n the icl2 - ion, how many valence electrons are there surrounding the central iodine atom? a. 4 b. 6 c. 8 d. 10 e. 12

Therefore, the answer is A) 1.6 x 10^18 electrons per second pass a given cross section of the wire.

To solve this problem, we need to use the equation that relates current, cross-sectional area, and electron flow:
I = nAvq
where I is the current, n is the number of electrons per unit volume, A is the cross-sectional area, v is the drift velocity of the electrons, and q is the charge of each electron.
First, we need to find the cross-sectional area of the gold wire:
diameter = 1.8 mm
radius = 0.9 mm
area = πr^2 = 3.14 x (0.9 mm)^2 = 2.54 mm^2 = 2.54 x 10^-6 m^2
Next, we need to convert the current from milliamperes to amperes:
260 mA = 0.26 A
Now we can rearrange the equation to solve for n, the number of electrons per unit volume:
n = I / Avq
Plugging in the values we have:
n = 0.26 / (2.54 x 10^-6 x 1.60 x 10^-19 x v)
We don't know the drift velocity, but we can assume that it is on the order of 10^-4 m/s for metallic conductors. Using this value, we get:
n = 0.26 / (2.54 x 10^-6 x 1.60 x 10^-19 x 10^-4) = 1.6 x 10^18

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The molecular formula of the gas is C2H4 (Option B). To determine the molecular formula of the gas, we need to compare the empirical formula (CH2) with the molar mass of the gas.

First, we calculate the molar mass of the empirical formula by adding the atomic masses of carbon (C) and hydrogen (H). The atomic mass of carbon is 12.01 g/mol, and hydrogen is 1.01 g/mol. Thus, the molar mass of the empirical formula (CH2) is 12.01 + 2(1.01) = 14.03 g/mol. Next, we calculate the number of empirical formula units in the given mass of the gas. The given mass is 0.72 g, and the molar mass of the empirical formula is 14.03 g/mol. Dividing the given mass by the molar mass, we find the number of empirical formula units to be 0.72 g / 14.03 g/mol = 0.0514 mol. Since the molecular formula represents the actual number of atoms in the molecule, we need to determine the ratio of empirical formula units to the actual number of molecules. To do this, we divide the molar mass of the gas (calculated from the volume and pressure data) by the molar mass of the empirical formula. Using the ideal gas law, PV = nRT, we can rearrange the equation to solve for n (the number of moles). Plugging in the given pressure (0.967 atm), volume (0.559 L), temperature (110°C = 383 K), and the ideal gas constant (0.0821 L·atm/(mol·K)), we find n to be approximately 0.0294 mol. Finally, we divide the number of moles of the empirical formula (0.0514 mol) by the number of moles determined from the gas volume and pressure (0.0294 mol). The ratio is approximately 1.75. Since the ratio is not close to a whole number, we multiply the subscripts of the empirical formula by 1.75 to obtain the molecular formula. This gives us C2H4, matching Option B. Therefore, the molecular formula of the gas is C2H4 (Option B).

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Pi bonds often have lower strength than sigma bonds. For instance, a carbon-carbon double bond with one sigma and one pi bond has double the bond energy of a carbon-carbon single bond (sigma bond).

Pi bonds are covalent chemical bonds in which two lobes of one atomic orbital are lateral overlapped by two lobes of an atomic orbital that belongs to a different atom. Pi bonds are frequently expressed as "bonds," where the Greek character alludes to the p orbital and the pi bond's shared symmetry.

Pi bonding frequently involves p orbitals. D orbitals can, however, also engage in other sorts of bonds, and these d orbital-based bonds can be seen in the numerous bonds that are created between two metals.

Here the given molecule consists of only two π bonds.

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To calculate the energy needed to melt 750 grams of gold starting at 24°C, we need to consider two components: the energy needed to raise the temperature of gold to its melting point and the energy needed for the actual phase change from solid to liquid.

First, let's calculate the energy required to raise the temperature of gold from 24°C to its melting point of 1,063°C using the specific heat capacity:

Energy = mass * specific heat * temperature change

For this calculation, we'll use the specific heat of gold, which is 0.128 kJ/kg°C:

Energy = 750 g * 0.128 kJ/kg°C * (1,063°C - 24°C)

Next, we calculate the energy needed for the phase change, known as the heat of fusion. The heat of fusion for gold is given as 66.6 kJ/kg.

Energy = mass * heat of fusion

Energy = 750 g * 66.6 kJ/kg

Finally, we add the two energies together to get the total energy needed:

Total Energy = Energy for temperature change + Energy for phase change

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Thallium is most likely to have a density similar to lithium, while lead is least likely.

To determine which element is most and least likely to have a density similar to lithium, we need to look at their respective atomic masses and densities. Lithium has an atomic mass of 6.94 g/mol and a density of 0.534 g/cm3.
Thallium has an atomic mass of 204.38 g/mol and a density of 11.85 g/cm3, which is closer to lithium's density than the other elements listed.
On the other hand, lead has an atomic mass of 207.2 g/mol and a density of 11.34 g/cm3, which is significantly higher than lithium's density. Oxygen has a much lower atomic mass and density, while rubidium has a higher atomic mass but lower density than lithium.
Therefore, based on their atomic masses and densities, thallium is most likely to have a density similar to lithium, while lead is least likely.

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The hydroxide ion concentration [OH-] in a 1.1×10−3 M solution of Sr(OH)2 can be calculated using the solubility product constant (Ksp) for the compound. The final concentration of [OH-] in the solution is 2.4×10−4 M.

1. The Ksp for Sr(OH)2 is 5.4×10−12, which represents the equilibrium constant for the dissolution of Sr(OH)2 in water. By assuming that the dissociation of Sr(OH)2 in water is complete, we can calculate the molar concentration of [OH-] from the stoichiometry of the reaction.

2. The solubility product constant (Ksp) is the equilibrium constant for the dissolution of a sparingly soluble salt in water. It represents the concentration of the ions produced when the solid salt dissolves. For Sr(OH)2, the Ksp is given as: Sr(OH)2 ⇌ Sr2+ + 2OH−

Ksp = [Sr2+][OH−]2 = 5.4×10−12

3. The stoichiometry of the reaction shows that for every one mole of Sr(OH)2 that dissolves, it produces one mole of Sr2+ ions and two moles of OH− ions. Therefore, if we assume that all of the Sr(OH)2 has dissociated completely, then the molar concentration of [OH−] is twice that of [Sr(OH)2]. [OH−] = 2[ Sr(OH)2]

[OH−] = 2 × 1.1×10−3 M

[OH−] = 2.2×10−3 M

4. However, we need to take into account the fact that [Sr2+] and [OH−] will recombine to form Sr(OH)2, which will affect the concentration of [OH−]. To calculate the concentration of [OH−] at equilibrium, we can use the quadratic equation to solve for x in the expression for the Ksp:

Ksp = [Sr2+][OH−]2 = (x)(2x)2 = 5.4×10−12

x = [OH−] = 2.4×10−4 M

5. Thus, the final concentration of [OH−] in the solution is 2.4×10−4 M, which is much smaller than the initial concentration of 2.2×10−3 M. This indicates that the reaction has reached equilibrium, with most of the Sr2+ and OH− ions combining to form solid Sr(OH)2.

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Gallium (Ga, atomic number = 31) has 3 valence electrons, as indicated by the 4s² 4p¹ configuration. Valence electrons are the electrons located in the outermost energy level or shell of an atom, and they play a crucial role in determining the chemical properties and reactivity of an element.

Gallium's electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p¹. Valence electrons are the electrons in the outermost energy level, which in this case is the 4th energy level (4s² 4p¹).

Having three valence electrons, gallium (Ga) belongs to group 13. Therefore, the complete amount of electrons in the 4s and 4p subshells three in all can be lost in a chemical process.

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A compound inequality is an inequality that includes two or more inequalities joined together by the words "and" or "or".

To determine which compound inequality can be used to solve a given inequality, we need to consider the operations involved and the desired outcome. For example, if we want to find the values of x that satisfy the inequality "3x - 4 < 7", we can add 4 to both sides to get "3x < 11", then divide by 3 to get "x < 11/3". This gives us a single inequality.

However, if we have an inequality like "2x + 5 < 9 or 4x - 3 > 5", we have two separate inequalities that need to be solved. We can write this as a compound inequality using the word "or": "2x + 5 < 9 or 4x - 3 > 5" becomes "2x < 4 and 4x > 8". This can be written more compactly as "2 < x < 2.5", since the values of x that satisfy both inequalities are between 2 and 2.5.

In general, we use "and" to represent the intersection of two sets (values that satisfy both inequalities), and "or" to represent the union of two sets (values that satisfy either one or both inequalities). So, to solve an inequality using a compound inequality, we need to identify the set of values that satisfy the given inequality, and then use "and" or "or" to combine the necessary inequalities.

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If the pH of the cell is increased by adding NaOH to the beaker containing chromium solution, the value of E, the standard electrode potential, would likely increase.

This is because an increase in pH typically results in a decrease in the concentration of hydrogen ions (H+) in the solution, which in turn affects the reduction potential of the half-cell reaction. As the concentration of H+ decreases, the reduction potential becomes more positive, leading to an increase in E.

However, it's important to note that the specific effect on E will depend on the details of the reaction and the concentrations of the various species involved.

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Balancing equations in grade 9 natural science involves ensuring that the number of atoms of each element is equal on both sides of the chemical equation. Here's a step-by-step process to balance equations:

Start by writing down the unbalanced equation, including the formulas of all reactants and products.

Count the number of atoms for each element on both sides of the equation.

Begin by balancing elements that appear in only one compound on each side. Adjust the coefficients (numbers in front of the formulas) to balance the number of atoms.

Next, balance elements that appear in multiple compounds. Remember that coefficients apply to the entire compound. Avoid changing subscripts, as they represent different substances.

Keep adjusting the coefficients until the number of atoms is the same on both sides.

Check your work by counting the atoms again to ensure they are balanced.

Remember, balancing equations requires practice. Be patient and persistent. It's helpful to start with simpler equations and gradually work your way up to more complex ones. Balancing equations is an essential skill in chemistry as it demonstrates the law of conservation of mass and allows for accurate predictions of chemical reactions.

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Job Instruction Training which is option.c, is the method involves breaking down a job into specific tasks and teaching employees how to perform each task through a series of steps.

This training is usually done by a supervisor or experienced employee who works closely with the trainee until they can perform the task independently. It is a highly effective way to train non-managerial employees as it provides hands-on experience, immediate feedback, and allows for customization to fit the needs of each individual.

On-the-Job Training and Employee Development Training are broader terms that may include various training methods, while Intensive Job Orientation typically refers to a shorter, introductory training period for new employees.

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When the name of the anion ends in -ide, the acid name begins with the prefix "hydro-" followed by the stem of the nonmetallic element of the anion and the suffix "-ic". For example, the anion chloride (Cl-) becomes hydrochloric acid (HCl) and the anion sulfide (S2-) becomes hydrosulfuric acid (H2S).

This naming convention is used for binary acids, which are compounds composed of hydrogen and a nonmetallic element. However, for oxyacids, which contain oxygen, the naming convention is different and depends on the number of oxygen atoms present in the molecule.

When the name of an anion ends in -ide, the acid name begins with the prefix "hydro-" and ends with the suffix "-ic acid."  Step-by-step explanation:1. Identify the anion with a name ending in -ide. 2. Add the prefix "hydro-" to the beginning of the anion's root name. 3. Add the suffix "-ic acid" to the end of the root name.
For example, if the anion is chloride (Cl-), the corresponding acid name would be hydrochloric acid (HCl).

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The volume of helium gas at 40°C and 1.20 atm will be approximately 20.4 L., which is option A.

Using the combined gas law:

(P1V1)/T1 = (P2V2)/T2

where P is pressure, V is volume, and T is temperature in Kelvin.

Converting 23°C and 40°C to Kelvin:

23°C + 273.15 = 296.15 K

40°C + 273.15 = 313.15 K

Plugging in the given values:

(0.956 atm)(14.7 L)/(296.15 K) = (1.20 atm)(V2)/(313.15 K)

Solving for V2:

V2 = (0.956 atm)(14.7 L)(313.15 K)/(1.20 atm)(296.15 K)

V2 = 20.4 L

Therefore, the answer is B) 20.4 L.

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Sodium hydroxide would most effectively deprotonate benzoic acid due to its strength as a base.

Benzoic acid (PhCOOH) is a weak acid with a pKa value of 4.2. To deprotonate benzoic acid, a strong base is required. The base should be able to remove the hydrogen ion (H+) from the carboxylic acid group. The strength of a base is determined by its ability to accept a proton. Therefore, a stronger base will be able to more effectively deprotonate benzoic acid.
There are several strong bases that can be used to deprotonate benzoic acid. Some of the most commonly used strong bases include sodium hydroxide (NaOH), potassium hydroxide (KOH), lithium hydroxide (LiOH), and sodium methoxide (NaOMe).
Among these strong bases, sodium hydroxide is the most commonly used base to deprotonate benzoic acid. This is because sodium hydroxide is a very strong base with a pKa value of 14. Therefore, it is highly effective in deprotonating benzoic acid.
In conclusion, sodium hydroxide would most effectively deprotonate benzoic acid due to its strength as a base.
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Cepheid variables and RR Lyrae stars are both pulsating stars that astronomers use as standard candles to measure cosmic distances.

However, Cepheids are brighter and have longer periods than RR Lyrae stars. This means that they can be seen farther away and used to measure distances to more distant galaxies than RR Lyrae stars. In fact, a typical Cepheid variable is about 100 times brighter than a typical RR Lyrae star. This means that Cepheids can be used to measure distances up to tens of millions of light-years, while RR Lyrae stars are only useful for distances within our own galaxy and nearby galaxies. The use of Cepheids as distance-measuring tools has been crucial for the study of the large-scale structure and evolution of the universe.

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The rate of reaction increases, we need to consider the given reaction and orders of reactants A and B the rate of reaction increases by a factor of 50.

A + 2B -> Products
The reaction is first-order with respect to A and second-order with respect to B. Therefore, the rate law for the reaction can be written as:

Rate = k[A]^1[B]^2
Now, let's consider the changes in the concentrations of A and B:- Concentration of A doubles, so the new concentration of A is 2[A]. - Concentration of B increases 5 times, so the new concentration of B is 5[B].

Now we can find the new rate of reaction:
New Rate = k(2[A])^1(5[B])^2
To determine the factor by which the rate of reaction increases, we need to divide the new rate by the original rate:
Factor = (k(2[A])^1(5[B])^2) / (k[A]^1[B]^2)
By simplifying this expression, we get:
Factor = (2^1)(5^2)
Factor = 2 * 25 = 50
So, the rate of reaction increases by a factor of 50.

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Answer:  Yes, elements in the 6th period have a greater atomic size than elements in the 1st period.

Explanation:

All the electrons of noble gas elements are paired (ns 2np 6 configurations), and paired electrons produce inter-electronic repulsions which weakens the effective nuclear force, so electrons tend to move away from the nucleus because of repulsions. So the size of noble gases is bigger than the other elements in their respective periods

Answer:

because in going down a column you are jumping up to the next higher main energy level

Explanation:

This is because in going down a column you are jumping up to the next higher main energy level (n) and each energy level is further out from the nucleus - that is, a bigger atomic radius. Atoms get smaller as you go across a row from left to right.

The pH of the resulting solution would depend on the identity and concentration of the added neutralizing solution.

When a solution is neutralized, it means that the pH of the solution has been adjusted to 7, which is considered neutral on the pH scale. The pH of the resulting solution after neutralization would depend on the identity and concentration of the neutralizing solution that was added.

For example, if an acidic solution with a pH of 3 was neutralized with a basic solution with a pH of 11, the resulting pH would be around 7. However, if a weaker basic solution with a pH of 9 was used instead, the resulting pH would be slightly acidic, around 6.5. It is important to note that the amount of the neutralizing solution added also plays a role in determining the final pH of the resulting solution.

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When an acidic or basic solution is neutralized, the pH of the resulting solution depends on the identity and concentration of the neutralizing solution.

If the neutralizing solution is a strong acid or base, it would completely dissociate in water and result in a pH closer to the pH of the added solution. For example, if a strong base like sodium hydroxide (NaOH) is added to an acidic solution, the hydroxide ions (OH-) from NaOH will react with the hydrogen ions (H+) from the acid to form water. The resulting solution will have a pH closer to 7, which is neutral.

On the other hand, if the neutralizing solution is a weak acid or base, the pH of the resulting solution would depend on the concentration and dissociation constant of the weak acid or base. The resulting solution will have a pH that is slightly higher than the initial pH, depending on the concentration and dissociation constant of acetic acid.

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The new volume at the given pressure and temperature would be approximately 10.3 L.

To solve this problem, we need to use the combined gas law equation:

(P1V1/T1) = (P2V2/T2)

where P1, V1, and T1 are the initial pressure, volume, and temperature, and P2 and T2 are the new pressure and temperature, respectively. We can rearrange the equation to solve for V2:

V2 = (P1V1T2)/(T1P2)

Plugging in the values given in the problem, we get:

V2 = (2.18 atm x 6.7 L x 264 K)/(239 K x 6.14 atm) ≈ 10.3 L

It's important to note that when solving gas law problems, we must make sure that all units are in the correct SI units (atm, L, and K) and that the temperature is always in Kelvin. Also, we assume that the gas behaves ideally and that there are no chemical reactions occurring.

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The best buffer in this experiment is the one with the highest buffering capacity, maintaining a stable pH when small amounts of acid or base are added.

To compare the buffer performance in this experiment, you need to evaluate the buffering capacity of each buffer, which is the ability to maintain a stable pH when small amounts of acid or base are added. A higher buffering capacity indicates better performance.

To do this, you can compare the pH changes of each buffer upon addition of the same amounts of acid or base. The buffer with the smallest pH changes demonstrates the highest buffering capacity and is considered the best. Factors that influence buffering capacity include the concentration of the buffer components and the pKa of the buffering agent, which should be close to the desired pH of the buffer solution.

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The buffer in this experiment that can maintain a steady pH when modest quantities of acid or base are introduced has the highest buffering capacity.

The ability of each buffer to maintain a stable pH when tiny additions of acid or base are made is its buffering capacity, which must be assessed in order to compare the buffer performance in this experiment. Better performance is indicated by a larger buffering capacity.

To achieve this, you may evaluate how each buffer's pH changes when the identical amounts of acid or base are added. The buffer that exhibits the least amount of pH change is said to have the maximum buffering capability. The concentration of the buffer components and the pKa of the buffering agent, which should be near to the intended pH of the buffer solution, are factors that affect buffering capacity..

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The helical structure (also known as the alpha helix) is part of the protein's secondary structure, along with beta-pleated sheet.

The reactions for which ΔH∘rxn is equal to ΔH∘f of the product(s) are:
- CaCO3(g)→CaO+CO2(g)
- C(s,graphite)+O2(g)→CO2(g)
- CO(g)+12O2(g)→CO2(g)

In a chemical equation, reactions are the chemical changes that occur when two or more substances, called reactants, are converted into new substances, called products. A chemical equation is a symbolic representation of a chemical reaction, and it provides information about the chemical identities of the reactants and products, as well as the stoichiometry of the reaction, or the ratios of the reactants and products.

The reactions in a chemical equation are represented by chemical formulas, which use symbols to represent the atoms and molecules of the reactants and products. For example, the balanced chemical equation for the reaction of hydrogen gas with oxygen gas to form water is:

2H2(g) + O2(g) → 2H2O(l)

This equation shows that two molecules of hydrogen gas react with one molecule of oxygen gas to form two molecules of liquid water. The coefficients in front of the chemical formulas indicate the stoichiometry of the reaction, and they are used to balance the equation so that the number of atoms of each element is the same on both sides of the equation.

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The correct answer is option a) 125,400 J.

To determine the amount of energy required to heat 500 g of ice from 0 °C to 60 °C, we must consider two processes: (1) heating the ice to its melting point and (2) heating the resulting water from 0 °C. C to 60 °C.

The first step involves heating the ice from its initial temperature of 0 °C to its melting point, which is also 0 °C. This requires energy to raise the temperature of the ice without changing its state. The amount of energy required for this process is calculated according to the formula:

Q1 = m * C * AT

Where:

- Q1 represents the energy required for the first step

- m is the mass of ice (500 g)

- C is the specific heat capacity of ice (2.09 J/g°C)

- ΔT is the temperature change (0°C - 0°C = 0°C)

Since there is no temperature change, the value of Q1 is zero for this step.

