When heat is removed from a substance, describe how the molecules are affected, what are causing these changes, describe what's happening in whole, and does the temp. increase or decrease?

The speed of a wave can be determined by multiplying its frequency by its wavelength. In this case, a wave with a frequency of 14 hertz and a wavelength of 3 meters will travel at a speed of 42 meters per second.

This is a question related to the physics of wave motion. The speed of a wave can be calculated using the formula:

Speed = Frequency x Wavelength

. Given the frequency of the wave is 14 hertz and the wavelength is 3 meters, you can plug these values into the formula. Therefore, the speed of the wave would be:

14 Hertz x 3 meters = 42 meters per second

. Hence, the wave will travel at a speed of 42 meters per second.

https://brainly.com/question/32331816

#SPJ6

Answer:

A) 0.225atm

Explanation:

P1V1 = P2V2

V1 = 450ml

P1 = 1.0atm

V2 = 2L = 2 X 100 = 2000ml

P2 =?

1.0 X 450 = P2 X 2000

P2 = (1.0 X 450)/2000

    = 0.225atm

Final answer:

The activation energy for the reverse reaction is calculated using the given activation energy for the forward reaction (100 kJ/mol) and the change in enthalpy of the reaction (-250 kJ/mol), resulting in an activation energy of 350 kJ/mol for the reverse reaction.

Explanation:

The question is about finding the activation energy for the reverse reaction based on the given activation energy and the change in enthalpy for the forward reaction. Using the provided data, Ea for the forward reaction is 100 kJ/mol and ΔH for the reaction is -250 kJ/mol.

To find the activation energy for the reverse reaction, we can use the concept that the sum of the activation energies for the forward and reverse reactions is equal to the difference in energy between the products and reactants. This relationship is derived from the potential energy diagram of a chemical reaction.

The activation energy for the reverse reaction can be calculated using the equation:
Ea(reverse) = Ea(forward) + ΔH

Substituting the given values:

Ea(reverse) = 100 kJ/mol - (-250 kJ/mol)

Ea(reverse) = 100 kJ/mol + 250 kJ/mol

Ea(reverse) = 350 kJ/mol

Therefore, the activation energy for the reverse reaction is 350 kJ/mol.

Final answer:

The subject of the sentence is 'horses,' as it is the main noun that the sentence is about, while 'Arabian,' 'Appaloosa,' and 'Morgan' are adjectives. Furthermore, horses can be classified as mammals in the animal kingdom. Option C

Explanation:

The subject in the sentence 'Arabian, Appaloosa, and Morgan horses will be at the county horse show this week.' is horses. When identifying the subject of a sentence, you are looking for the main noun or noun phrase that the sentence is about.

In this case, 'Arabian,' 'Appaloosa,' and 'Morgan' serve as adjectives describing the kinds of horses that will be present at the show. Therefore, the correct answer is C. horses.

As for classification, it's easy enough to classify the horse in the animal kingdom. That's one level of classification. But horses also belong to other groups; one important group is the mammals. These animals all have fur and nurse their young, which are key characteristics of mammals. Option C

Answer :

(1) The time passed by the sample is [tex]2.4\times 10^3\text{ years}[/tex]

(2) The amount left after decay process is 1.14 mmol.

Explanation :

Part 1 :

Half-life = 5730 years

First we have to calculate the rate constant, we use the formula :

[tex]k=\frac{0.693}{t_{1/2}}[/tex]

[tex]k=\frac{0.693}{5730\text{ years}}[/tex]

[tex]k=1.21\times 10^{-4}\text{ years}^{-1}[/tex]

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]

where,

k = rate constant  = [tex]1.21\times 10^{-4}\text{ years}^{-1}[/tex]

t = time passed by the sample  = ?

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process = 100 - 25 = 75 g

Now put all the given values in above equation, we get

[tex]t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{100}{75}[/tex]

[tex]t=2377.9\text{ years}=2.4\times 10^3\text{ years}[/tex]

Therefore, the time passed by the sample is [tex]2.4\times 10^3\text{ years}[/tex]

Part 2 :

Now we have to calculate the amount left.

