydrogen phosphate ions (HPO4^2-). This reaction reduces the concentration of phosphate ions, decreasing the formation of insoluble Ba3(PO4)2 and enhancing its solubility. On the other hand, ZnS (zinc sulfide), BiI3 (bismuth triiodide), and AgCN (silver cyanide) do not show a significant change in solubility in an acidic solution compared to pure water. Their solubilities are primarily governed by factors such as the lattice energy and the ion-ion interactions within the crystal lattice, which are less influenced by changes in pH.

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The hydroxide ion concentration in the given aqueous solution is 2.86 × 10-12 M.

The given aqueous solution has a hydronium ion concentration of 3.50 × 10-3 M. To calculate the hydroxide ion concentration, the following steps need to be followed:

Step 1: Write the balanced chemical equation for the dissociation of water:

H2O(l) ⇌ H+(aq) + OH-(aq)

Step 2: Write the expression for the equilibrium constant for this reaction:

Kw = [H+(aq)][OH-(aq)]

Step 3: Substitute the value of

Kw (1.0 × 10-14 M2 at 25°C) and the given hydronium ion concentration (3.50 × 10-3 M) in the expression to solve for hydroxide ion concentration:

[OH-(aq)] = Kw/[H+(aq)] = (1.0 × 10-14 M2) ÷ (3.50 × 10-3 M) = 2.86 × 10-12 M

Therefore, the hydroxide ion concentration in the given aqueous solution is 2.86 × 10-12 M.

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the alkyl halide (CH3)2CHCH2CH2C! is likely the least reactive in an E2 reaction among the provided options.

The reactivity of alkyl halides in an E2 (elimination bimolecular) reaction is influenced by the stability of the carbocation intermediate formed during the reaction. In general, more substituted alkyl halides form more stable carbocations, resulting in increased reactivity in E2 reactions.

Let's analyze the given options:

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In the presence of 0.10 M NH₄NO₃ or 0.10 M NaCH₃COO, the solubility of barium phosphate is expected to be similar to its solubility in pure water. In the presence of 0.10 M Na₃PO₄ or 0.10 M Ba(CH₃COO)₂, the solubility of barium phosphate is likely to be reduced compared to its solubility in pure water.

To determine the effect of each aqueous solution on the solubility of barium phosphate, we need to consider the common ion effect and the solubility product constant (KSP) of barium phosphate (Ba₃(PO₄)₂).

0.10 M NH₄NO₃

NH₄NO₃ does not contain any common ions with barium phosphate. Therefore, the solubility of barium phosphate is likely to be similar to its solubility in pure water. The correct answer is 2) similar solubility as in pure water.

0.10 M Na₃PO₄

Na₃PO₄ introduces PO₄⁻ ions, which are common ions with barium phosphate. According to the common ion effect, the presence of a common ion reduces the solubility of a compound. Therefore, the solubility of barium phosphate is expected to be reduced in the presence of Na₃PO₄. The correct answer is 3) less soluble than in pure water.

0.10 M Ba(CH₃COO)₂

Ba(CH₃COO)₂ introduces Ba₂⁺ ions, which are common ions with barium phosphate. Again, according to the common ion effect, the solubility of barium phosphate is likely to be reduced in the presence of Ba(CH₃COO)₂. The correct answer is 3) less soluble than in pure water.

0.10 M NaCH₃COO

NaCH₃COO does not contain any common ions with barium phosphate. Therefore, the solubility of barium phosphate is likely to be similar to its solubility in pure water. The correct answer is 2) similar solubility as in pure water.

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☁️ Answer ☁️

Yes.

"In so doing, he became the first scientist to explain the behavior of atoms in terms of the measurement of weight. He also uncovered the fact that atoms couldn't be created or destroyed. Dalton's theory additionally examined the compositions of compounds, explaining that the tiny particles (atoms) in a compound were compound atoms"

Hope it helps.

Have a nice day noona/hyung.

The correct expression relating the rates of consumption of reactants and the rates of formation of products for the given reaction is A[[tex]LaCl_{3}[/tex]] / 2Δt = A[La_{2}(CO_3){3}] / 3Δt = A[[tex]Na_{2}CO_{3}[/tex]] / Δt.

The balanced chemical equation for the reaction is 2LaCl_{3}+ 3[tex]Na_{2}CO_{3}[/tex]→ La_{2}(CO_3){3} + 6NaCl. To determine the expressions relating the rates of consumption of reactants and the rates of formation of products, we can use the stoichiometric coefficients from the balanced equation. According to the stoichiometry of the reaction, for every 2 moles of [tex]LaCl_{3}[/tex]consumed, 1 mole of La_{2}(CO_3){3}is formed. Therefore, the expression relating the rates of consumption of LaCl3 and formation of La_{2}(CO_3){3}is A[LaCl_{3}] / 2Δt = A[[tex]La_{2}(CO_3){3}[/tex]] / Δt. Similarly, for every 3 moles of [tex]Na_{2}CO_{3}[/tex]consumed, 1 mole of La_{2}(CO_3){3} is formed. Therefore, the expression relating the rates of consumption of [tex]Na_{2}CO_{3}[/tex] and formation ofLa_{2}(CO_3){3}is A[] / 3Δt = A[La_{2}(CO_3){3}] / Δt.