The second step involves heating the water from 0°C to 60°C. The amount of energy required for this process is calculated according to the formula:

Q2 = m * C * AT

Where:

- Q2 represents the energy needed for the second step

- m is the mass of water (500 g)

- C is the specific heat capacity of water (4.18 J/g°C)

- ΔT is the temperature change (60 °C - 0 °C = 60 °C)

By substituting the values ​​into the equation, we have:

Q2 = 500 g * 4.18 J/g °C * 60 °C = 125,400 J

The total energy required to heat 500 g of ice from 0 °C to 60 °C is therefore given by:

Total energy = Q1 + Q2 = 0 J + 125,400 J = 125,400 J

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none of the given option is correct

sp² hybridization is the process by which one s orbital and two p orbitals from the same atomic shell combine to create a new equivalent orbital. Here the hybridization of terminal carbons in h₂c=c=ch₂. The correct option is E.

In order to produce the same number of a new type of hybrid orbitals, atomic orbitals with the same energy levels are mixed together in a process known as hybridization. Typically, this mixing creates hybrid orbitals with completely distinct energies, morphologies, etc.

Hybridization of 'C' can be find out by counting the number of σ bonds and Π bonds present on 'C' atom. In the terminal carbon atoms there are 3σ bonds and it is sp² hybridized.

Thus the correct option is E.

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The closest answer choice is A) 122 mL.  The key idea here is to use the combined gas law to relate the initial conditions to STP (standard temperature and pressure). The combined gas law is:

(P1V1) / (T1) = (P2V2) / (T2)

where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively. At STP, P2 = 1 atm, T2 = 273 K, and we want to find V2.

We can use the vapor pressure of water at 22°C to find the partial pressure of the dry gas:

Pdry = Ptotal - PH2O = 753 torr - 20 torr = 733 torr

Now we can plug in the values we know:

(P1V1) / (T1) = (P2V2) / (T2)

(733 torr)(142 mL) / (295 K) = (1 atm)(V2) / (273 K)

Solving for V2, we get:

V2 = (733 torr)(142 mL)(273 K) / (295 K)(1 atm) = 123 mL

So the volume of the dry gas at STP is 123 mL (rounded to 3 significant figures). The closest answer choice is A) 122 mL.

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Answer: (a) 4s, (b) 3p, (c) 3p, (d) 4s

Explanation:

The orbital from which an electron will be removed from will be the highest energy orbital in the valence shell. The valence shell for Zn and Cu is the 4th shell, and they only have electrons in the 4s orbitals, so removed electrons will come from their 4s orbital electrons. Cl and Al have the 3rd shell as their valence shell, and they both have electrons in 3p orbitals, which are the highest energy orbitals they have in the valence shell. Thus, removed electrons will come from their 3p orbital electrons.

Answer:

What is photosynthesis

The melting of ice is a spontaneous reaction at room temperature and pressure because it is accompanied by an increase of entropy.

The spontaneity of a reaction is determined by the change in Gibbs free energy (∆G), which is given by the equation:

∆G = ∆H - T∆S

where ∆H is the change in enthalpy, T is the temperature in Kelvin, and ∆S is the change in entropy. A reaction is spontaneous if ∆G is negative.

In the case of ice melting at room temperature and pressure, the process is accompanied by an increase in entropy because the solid phase (ice) has a more ordered arrangement than the liquid phase (water).

This increase in entropy (∆S) contributes a negative term to the ∆G equation, making ∆G negative and the reaction spontaneous.

Therefore, the correct option is (b) Melting is accompanied by an increase of entropy, which explains why the melting of ice is a spontaneous reaction at room temperature and pressure.

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The rate of reaction decreased by a factor of 1/2 or 50%.

The rate law for the given reaction can be represented as,

Rate = k[A]^2[B]^1

If [A]° is halved, the new concentration of A is (1/2)[A]°. If [B] is doubled, the new concentration of B is 2[B]°. Let's plug these values into the rate law:

New Rate = k[(1/2)[A]°]^2[2[B]°]^1

Simplifying, we get:

New Rate = k(1/4)[A]°^2[2[B]°]

Now, let's find the factor by which the rate of the reaction decreased. Divide the New Rate by the original Rate:

Factor = (New Rate) / (Rate) = (k(1/4)[A]°^2[2[B]°]) / (k[A]^2[B]^1)

The k, [A]^2, and [B] terms cancel out, leaving us with:

Factor = (1/4) * 2 = 1/2

So, the rate of the reaction decreased by a factor of 1/2 or 50%.

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