Expression for rate law for first order kinetics is given by:

[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]

where,

k = rate constant  = [tex]1.21\times 10^{-4}\text{ years}^{-1}[/tex]

t = time passed by the sample  = 2255 years

a = let initial amount of the reactant  = 1.5 mmol

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

[tex]2255=\frac{2.303}{1.21\times 10^{-4}}\log\frac{1.5}{a-x}[/tex]

[tex]a-x=1.14mmol[/tex]

Therefore, the amount left after decay process is 1.14 mmol.

Answer: Option (A) is the correct answer.

Explanation:

When a gas is heated then its molecules gain more amount of kinetic energy. And, as kinetic energy is the energy obtained by an object due to its motion.

Therefore, with increase in kinetic energy of molecules of the gas there will occur more number of collisions. Hence, the gas will move more rapidly from one place to another.

Therefore, upon heating of a gas the energy absorbed by the gas will get converted into kinetic energy due to which gas move much more rapidly.

Potential energy is the energy obtained by an object due to its position and not because of its movement.

Thus, we can conclude that when a gas is heated all of the absorbed energy is converted to kinetic energy.

The pH and pOH of 6.6g of NaOH in 100mL solution are -0.217 and 14.217 respectively

Data;

To find the pOH and pH of this solution, we have to know the molarity of this solution.

Molarity = number of moles of solute / volume of the solution

number of moles of the solute = mass / molar mass

molar mass of NaOH = 40g/mol

number of moles = 6.6/40 = 0.165moles

Molarity of this solution is

[tex]M = \frac{number of moles }{volume of solution}\\M = \frac{0.165}{0.1}\\ Molarity = 1.65M[/tex]

[tex]pOH= -log[OH^-]\\pOH = -log[1.65]\\pOH = -0.217[/tex]

At 25°C, the pOH of NaOH is -0.217, let's calculate the pH

[tex]pOH+pH=14\\-0.217+pH=14\\pH=14-(-0.217)\\pH=14.217[/tex]

From the calculations above, the pH and pOH of 6.6g of NaOH in 100mL solution are -0.217 and 14.217 respectively.

Learn more on molarity, pH and pOH of a solution here;

https://brainly.com/question/13557815

https://brainly.com/question/13423425

The number of moles of NaOH in the solution is 0.165 mol and its molarity is 1.65 M. The pOH of the solution is -0.217 and the pH is 14.217.

The mass of Sodium Hydroxide present in the solution is 6.6 g. The molar mass of Sodium Hydroxide (NaOH) is approximately 40 g/mol. Therefore, the number of moles present in the solution can be calculated by dividing the mass by the molar mass. So, the number of moles = 6.6/40 = 0.165 mol.

The volume of the solution is 100 ml or 0.1 L. The molarity of the solution can be found by dividing the number of moles by the volume in liters which gives us, Molarity = 0.165/0.1 = 1.65 M.

Because NaOH is a strong base, in water it dissociates completely to form hydroxide ions (OH-). Hence, the molarity of OH- is the same as the molarity of NaOH i.e., 1.65 M. In order to find the pOH we can use the formula -log[OH-], hence the pOH= -log(1.65) = -0.217.

The relationship between pH and pOH at 25 °C is given by the expression, pH + pOH = 14. Therefore, the pH can be calculated as follows, pH = 14 - pOH = 14 - (-0.217) = 14.217.

https://brainly.com/question/36153729

#SPJ3

Final answer:

The percent ionization of HA in a 0.10 M solution, with a Ka of 3.6×10⁻⁷, is approximately 0.19%.

Explanation:

The percent ionization of a weak acid can be calculated using its Ka value and the initial concentration of the acid. For HA, with a Ka of 3.6×10⁻⁷ and an initial concentration of 0.10 M, the percent ionization is determined as follows: First, set up the reaction as HA → H+ + A-. The equilibrium expression is Ka = [H+][A-] / [HA]. Assuming x is the amount ionized, we have Ka = x² / (0.10 - x). Solving this equation for x, we approximate that x is small compared to the initial concentration, so 0.10 - x is nearly 0.10. Therefore, Ka ≈ x² / 0.10 M. Solving for x gives x = sqrt(Ka × 0.10 M), and the percent ionization = (x / 0.10 M) × 100%. Substituting in the given values, we get percent ionization ≈ sqrt(3.6×10⁻⁷ × 0.10 M) / 0.10 M × 100% = 0.19%.

The solution is basic due to its high hydroxide ion concentration. The pOH obtained should be around -0.748.

To determine the pOH of a 2.80 M Ba(OH)₂ solution, calculate the hydroxide ion concentration and find its negative logarithm.

To find the pOH of a 2.80 M Ba(OH)₂ solution, follow these steps:

Determine the concentration of OH- ions produced by Ba(OH)₂. Since each Ba(OH)₂ molecule dissociates into one Ba²⁺ ion and two OH⁻ ions, the concentration of OH⁻ ions will be twice the concentration of Ba(OH)₂.

Therefore, [OH⁻] = 2 × 2.80 M = 5.60 M.

Calculate the pOH by taking the negative logarithm of the hydroxide ion concentration: pOH = -log(5.60).

Using the logarithm function, pOH = -log(5.60) ≈ -0.748.

However, the pOH value less than zero is not realistic, indicating very high basicity. Generally, the practical range considers the pOH to lie between 0 and 14.

Since a high [OH⁻] concentration results in a very low pOH, we can conclude that the solution is basic.

The pH of a sodium acetate (NaC2H3O2) solution can be calculated using the equilibrium expression for the base hydrolysis reaction of the acetate ion.

The pH of a sodium acetate (NaC2H3O2) solution can be calculated using the equilibrium expression for the base hydrolysis reaction of the acetate ion:

CH3CO2⁻ + H2O ⇌ CH3CO2H + OH⁻

This reaction is the reverse of the ionization reaction for acetic acid. The Kb value for the acetate ion is calculated as Kw/Ka, where Ka is the acid dissociation constant for acetic acid, given as 1.8 × 10⁻⁵ at 25.0 °C. To find the pH, we need to calculate the concentration of hydroxide ions (OH⁻) in the solution.

The hydroxide ion concentration can be calculated using the equation [OH⁻] = √(Kb × [CH3CO2H]). Given that the concentration of acetate ion [CH3CO2⁻] is equal to the initial concentration of sodium acetate, and assuming complete dissociation of sodium acetate in water, the concentration of acetic acid [CH3CO2H] will be equal to the initial concentration of sodium acetate. Therefore, [OH⁻] = √(Kb × [NaC2H3O2]). Finally, the pH of the solution can be calculated as -log[OH⁻].

https://brainly.com/question/32252831

#SPJ6

The correct formula for the compound formed between the iron(III) ion and the oxide ion is Fe2O3.

The compound formed between the iron(III) ion and the oxide ion is called iron(III) oxide. The formula for this compound can be determined by balancing the charges of the ions. The iron(III) ion has a charge of +3 and the oxide ion has a charge of -2. To balance the charges, we need two iron(III) ions for every three oxide ions. Therefore, the correct formula for the compound is Fe2O3.

https://brainly.com/question/32503849

#SPJ6

Answer: The given statement is false.

Explanation:

Oxidizing agents are defined as the agents that helps in the oxidation of other substance and itself gets reduced. These agents undergo reduction reactions and reduction reaction is the reaction in which an atom gains electrons.

Reducing agents are defined as the agents that helps in the reduction of other substance and itself gets oxidized. These agents undergo oxidation reactions and oxidation reaction is the reaction in which an atom looses electrons.

So, a strong oxidizing agent will gain electrons easily

Hence, the given statement is false.