From the balanced equation, we can also see that 6 moles of NaCl are formed for every 2 moles of LaCl3 consumed. However, the question does not provide options involving the rate of consumption of NaCl. In conclusion, the correct expression relating the rates of consumption of reactants and the rates of formation of products for the given reaction is A[LaCl_{3}] / 2Δt = A[La_{2}(CO_3){3}] / 3Δt = A[[tex]Na_{2}CO_{3}[/tex]] / Δt.

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The correct answer is option A) 28%. The volume percentage of ethanol = (volume of ethanol / total volume of solution) x 100Volume percentage of ethanol = (38 mL / 138 mL) x 100Volume percentage of ethanol = 0.2754 x 100Volume percentage of ethanol = 27.54 % ≈ 27.6 %

The volume percentage of ethanol in the solution is 27.6%. The volume percentage is the ratio of the volume of solute to the volume of the solution multiplied by 100. It is denoted by (v/v)% and it is used in chemistry to measure the volume of a solute dissolved in a solution. To find the volume percentage of ethanol in the solution, we first need to calculate the total volume of the solution. The total volume of the solution = volume of ethanol + volume of toluene= 38 mL + 100 mL= 138 mL.

Now, we can calculate the volume percentage of ethanol in the solution. The volume percentage of ethanol = (volume of ethanol / total volume of solution) x 100. The volume percentage of ethanol = (38 mL / 138 mL) x 100. The volume percentage of ethanol = 0.2754 x 100Volume percentage of ethanol = 27.54 % ≈ 27.6 %Therefore, the correct answer is option A) 28%.

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Answer:

As we know that

At STP

1 mole of any gas occupies 22.4L  

so

15.7L / 22.4L/mole = 0.7moles Cl2  

0.7moles x 6.022x10^23molecules/mole = 4.22x10^23molecules

hope it helps

Answer:

"Symbolic representation of a chemical reaction in the form of symbols and formulae"

Answer:

A chemical equation is the symbolic representation of a chemical reaction in the form of symbols and formulas.

Explanation:

Example- Zinc metal reacts with dilute sulphuric acid to form zinc sulphate and hydrogen gas. This equation is written as: Zn+H2SO4→ZnSO4+H2.

To calculate the required flow rate of the entering steam and the rate of heat transfer from water to methanol, we can use the principles of energy balance and heat transfer.

Given data:

Temperature of saturated steam entering the heat exchanger (T1): 273.3°C

Temperature of methanol vapor entering the heat exchanger (T2): 70.0°C

Temperature of methanol vapor leaving the heat exchanger (T3): 252.9°C

Temperature of water leaving the heat exchanger (T4): 90.0°C

Flow rate of methanol vapor (Q2): 6530 standard liters per minute

Step 1: Calculate the required flow rate of entering steam (Q1) in m³/min.

We need to determine the mass flow rate of methanol vapor (m2) and the mass flow rate of water (m4) to perform the energy balance.

To convert the flow rate of methanol vapor from standard liters per minute to m³/min, we use the ideal gas law:

PV = nRT

where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.

Using the ideal gas law, we can calculate the volume of methanol vapor (V2) in m³/min:

V2 = (Q2 * R * T2) / (P * 1000)

Assuming the pressure is constant, we can write the equation for the methanol vapor mass flow rate (m2) as:

m2 = V2 * ρ2

where ρ2 is the density of methanol vapor at temperature T2.

Step 2: Calculate the mass flow rate of water (m4) in kg/min.

We can determine the mass flow rate of water based on its specific heat capacity (Cp) and the temperature change (ΔT = T3 - T4).

Using the equation:

Q = m4 * Cp * ΔT

where Q is the heat transferred from water to methanol, we can solve for m4:

m4 = Q / (Cp * ΔT)

Step 3: Calculate the required flow rate of entering steam (Q1) in m³/min.

To perform the energy balance, we assume that the heat transferred from the steam to the methanol is equal to the heat transferred from the water to the methanol:

m1 * H1 = m2 * H2 + m4 * H4

where m1 is the mass flow rate of entering steam, H1 is the enthalpy of saturated steam at temperature T1, H2 is the enthalpy of methanol vapor at temperature T2, and H4 is the enthalpy of water at temperature T4.

Rearranging the equation to solve for m1:

m1 = (m2 * H2 + m4 * H4) / H1

Step 4: Calculate the rate of heat transfer from water to methanol (Q) in kW.

We can now substitute the values of m2, m4, and H2, H4 into the energy balance equation to calculate Q:

Q = m2 * (H2 - H3)

where H3 is the enthalpy of methanol vapor at temperature T3.

Finally, we can convert the flow rates from kg/min to m³/min by dividing by the density of water and methanol vapor, respectively.

Conclusion:

The required flow rate of entering steam (Q1) can be calculated using the mass flow rates and enthalpies, and the rate of heat transfer from water to methanol (Q) can be determined using the enthalpies.

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Answer:

Thank youuuuu bestie, you really pulling through

Answer:

yessss

Explanation: