What effect does the cation of an ionic compound have on the appearance of the solution?.

The atom economy of reaction 2 is  57.8%.

We have to note that the atom economy would have to do with the atoms that are utilized and the atoms that are not utilized in the process. We have to consider the left and the right hand sides of the reaction.

Mass of the desired product = 63.5 + 2(35.5) = 134.5

Mass of the reactants = 2(63.5 + 16) + 2(35.5 + 1) = 159 + 73 = 232

Then atom economy = Mass of atoms in desired product/Mass of atom in reactant * 100/1

= 134/232 * 100 = 57.8%

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Answer:

six valence electrons

The outer energy level in this atom is n = 3. It holds six electrons, so sulfur has six valence electrons.

Explanation:

Thank you

Answer:

Li

Explanation:

Li has the highest IE2 , because to remove the second electron we must break the stable 1s2 noble gas shell.

Answer:

Li has the highest IE2 , because to remove the second electron we must break the stable 1s2 noble gas shell.

Answer:

the first statement is correct...

it only supports the main idea of the law of conservation of mass...

hope I can help you...

Ammonia can form nitrite, which can then be converted into nitrate. Both can be taken in by plants. This process is known as nitrification.

The nitrogen cycle refers to the processes by which nitrogen is cycled between the living and non-living components of the environment.

The nitrogen cycle consists of processes that add as well as remove nitrogen from the soil and the atmosphere.

Plants require nitrogen for growth. This nitrogen is sourced mainly from the atmosphere. Nitrogen is the most abundant gas in the atmosphere, however, it exists in a form that is not readily available to plants.

Hence several processes such as nitrogen fixation, nitrification, and ammonia conversion are required for plants to obtain nitrogen.

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Calculating probabilities in R

(a) The probability that exactly 20 fish survive can be calculated using the binomial probability formula in R. Let's denote this probability as P(X = 20), where X follows a binomial distribution with n = 29 (total number of fish) and p = 0.74 (probability of survival, calculated as 1 - 0.26). The calculation in R would be:

```R

dbinom(20, size = 29, prob = 0.74)

```

The resulting probability is the answer to part (a).

(b) The probability that at least 14 fish survive can be calculated by summing the probabilities of 14, 15, 16, and so on, up to 29. Denoting this probability as P(X ≥ 14), the calculation in R would be:

```R

sum(dbinom(14:29, size = 29, prob = 0.74))

```

The resulting probability is the answer to part (b).

(c) The probability that at most 23 fish survive can be calculated by summing the probabilities of 0, 1, 2, and so on, up to 23. Denoting this probability as P(X ≤ 23), the calculation in R would be:

```R

sum(dbinom(0:23, size = 29, prob = 0.74))

```

The resulting probability is the answer to part (c).

(d) To find the mean and variance of the number of fish that survive, we can use the formulas for the binomial distribution. The mean (μ) is calculated as n * p, and the variance (σ²) is calculated as n * p * (1 - p). Using the given values of n = 29 and p = 0.74, we can calculate the mean and variance in R:

```R

n <- 29

p <- 0.74

mean <- n * p

variance <- n * p * (1 - p)

```

The resulting mean and variance values are the answer to part (d).

What are the probabilities and statistics calculated in R for fish survival in polluted water?

(a) To calculate the probability that exactly 20 fish survive, we use the binomial probability formula in R, which considers the total number of fish (n = 29) and the probability of survival (p = 0.74). The resulting probability represents the likelihood of observing exactly 20 surviving fish.

(b) The probability that at least 14 fish survive is calculated by summing the probabilities of 14, 15, 16, and so on, up to the total number of fish (29). This cumulative probability indicates the likelihood of observing 14 or more surviving fish.

(c) The probability that at most 23 fish survive is calculated by summing the probabilities of 0, 1, 2, and so on, up to 23. This cumulative probability represents the likelihood of observing 23 or fewer surviving fish.

(d) The mean and variance of the number of fish that survive can be found using the formulas for the binomial distribution. The mean (μ) is equal to the product of the total number of fish (n) and the probability of survival (p).

The variance (σ²) is calculated as the product of n, p, and (1 - p). These values provide insights into the average and variability of fish survival in the given conditions.

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Answer:

4:D 5:D

Explanation:

Explanation:

Review and Study Material Before Going to Class. ...

Take Good Notes. ...

Practice Daily. ...

Take Advantage of Lab Time. ...

Use Flashcards. ...

Use Study Groups. ...

Break Large Tasks Into Smaller Ones.

reading a variety of chemistry books

Answer:

An ion is an atom or molecule that carries an electric charge.

Ions are identified by the use of a superscript to show the sign and size of their electric charge.

Negatively charged ions are called anions, and positively charged ions are called cations.

Explanation:

An atom is the smallest unit of ordinary matter that forms a chemical element. Every solid, liquid, gas, and plasma is composed of neutral or ionized atoms. Atoms are extremely small, typically around 100 picometers across.

We do not have enough information to deduce the isotope that would have nine neutrons. Option 5

The atom is composed of subatomic particles. The subatomic particles that are in the atom are such that we have the electrons that are negatively charged, the protons that are positively charged and the neutrons that are uncharged.

Now, we know that the number of neutrons is obtained as the difference between the mass number and the atomic number of the atom. In this case, we only have the mass numbers and not the atomic numbers thus we do not have sufficient information.

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Answer:

Ionic compounds conduct electricity when molten (liquid) or in aqueous solution (dissolved in water), because their ions are free to move from place to place. Ionic compounds cannot conduct electricity when solid, as their ions are held in fixed positions and cannot move.

Explanation:

Sodium and potassium are s-block alkali metals. Because potassium and sodium are metals, they each lose one electron.

Reactivity is indeed a measure of how easily a material conducts a chemical reaction in chemistry. The reaction might include the chemical alone or in combination with those other atoms or molecules, and it is usually followed by the release of energy. The most reactive components and elements have the potential to spontaneously or explosively ignite.

Because potassium and sodium are metals, they each lose one electron. They create cations and react with nonmetals to generate salt. Sodium and potassium are s-block alkali metals.

Therefore, because potassium and sodium are metals, they each lose one electron.

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saturated fats are are mainlyderived from animals eg butter and lard and they are solid at room temperature

Among the given statements, the correct statement is: B. All three are three-membered rings bearing a positive charge that occur as intermediates.

A protonated epoxide, a bromonium ion, and a mercurinium ion are all three-membered rings bearing a positive charge. However, their roles and reactivities differ.

A protonated epoxide is formed by the addition of a proton to an epoxide, resulting in the formation of a three-membered ring with a positive charge. It can be attacked by nucleophiles, including water, from the back side in an SN2 reaction.

A bromonium ion is formed during the halogenation of an alkene with a bromine molecule. It is a three-membered ring with a positive charge, and it is highly reactive. Nucleophiles can attack the bromonium ion from either side, leading to the formation of a vicinal dihalide.

A mercurinium ion is formed during the oxymercuration-demercuration of an alkene, where a mercury acetate complex adds across the double bond. The resulting mercurinium ion is a three-membered ring with a positive charge. Nucleophiles can attack the mercurinium ion, leading to the addition of the nucleophile across the double bond.

Therefore, the correct statement is that all three, the protonated epoxide, bromonium ion, and mercurinium ion, are three-membered rings bearing a positive charge that occur as intermediates in different reactions.

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Answer:

See explanation

Explanation:

In the section A,B

C triple bond O has a higher bond energy. A triple bond is shorter and more difficult to break hence they have a higher bond energy.

For the same reason in the C,D section; C triple bond N has a higher bond energy.

In the next section, the higher the bond order the shorter the bond. Hence C triple bond C is the shortest in the A,B section while C triple bond O is the shortest in the C,D section. Both have a bond order of 3.

In the A,B section; I is larger than Br hence H-Br has the shortest bond length.

In the C,D section; Br is larger than Cl hence P-Cl has the shortest bond length.

Using the ideal gas law PV =nRTPV=nRT , we find that the pressure will be P =\frac{nRT}{V}P=

V

nRT

​

 . Then, we'll substitute and find the pressure, using T = -25 °C = 248.15 K and R = 0.0821 \frac{atm\cdot L}{mol \cdot K}

mol⋅K

atm⋅L

​

 :

P =\frac{nRT}{V} = \frac{(0.33\,\cancel{mol})(0.0821\frac{atm\cdot \cancel{L}}{\cancel{mol \cdot K}})(248.15\,\cancel{K})}{15.0\,\cancel{L}} = 0.4482\,atmP=

V

nRT

​

=

15.0

L

​

(0.33

mol

)(0.0821

mol⋅K

atm⋅

L

​

​

)(248.15

K

​

)

​

=0.4482atm

In conclusion, the pressure of this gas is P=0.4482 atm.

Reference:

Chang, R. (2010). Chemistry. McGraw-Hill, New York.

The solubility product constant expression for copper(II) oxalate is:

CuC₂O₄(s) ⇌ Cu²⁺(aq) + C₂O₄²⁻(aq)

The Ksp value given is 2.9 x 10⁻⁸ at 25°C.

Let's assume that "x" is the molar solubility of C₂O₄²⁻   in water, then the equilibrium concentrations of Cu²⁺ and C₂O₄²⁻ ions are also "x" because 1 mol of C₂O₄²⁻ produces 1 mol of Cu²⁺ and 1 mol of C₂O₄²⁻

So, it can be written as:

Ksp = [Cu²⁺ ][C₂O₄²⁻ ] = x²

Substituting the given Ksp value, we get:

2.9 x 10⁻⁸ = x²

Taking the square root of both sides gives:

x = 1.7 x 10⁻⁴ M

Therefore, the solubility of copper(II) oxalate at 25°C is 1.7 x 10⁻⁴ moles/litre, rounded to two decimal places.

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The concentration of A decreases and then levels out. Option A

In many chemical reactions, a reactant is consumed as the reaction progresses, leading to a decrease in its concentration over time. The reactant molecules are transformed into products, and as the reaction proceeds, the concentration of the reactant gradually diminishes.

At equilibrium, the concentrations of both reactants and products remain relatively constant over time, although they can coexist.

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The best option that express the results that occur in exothermic reaction is Light as well as heat is released into the environment.

Therefore, forth option is correct.

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Answer:

(Question) Which result occurs during an exothermic reaction?

(Answer) Light or heat are released into the environment.

(Question) Which event is an example of an endothermic reaction?

(Answer) photosynthesis

(Question) How is total reaction energy calculated?

(Answer) reactant bond energy – product bond energy

(Question) When producing hydrogen iodide, the energy of the reactants is 581 kJ/mol, and the energy of the products is 590 kJ/mol. The equation is shown.

H2 + I2 → 2HI

What is the total energy of the reaction? Is this an endothermic or exothermic reaction?

(Answer) –9 kJ/mol, exothermic

(Question) Hydrogen bromide breaks down into diatomic hydrogen and bromine in the reaction shown.

2HBr → H2 + Br2

The energy of the reactant is 732 kJ/mol, and the energy of the products is 630 kJ/mol.

What is the total energy of the reaction? Is this reaction endothermic or exothermic?

(Answer) 102 kJ/mol, endothermic

Explanation:

just finished the quick check, enjoy

Answer:

The increased electron density along the internuclear axis is called a n bond. a. True b. False

B. False

Explanation:

The lanthanides are the 14 elements from cerium (atomic number 58) to lutetium (atomic number 71). The word comes from the Greek “lanthanein” meaning “to be hidden.”

Typical DMI (Dry Matter Intake) for dairy cattle during the dry period is around 2.0 to 2.2% of their body weight. This means that a dairy cow will consume approximately 2.0 to 2.2% of its body weight in dry feed each day during the dry period.

Dairy cows are not producing milk during the dry period, which is the time frame of about 60 days prior to calving, and are instead getting ready for the upcoming lactation cycle. Their dietary needs at this period are different from those they have when they are actively making milk.

There are a number of reasons why DMI is lower during the dry season. First of all, during the dry phase, the cow's udder involutes (regresses) and gets ready for the impending lactation. The cow's metabolic activity decreases as a result, and she requires less energy.

Second, during the dry season, the cow's rumen capacity drops. Due to the growing calf and the shrinking amount of the udder, the capacity of the rumen, the main component of the cow's stomach, has been diminished. The amount of feed that the cow can eat is constrained by the small rumen space.

The cow's physiological state during the dry period and fluctuations in hormone levels can also have an impact on the cow's appetite. A decrease in feed intake may result from hormonal changes and rumen motility changes that the cow may encounter.

It is crucial to keep in mind that the precise DMI during the dry period can change depending on elements including cow genetics, physical condition, and management techniques. For dairy farmers to maintain ideal cow health and get their animals ready for a successful lactation phase, it's critical to track each cow's consumption and modify feeding techniques accordingly.

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Answer:

antimony

Explanation:

it has 51 protons, 4 more than silver and two less than xenon

An exothermic reaction is spontaneous if the overall Gibbs free energy change (ΔG) is negative, indicating that the reaction is energetically favorable and will proceed without an external energy input. However, an exothermic reaction can become non-spontaneous under certain circumstances.

One such circumstance is when the entropy change (ΔS) is negative. If ΔH is negative (exothermic) but ΔS is also negative (decrease in disorder), the value of ΔG could still be positive (non-spontaneous) or close to zero (at equilibrium) at temperatures where ΔH is not sufficiently large to overcome the negative ΔS.

This means that even though energy is released during the reaction, the decrease in disorder can make the reaction unfavorable.

Another circumstance is when the reactants are in a highly ordered or low-energy state, and the products are in a highly disordered or high-energy state. In such cases, the enthalpy change (ΔH) may be negative (exothermic), but the entropy change (ΔS) is also positive, and the resulting ΔG value may still be positive, making the reaction non-spontaneous.

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The major elimination product when I-bromobutane is reacted with the bulky base, potassium t-butoxide, in t-butyl alcohol is cis-2-butene.

This is because when a base is added to a molecule with a double bond, the electrons in the double bond are shifted to the more substituted carbon, resulting in the formation of a new single bond and a new double bond. Since the bulky base, potassium t-butoxide, is larger than the hydrogen atom, it will be shifted to the more substituted carbon, forming a cis-2-butene. This is because the larger size of the base causes it to be more sterically hindered, thus making it more difficult for the electrons to move to the more substituted carbon. As a result of this reaction, the mass of the reactants and products remains constant, as per the law of conservation of mass.

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Answer:

128gCaCl2 per 200gH2O

Explanation:

I might be wrong rlly srry if I am:/ Hope this helps doe have a wonderful day

A down arrow (↓) or (s) are symbols that are used to represent the precipitation product produce in the reaction.

AgNO3 (aq) + KCl (aq) → AgCl (s) + KNO3(aq)

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Decreasing the TBCl concentration will not affect the rate constant in this experiment. The rate constant is determined by the specific reaction and temperature conditions and is independent of the reactant concentrations.

The rate constant (k) is a proportionality constant that relates the rate of a reaction to the concentrations of the reactants. However, the rate constant itself is not affected by the concentrations of the reactants. It is determined by the specific reaction and temperature conditions.

The rate of a chemical reaction can be expressed using the rate equation, which typically includes the concentration terms for the reactants raised to certain powers.

These powers, known as reaction orders, can be determined experimentally. However, the rate constant is a separate factor in the rate equation and is not dependent on the reactant concentrations.

By decreasing the TBCl concentration, the rate of the reaction may be affected, as the rate is directly proportional to the reactant concentrations.

However, the rate constant itself remains unchanged. The rate constant is influenced by factors such as temperature, presence of catalysts, and the nature of the reacting species, but not by the concentrations of the reactants.

Therefore, decreasing the TBCl concentration will not affect the rate constant in this experiment.

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