Rank the ions in each set in order of decreasing size, and explain your ranking:
(a) Se²⁻, S²⁻, O²⁻

Answer:

The correct answer is coats the pathogen exterior so that it is recognized by the host's phagocytes. :)

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Answer:

File down there

Explanation:

Answer:

2.232 g/L

Explanation:

Assuming 1 mol, volume at STP is 22.4 L so you simply divide 50g by 22.4 L to get density

The density of the given gas is required.

The density of the gas at STP is 2.232 g/L.

M = Molar mass of gas = 50 g/mol

1 mole of any gas at STP occupies 22.4 L.

V = Volume per mole = 22.4 L/mol

Density is given by

\(\rho=\dfrac{M}{V}\\\Rightarrow \rho=\dfrac{50}{22.4}\\\Rightarrow \rho=2.232\ \text{g/L}\)

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Answer:

0.30 ug of U-235

Explanation:

N = N0 * (1/2)^(1/2)

0.15 ug = N0 * (1/2)^(1/2)

N0 = 0.15 ug / (1/2)^(1/2)

N0 = 0.30 ug

Answer:

B

Explanation:

The decay of U-235 follows an exponential decay model, which can be described by the equation:

N = N0 * (1/2)^(t/T)

where N is the amount of U-235 at a given time t, N0 is the initial amount of U-235, t is the elapsed time, and T is the half-life of U-235.

In this case, we know that the sample has undergone two half-lives, which means that the elapsed time is:

t = 2 * T

We also know that the amount of U-235 in the sample is 0.15 ug. We can use this information to solve for the initial amount of U-235 (N0):

N = N0 * (1/2)^(t/T)

0.15 = N0 * (1/2)^(2)

0.15 = N0 * (1/4)

N0 = 0.15 / (1/4)

N0 = 0.60 ug

Therefore, the amount of U-235 that was originally present in the sample before it decayed was 0.60 ug. The answer is B) 0.60 ug.

Even more:

It's not A) 0.30 ug because if the sample originally contained 0.30 ug of U-235 and underwent two half-lives, the amount of U-235 remaining would be:N = N0 * (1/2)^(t/T)

N = 0.30 * (1/2)^(2)

N = 0.30 * (1/4)

N = 0.075 ug

This means that the amount of U-235 in the sample after two half-lives would be 0.075 ug, which is not consistent with the given information that the sample contains 0.15 ug of U-235.

Therefore, the correct answer is B) 0.60 ug, which is the initial amount of U-235 that would have been present in the sample before it decayed.

Answer:

Q1. B

Q2. D

Q3. C

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The resulting NaCl concentration when 274 mL of a 0.740 M NaCl solution is mixed with 597 mL of a 0.390 M NaCl solution can be calculated using the principles of dilution and the volume ratios of the two solutions.

To find the resulting concentration, we can use the equation for dilution:

C1V1 = C2V2

Where C1 and V1 are the initial concentration and volume of the first solution, and C2 and V2 are the final concentration and volume of the diluted solution.

In this case, we have:

C1 = 0.740 M

V1 = 274 mL

C2 = ?

V2 = 274 mL + 597 mL = 871 mL

Substituting the values into the dilution equation:

(0.740 M)(274 mL) = C2(871 mL)

Solving for C2:

C2 = (0.740 M)(274 mL) / (871 mL)

C2 = 0.232 M

Therefore, the resulting NaCl concentration when the two solutions are mixed is 0.232 M.

In summary, by applying the principle of dilution and using the volume ratios of the two solutions, we can determine that the resulting NaCl concentration after mixing 274 mL of a 0.740 M NaCl solution with 597 mL of a 0.390 M NaCl solution is 0.232 M.

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The process of a substance moving from an area of low concentration to an area of high concentration while using energy is termed as Active transport.

Active transport is the movement of substances against the concentration gradient (from an area of low concentration to an area of high concentration) across a cell membrane using energy from ATP (Adenosine Triphosphate). Active transport requires a carrier protein in the membrane to assist the molecule's passage, and energy is needed for the conformational change of the carrier protein, so it's a coupled process.

Active transport is used by cells to bring in nutrients such as glucose, amino acids, and ions like sodium, calcium, and potassium. The Sodium-Potassium pump is an excellent example of active transport. It works by pumping sodium ions out of the cell and potassium ions into the cell. This process is essential in maintaining the concentration gradient necessary for nerve impulse transmission, muscle contraction, and other essential physiological processes.

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Δ'∘ for the reaction is approximately -1.1 kJ/mol.

Δ for the reaction is approximately -1.69 kJ/mol.