Final answer:

Carbon monoxide and carbon dioxide both contain carbon and oxygen atoms. CO has one oxygen atom with a triple bond to carbon, whereas CO2 has two oxygen atoms, each with a double bond to carbon. CO is toxic while CO2 contributes to global climate change.

Explanation:

The chemical composition of carbon monoxide (CO) is similar to that of carbon dioxide (CO2) in that both compounds consist of carbon (C) and oxygen (O) atoms. However, the key difference is in the number of oxygen atoms. CO has one oxygen atom, while CO2 has two oxygen atoms. The molecular structures of both compounds reveal these differences. CO has a triple bond between the carbon and the oxygen atom, which includes two covalent bonds and one dative covalent bond. In contrast, the CO2 molecule has a linear structure with a double bond to each oxygen atom, forming an O=C=O configuration.

In terms of Lewis structures, CO's Lewis structure consists of a carbon atom triple-bonded to an oxygen atom with a lone pair on the oxygen, while CO2's Lewis structure displays the carbon atom with two double bonds, each connected to an oxygen atom with two sets of lone pairs.

Both CO and CO2 are important in context as they have significant environmental and health impacts. Carbon monoxide is a toxic gas with the potential to bind to hemoglobin, making it a competitive inhibitor for oxygen transport in the bloodstream. Carbon dioxide, while non-toxic at normal concentrations, is a significant greenhouse gas contributing to global climate change.

The specific heat capacity of the metal is calculated using the formula q=mcΔT. With the provided information (1495 J of heat, 337 g of metal, temperature change from 55.0°C to 66.0°C), the specific heat capacity is found to be 0.399 J/g°C.

To calculate the specific heat capacity of the metal, we can use the formula q = mcΔT, where q is the heat absorbed or released (in joules), m is the mass of the substance (in grams), c is the specific heat capacity (in J/g°C), and ΔT is the change in temperature (in °C).

Given that 1495 J of heat is needed to raise the temperature of a 337 g sample of a metal from 55.0°C to 66.0°C, we have:

ΔT = 66.0°C - 55.0°C = 11.0°C
q = 1495 J
m = 337 g

Plugging these values into the formula, we can solve for c:

1495 J = (337 g)(c)(11.0°C)
c = 1495 J / (337 g × 11.0°C)
c = 0.399 J/g°C

Thus, the specific heat capacity of the metal is 0.399 J/g°C.

c₀(CH₃NH₂) = 0,57 M.

c(CH₃NH₃⁺) = c(OH⁻) = x.

c(NH₂OH) = 0,57 M - x.

Kb = c(CH₃NH₃⁺) · c(OH⁻) / c(CH₃NH₂).

0,00044 = x² /  (0,57 M - x). 

Solve quadratic equation: x = c(OH⁻) = 0,0156 mol/L.

pOH = -log(0,0156 mol/L.) = 1,80.

pH = 14 - 1,80 = 12,2.

The pH value is 12.2

Given:

Question:

The pH value of methylamine

The Process:

Methylamine  is a weak base. When a weak base reacts with water, it produces its conjugate acid and hydroxide ions.

[tex]\boxed{ \ CH_3NH_2_{(aq)} + H_2O_{(l)} \rightleftharpoons CH_3NH_3_{(aq)} + OH^-_{(aq)} \ }[/tex]

Let's prepare the equilibrium system to calculate the concentration of hydroxide ions. In chemical equilibrium, the liquid phase has no effect.

[tex]\boxed{ \ K_b = \frac{ [CH_3NH_3] [OH^-] }{ [CH_3NH_2] } \ }[/tex]

Here Kb acts as Kc or equilibrium constant.

[tex]\boxed{ \ 4.4 \times 10^{-4} = \frac{ x \cdot x }{ 0.57 - x } \ }[/tex]

[tex]\boxed{ \ 4.4 \times 10^{-4} = \frac{x^2}{0.57 - x} \ }[/tex]

[tex]\boxed{ \ 2.508 \times 10^{-4} - 4.4 \times 10^{-4}x = x^2 \ }[/tex]

[tex]\boxed{ \ x^2 + 4.4 \times 10^{-4}x - 2.508 \times 10^{-4} = 0 \ }[/tex]

The solution is obtained through the formula of quadratic equations, i.e., [tex]\boxed{ \ x = [OH^-] = 0.0156 \ M \ }[/tex]

Next, we calculated the pOH value followed by the pH value.

[tex]\boxed{ \ pOH = -log [OH^-] \ }[/tex]

[tex]\boxed{ \ pOH = -log [0.0156] \ }[/tex]

We get [tex]\boxed{ \ pOH = 1.81 \ }[/tex]

[tex]\boxed{ \ pH + pOH = 14 \ }[/tex]

[tex]\boxed{ \ pH = 14 - pOH \ }[/tex]

[tex]\boxed{ \ pH = 14 - 1.81 \ }[/tex]

Thus [tex]\boxed{\boxed{ \ pH = 12.19 \ rounded \ to \ 12.2 \ }}[/tex]

- - - - - - -

Quick Steps

0.57 M methylamine (CH₃NH₂)

[tex]K_b = 4.4 \times 10^{-4}[/tex]

We immediately use the formula to calculate the concentration of hydroxide ions for weak bases.

[tex]\boxed{\boxed{ \ [OH^-] = \sqrt{K_b \times base \ concentration} \ }}[/tex]

[tex]\boxed{ \ [OH^-] = \sqrt{4.4 \times 10^{-4} \times 0.57} \ }[/tex]

[tex]\boxed{ \ [OH^-] = 0.0158 \ M \ }[/tex]

Like the steps above, we calculated the pOH value followed by the pH value.

[tex]\boxed{ \ pOH = -log [OH^-] \ }[/tex]

[tex]\boxed{ \ pOH = -log [0.0158] \ }[/tex]

[tex]\boxed{ \ pOH = 1.8 \ }[/tex]

[tex]\boxed{ \ pH = 14 - pOH \ }[/tex]

[tex]\boxed{ \ pH = 14 - 1.8 \ }[/tex]

Thus [tex]\boxed{\boxed{ \ pH = 12.2 \ }}[/tex]

Keywords: determine, the pH, 0.57 M, methylamine, CH₃NH₂, CH₃NH₃, OH⁻, Kb, Kc, equilibrium constant, weak base

Answer : The number of moles of nitrogen present in nitrous oxide is 3.32 moles.

Explanation : Given,

Mass of nitrous oxide = 73.0 g

Molar mass of nitrous oxide = 44 g/mole

Now we have to calculate the moles of [tex]N_2O[/tex].

Formula used :

[tex]\text{ Moles of }N_2O=\frac{\text{ Mass of }N_2O}{\text{ Molar mass of }N_2O}[/tex]

[tex]\text{ Moles of }N_2O=\frac{73.0g}{44g/mole}=1.66moles[/tex]

Now we have to calculate the moles of nitrogen in nitrous oxide.

In [tex]N_2O[/tex] molecule, there are 2 moles of nitrogen atoms and 1 mole of oxygen atom.

As, 1 mole of [tex]N_2O[/tex] contains 2 moles of nitrogen

So, 1.66 moles of [tex]N_2O[/tex] contains [tex]1.66\times 2=3.32[/tex] moles of nitrogen.

Therefore, the number of moles of nitrogen present in nitrous oxide is 3.32 moles.

The value of [H+] for the solution with [OH−] = 4.0×10−8 is calculated using the formula [H+] = Kw / [OH−], yielding a hydronium ion concentration of 2.5×10−7 M.

To find the value of the hydronium ion concentration ([H+]) for a solution with a given hydroxide ion concentration ([OH−]), you can use the ion product constant for water (Kw), which is always 1.0 × 10−14 M2 at 25°C. The formula is [H+] = Kw / [OH−]. When the [OH−] is 4.0 × 10−8, we can calculate the [H+] as follows:

Therefore, the hydronium ion concentration of the solution is 2.5 × 10−7 M.

https://brainly.com/question/13387755

#SPJ3