The initial concentrations of reactant and product being 1 M is not consistent with the conditions at which Δ'∘ is measured, as the concentrations used for the calculation are different (1.2 M and 0.65 M).

To determine the values of Δ'∘ and Δ for the given interconversion reaction, we need to use the equation:

Δ'∘ = -RT ln(K'eq)

Where:

Δ'∘ is the standard Gibbs free energy change for the reaction at 25 °C,

R is the gas constant (8.314 J/(mol·K)),

T is the temperature in Kelvin,

ln represents the natural logarithm,

K'eq is the equilibrium constant at the given temperature.

Given that K'eq is 1.97, we can calculate Δ'∘ as follows:

Δ'∘ = -(8.314 J/(mol·K))(298 K) ln(1.97)

Δ'∘ ≈ -1.1 kJ/mol

Therefore, Δ'∘ for the reaction is approximately -1.1 kJ/mol.

To calculate Δ, the Gibbs free energy change under the given conditions with adjusted concentrations, we can use the equation:

Δ = Δ'∘ + RT ln(Q)

Where:

Δ is the Gibbs free energy change,

Δ'∘ is the standard Gibbs free energy change (previously calculated),

R is the gas constant (8.314 J/(mol·K)),

T is the temperature in Kelvin,

ln represents the natural logarithm,

Q is the reaction quotient.

Given the concentrations of fructose 6-phosphate (1.2 M) and glucose 6-phosphate (0.65 M), the reaction quotient (Q) is:

Q = ([glucose 6-phosphate]^n)/([fructose 6-phosphate]^m)

Q = (0.65 M)^1 / (1.2 M)^1

Q ≈ 0.54

Now, we can calculate Δ using the equation:

Δ = -1.1 kJ/mol + (8.314 J/(mol·K))(298 K) ln(0.54)

Δ ≈ -1.1 kJ/mol - 0.59 kJ/mol

Δ ≈ -1.69 kJ/mol

Therefore, Δ for the reaction is approximately -1.69 kJ/mol.

Among the given statements, the ones that are consistent with the conditions at which Δ'∘ is measured are:

The temperature is 273 K.

The pH is 7.

The pressure is 101.3 kPa (1 atm).

The initial concentrations of reactant and product being 1 M is not consistent with the conditions at which Δ'∘ is measured, as the concentrations used for the calculation are different (1.2 M and 0.65 M).

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Answer:

5.02 mol C

Explanation:

Step 1: Define

Avagadro's Number - 6.02 × 10²³ atoms, molecules, formula units, etc.

Step 2: Use Dimensional Analysis

\(3.02(10)^{24} \hspace{3} molecules \hspace{3} C(\frac{1 \hspace{3} mol \hspace{3} C}{6.02(10)^{23} \hspace{3} molecules \hspace{3} C} )\) = 5.01661 mol C

Step 3: Simplify

We have 3 sig figs.

5.01661 mol C ≈ 5.02 mol C

The given ionization constant for water (Kw) is 2.9 × 10−14 at 40 °C.  The [H3O+], [OH-], pH, and pOH values of pure water at 40 °C are [H3O+] = [OH-] = 1.7 x 10^-7, pH = 6.77, and pOH = 7.23

We are to calculate [H3O+], [OH−], pH, and pOH for pure water at 40 °C.

The dissociation of water into hydroxyl ions and hydronium ions can be represented as follows: H2O(l) ⇌ H+ (aq) + OH− (aq)The equilibrium constant, Kc, for this reaction is given by the expression :

Kc = [H+][OH−]/[H2O]

Since water is a pure liquid, its concentration is considered to be constant and its value is

1.00 x 10^-7 at 40°C.

Kc = [H+][OH−]/1.00 x 10^-7Kw

= [H+][OH−]Kw

= 2.9 x 10^-14[H+][OH−]

= 2.9 x 10^-14[H3O+] = [OH-]

= √(2.9 x 10^-14) = 1.7 x 10^-7pH

= -log[H3O+] = -log(1.7 x 10^-7)

= 6.77pOH = -log[OH-]

= -log(1.7 x 10^-7)

= 6.77pH + pOH

= 14.00 (constant at 25°C)

Therefore, pH + pOH at 40°C is given by

pH + pOH = 14.00p

OH = 14.00 - pHpOH

= 14.00 - 6.77

pOH = 7.23

Hence, the [H3O+], [OH-], pH, and pOH values of pure water at 40 °C are [H3O+] = [OH-] = 1.7 x 10^-7, pH = 6.77, and pOH = 7.23.

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Answer:

Ca2+ I'm not sure hahahahhahah

Answer:

Ca

hope it is helpful..

Answer:

The answer is a constellation

Answer:

i don't know but you can get this app called Socratic it scan questions  and tell you the answer to it

Explanation:

Answer:

to increase their number of species on the Earth and want to go long and long

Answer:

the second one

Explanation:

im positive bc ur gaining 2 ions

According to calculations, the cell's potential is \(0.118 V\).

What is two half cells of electrodes ?

A half-cell is a structure used in electrochemistry that has a conducting electrode and conductive electrolyte around it that are separated by a naturally occurring Helmholtz double layer. One of the two electrodes in a galvanic cell or basic battery is known as a half cell. For instance, the two half cells in the Zn–Cu batteries form an oxidizing–reducing pair. A half cell is created by putting a portion of reactant in an electrolyte solution.

Two electrodes make up a basic battery or galvanic cell. Half cells refer to each of a galvanic cell's electrodes. The two half cells in a battery create an oxidizing-reducing pair.

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Answer:

System

Explanation:

I got 100 on Edge

A system is a group of related objects which interact with each other and form a complex whole.

A system can be described as a group of interacting elements that act according to a set of rules to create a unified whole. A system is surrounded by its environment and is described by its boundaries, structure, and purpose.

A system in physics can be described as a collection of objects that can be identified. A system refers to a collection that makes thinking about a problem more convenient.

The surrounding is everything else that is not included in the system. An isolated system is a system in which no energy or matter is exchanged with the surroundings.

A closed system is one in which only energy can be exchanged with the surroundings. The open system is one in which both matter and energy can be exchanged with the surroundings.

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The method cbo.additem(s) adds an item s into a combobox cbo. Option C

The method that adds an item 's' into a ComboBox 'cbo' depends on the programming language or framework being used. However, based on common naming conventions and methods used in various programming languages, the most likely correct option is (c) cbo.addItem(s).

In many programming languages and frameworks, the method to add an item to a ComboBox is typically named 'addItem' or 'add' followed by the item's name or value. Let's analyze the given options to determine the most appropriate choice:

(a) cbo.addChoice(s):

This option uses the term 'addChoice,' which is not commonly used for adding items to ComboBoxes. It is less likely to be the correct method name.

(b) cbo.addObject(s):

Similar to option (a), 'addObject' is not a common method name for adding items to ComboBoxes. It is often used for adding objects to other data structures but not ComboBoxes specifically.

(c) cbo.addItem(s):

This option is the most commonly used method name for adding items to a ComboBox. It follows standard naming conventions and accurately describes the action of adding an item to the ComboBox.

(d) cbo.add(s):

This option is less specific and might be used in some cases, but 'addItem' is a more appropriate and descriptive method name for ComboBoxes.

(e) cbo.getItems():

This option retrieves the items from the ComboBox rather than adding an item. It is used to get the existing items in the ComboBox and not to add new ones.

In summary, based on standard naming conventions and commonly used methods in programming languages, the most appropriate method for adding an item 's' to a ComboBox 'cbo' is (c) cbo.addItem(s).

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Answer:

Explanation:

The density of a substance can be found by using the formula

\(density = \frac{mass}{volume} \\\)

From the question

mass of metal = 33.7 g

volume = 8.49 mL

So we have

\(density = \frac{33.7}{8.49} \\ = 3.969375736...\)

We have the final answer as

Hope this helps you

Answer: The number of moles of gas which are present in the sample are 1.49 mol.

Explanation:

Given: Volume = 55.3 L

Temperature = \(23.3^{o}C\) = (23.3 + 273) K = 296.3 K

Pressure = 0.658 atm

Formula used is as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

\(PV = nRT\\0.658 atm \times 55.3 L = n \times 0.0821 L atm/mol K \times 296.3 K\\n = \frac{0.658 atm \times 55.3 L}{0.0821 L atm/mol K \times 296.3 K}\\= \frac{36.3874}{24.32623}\\= 1.49 mol\)

Thus, we can conclude that the number of moles of gas which are present in the sample are 1.49 mol.

Answer:

A. 7

(assuming the water is neutral)

Answer:

sodium has a lower electronegativity:)

Explanation:

Answer:

Explanation:

Here, we want to check if the equation of the reaction is balanced or not

Now, for a balanced equation of reaction, the number of atoms of the individual elements on both sides of the equation is equal

Now, let us check the atoms:

We have 1 Sr on both sides

We have 1 SO4 on both sides

We have 1 Mg on both sides

We have 2 OH on both sides

From what we can see, the equation of the reaction is balanced

Thus, the correct answer choice is B

Answer:B

Explanation:

Wavelength (L) is calculated using: L = gT²/2π, here g=9.8 m/s2 and T is wave period in seconds.

Wavelength of a wave describes how long the wave is and the distance from the "crest" (top) of one wave to the crest of next wave is called wavelength. We can also measure from the "trough" (bottom) of one wave to trough of next wave and get the same value for the wavelength.

We measure wavelength in following ways:

Use photometer to measure the energy of wave.

Convert energy into joules (J).

Divide energy by Planck's constant, 6.626 x 10⁻³⁴, to get the frequency of  wave.

Divide speed of light, ~300,000,000 m/s, by frequency to get wavelength.

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Answer:

False

Explanation:

Answer:

False

Explanation:

The number of electrons determines the chemical properties.  

The balanced redox reaction in acidic solution:

6H₂O₂(aq) + 2Cr₂O₇⁻²(aq) + 14H⁺(aq) → 3O₂ + 4Cr³⁺(aq) + 20H₂O(l)

To balance the redox reaction in acidic solution:

H₂O₂(aq) + Cr₂O₇⁻²(aq) → O₂(g) + Cr³⁺(aq)

We will follow the steps for balancing redox reactions in acidic solution:

Step 1: Assign oxidation numbers to all elements in the equation:

H₂O₂(aq): H has an oxidation state of +1, O has an oxidation state of -1.

Cr₂O₇⁻²(aq): Cr has an oxidation state of +6, O has an oxidation state of -2.

O₂(g): O has an oxidation state of 0.

Cr³⁺(aq): Cr has an oxidation state of +3.

Step 2: Identify the elements that are being oxidized and reduced:

Oxidation: Cr is being reduced from +6 to +3.

Reduction: H₂O₂ is being oxidized from -1 to 0.

Step 3: Write the half-reactions for oxidation and reduction:

Oxidation half-reaction: H₂O₂(aq) → O2(g)

Reduction half-reaction: Cr₂O₇⁻²(aq) → Cr³⁺(aq)

Step 4: Balance the atoms other than H and O in each half-reaction:

Oxidation half-reaction: 2H₂O₂(aq) → O₂(g)

Reduction half-reaction: Cr₂O₇⁻²(aq) → 2Cr³⁺(aq)

Step 5: Balance the oxygen atoms by adding water molecules (H2O) to the side that lacks oxygen:

Oxidation half-reaction: 2H₂O₂(aq) → O₂(g) + 2H₂O(l)

Reduction half-reaction: Cr₂O₇⁻²(aq) → 2Cr³⁺(aq)

Step 6: Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that lacks hydrogen:

Oxidation half-reaction: 2H₂O₂(aq) → O₂(g) + 2H₂O(l)

Reduction half-reaction: Cr₂O₇⁻²(aq) + 14H+(aq) → 2Cr³⁺(aq) + 7H₂O(l)

Step 7: Balance the charges by adding electrons (e-) to the appropriate side of each half-reaction:

Oxidation half-reaction: 2H₂O₂(aq) → O₂(g) + 2H₂O(l) + 4e-

Reduction half-reaction: Cr₂O₇⁻²(aq) + 14H+(aq) + 6e- → 2Cr³⁺(aq) + 7H₂O(l)

Step 8: Multiply each half-reaction by a factor that will equalize the number of electrons in both half-reactions:

Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:

6H₂O₂(aq) → 3O₂(g) + 6H₂O(l) + 12e-

Cr₂O₇⁻²(aq) + 14H+(aq) + 12e- → 4Cr³⁺(aq) + 14HvO(l)

Step 9: Combine the two half-reactions, canceling out the electrons on both sides:

6H₂O₂(aq) + 2Cr₂O₇⁻²(aq) + 14H⁺(aq) → 3O₂ + 4Cr³⁺(aq) + 20H

Step 10: Combine all the species to form the balanced redox reaction:

6H₂O₂(aq) + 2Cr₂O₇⁻²(aq) + 14H⁺(aq) → 3O₂ + 4Cr³⁺(aq) + 20H₂O(l)

Step 11: Simplify the equation by canceling out common species:

6H₂O₂(aq) + 2Cr₂O₇⁻²(aq) + 14H⁺(aq) → 3O₂ + 4Cr³⁺(aq) + 20H₂O(l)

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2 C4H10 + 13 O2 → 8 CO2 + 10 H2O.

Answer: 91.96 grams

Explanation:

Answer:

a tendency to do nothing or to remain unchanged.

Explanation: