One hundred fifty joules of heat are removed from a heat reservoir at a temperature of 150 K. What is the entropy change of the reservoir (in J/K)?

Answer:

yes it is

Explanation:

because there are some acid which really harm skin.

Explanation:

1. Nucleus is a memberane bound organelle that contains cell,s chromosomes.

Kc require (aqueous/gaseous) products to be on the numerator and (aqueous/gaseous) reactants to be in the denominator, whereas Ksp will require (aqueous) products to be on the numerator and (aqueous) reactants to be in the denominator. Both require products on top and reactants in the bottom.

K = [products] / [reactants]

Kc is used when a reaction reaches dynamic equilibrium, whereas Ksp is used when an insoluble ionic solid dissolved by a tiny amount in a solution, as well as in determining whether or not a precipitate will form.

Kc can be used to measure equilibrium concentration for all reactions, whereas Ksp is limited to only ionic compounds' solubility.

The decomposition of HI (g) will required the use of Kc since the species are all gaseous, and gases cannot be ionic.

Answer:

endet nach selam nw

4gh7

Answer:

The median is also the number that is halfway into the set. To find the median, the data should be arranged in order from least to greatest. If there is an even number of items in the data set, then the median is found by taking the mean (average) of the two middlemost numbers.

I hope it helps

Answer:

Explanation:

a) 4-nitrobenzoic acid         pKa= 3.41

    benzoic acid                   pKa= 4.19

    4-chlorobenzoic acid     pKa= 3.98

b) benzoic acid                    pKa= 4.19    

   cyclohexanol                   pKa= 18.0

   phenol                              pKa= 9.95

c) 4-Nitrobenzoic acid             pKa= 3.41

   4-nitrophenol                       pKa= 7.15

   4-nitrophenylacetic acid     pKa= 3.85

Answer:

ΔHrxn = -635.14kJ/mol

Explanation:

We can make algebraic operations of reactions until obtain the desire reaction and, ΔH of the reaction must be operated in the same way to obtain the ΔH of the desire reaction (Hess's law). Using the reactions:

(1)Ca(s) + 2 H+(aq) → Ca2+(aq) + H2(g) ΔH = 1925.9 kJ/mol

(2) 2H2(g) + O2 g) → 2 H2O(l) ΔH = −571.68 kJ/mole

(3) CaO(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) ΔH = 2275.2 kJ/mole

Reaction (1) - (3) produce:

Ca(s) + H2O(l) → H2(g) + CaO(s)

ΔH = 1925.9kJ/mol - 2275.2kJ/mol = -349.3kJ/mol

Now this reaction + 1/2(2):

Ca(s) + ½ O2(g) → CaO(s)

ΔH = -349.3kJ/mol + 1/2 (-571.68kJ/mol)

Answer:

Atomic mass = 127.198 amu

Explanation:

The average atomic mass is obtained by summing the masses of the isotopes each multiplied by its abundance.

Atomic mass = (97.62 * 0.0825) + (109.3 * 0.2671) + (138.3 * 0.6504)

Atomic mass = 8.05365 +  29.19403 + 89.95032

Atomic mass = 127.198 amu

Answer:

[tex]V_2=1.17L[/tex]

Explanation:

Hello,

In this case, for dilution processes, we must remember that the amount of solute remains the same, therefore, we can write:

[tex]V_1C_1=V_2C_2[/tex]

Whereas V accounts for volume and C for concentration that in this case is %(m/v). In such a way, the final volume V2 turns out:

[tex]V_2=\frac{V_1C_1}{C_2}= \frac{0.551L*50.0\%}{23.5\%}\\ \\V_2=1.17L[/tex]

Best regards.

Answer:

The pressure law states that for a constant volume of gas in a sealed container the temperature of the gas is directly proportional to its pressure. ... This means that they have more collisions with each other and the sides of the container and hence the pressure is increased.

The temperature of the gas is directly proportional to its pressure.

The ideal gas equation can be given as:

PV = nRT

where P =pressure, V = volume, n = moles of gas, R = rydberg constant, and T = temperature.

The pressure law states that for a constant volume of gas in a container, the temperature of the gas is directly proportional to its pressure which means there are more collisions with each other and the sides of the container and hence the pressure is increased.

Hence, the temperature of the gas is directly proportional to its pressure.

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Answer:

Explanation:

We shall apply Arrhenius equation which is given below .

[tex]ln\frac{k_2}{k_1} = \frac{E_a}{R} [\frac{1}{T_1} -\frac{1}{T_2} ][/tex]

K₂ and K₁ are rate constant at temperature T₂ and T₁ , Ea is activation energy .

Putting the given values

[tex]ln\frac{5.9}{1} = \frac{39500}{8.3} [\frac{1}{298} -\frac{1}{T_2} ][/tex]

[tex].000373= [\frac{1}{298} -\frac{1}{T_2} ][/tex]

T₂ = 335.27 K

= 62.27 °C

The higher temperature is 62.27°C.

Given that the activation energy of the reaction is:

Eₐ =  39.5 kJ/mol

initial temperature T₁ = 25°C = 298K

Let the final temperature be T₂

The rate constant at temperature T₁ be K₁, and at a temperature T₂ be K₂.

According to the question: K₂/K₁ = 5.9

Now, applying the Arrhenius equation we get:

[tex]\ln\frac{K_2}{K1}=\frac{E_a}{R}[\frac{1}{T_1} -\frac{1}{T_2}]\\\\\ln(5.9)= \frac{39.5}{8.3}[\frac{1}{298} -\frac{1}{T_2}]\\\\0.000373=\frac{1}{298} -\frac{1}{T_2}[/tex]

T₂ = 335.27K

T₂ =  335.27 -273

T₂ = 62.27°C

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Answer:

THE FINAL TEMPERATURE OF WATER IS -4.117 °C

Explanation:

Mass of the aluminium = 50 g

c = 0.88 J/g C

Initial temperature of aluminium = 225 °C

Volume of water = 100 ml

Density of water = 1 g/ml

Mass of water = density * volume of water

Mass of water = 1 * 100 = 100 g of water

Initial temperature of water = 20 C

It is worthy to note that the heat of a system is constant and conserved as no heat is lost or gained by a closed system,

So therefore,

heat lost by aluminium = heat gained by water

H = mass * specific heat capacity * temeprature change

So:

m c ( T2- T1) = m c (T2-T1)

50 * 0.88 * ( T2 - 225) = 100 * 4.18 *( T2 - 20)

44 ( T2 - 225 ) = 418 ( T2 - 20)

44 T2 - 9900 = 418 T2 - 8360

-9900 + 8360 = 418 T2 - 44 T2

-1540 = 374 T2

T2 = - 4.117

So therefore the final temperature of water is -4.117 °C

Answer:

Half-life at 629K = 252.4min

Explanation:

Using Arrhenius equation:

[tex]ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})[/tex]

And as Half-life in a first order reaction is:

[tex]t_{1/2}=\frac{ln2}{K}[/tex]

We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:

[tex]58.0min=\frac{ln2}{K}[/tex]

K = 0.01195min⁻¹ = K₁

[tex]ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})[/tex]

[tex]ln\frac{0.01195min^{-1}}{K_2} =1.47[/tex]

[tex]\frac{0.01195min^{-1}}{K_2} =4.35[/tex]

K₂ = 2.75x10⁻³ min⁻¹

And, replacing again in Half-life expression:

[tex]t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}[/tex]

The half-life of the first-order reaction of ethylene oxide decomposition at 629 K is 251.1 min when the half-life at 652 K is 58.0 min and the activation energy is 218 kJ/mol.   

The activation energy of a reaction is related to its rate constant as follows:  

[tex] k = Ae^{-\frac{E_{a}}{RT}} [/tex]   (1)

Where:

We can find the rate constant of the first-order reaction at 652 K with the half-life as follows:

[tex]k_{652} = \frac{ln(2)}{t_{1/2}_{(652)}}[/tex]   (2)

Where [tex]t_{1/2}_{(652)}[/tex] is the half-life at 652 K= 58.0 min

Hence, the rate constant at 652 K is:                            

[tex] k_{652} = \frac{ln(2)}{58.0 min} = 0.012 min^{-1} [/tex]

Now, from equation (1) we can find the pre-exponential factor (A):

[tex]A = \frac{k_{652}}{e^{(-\frac{E_{a}}{RT_{1}})}} = \frac{0.012 \:min{-1}}{e^{(-\frac{218\cdot 10^{3} \:J/mol}{8.314 \:J/(K*mol)*652 \:K})}} = 3.51 \cdot 10^{15} min^{-1}[/tex]  

With the pre-exponential factor we can calculate the rate constant at 629 K (eq 1):

[tex]k_{629} = 3.51 \cdot 10^{15} min^{-1}*e^{(-\frac{218 \cdot 10^{3} J/mol}{8.314 J/(K*mol)*629 K})} = 2.76 \cdot 10^{-3} min^{-1}[/tex]

Finally, the half-life at 629 K is (eq 2):

[tex] t_{1/2}_{629} = \frac{ln(2)}{2.76\cdot 10^{-3} min^{-1}} = 251.1 min [/tex]

Therefore, the half-life at 629 K is 251.1 min.

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I hope it helps you!

Answer:

Explanation:

50 HP = 50 x 746 watt

= 37300 watt

= 37300 J /s

heat produced in one hour = 60 x 60 x 37300 J

= 134280 x 10³ J

latent heat of vaporization = 2260 x 10³ J / kg .

for evaporation of water of mass 1 kg , 2260 x 10³ J of heat is required .

kg of water being evaporated by boiler per hour

= 134280 x 10³ / 2260 x 10³

= 59.41 kg

rate of production of steam

= 59.41 kg / hr .

Answer:

Atomic mass is calculated by adding protons and neutrons.

Explanation:

Atomic mass is the sum of protons and neutrons in an atomic nucleus. For example, the element Oxygen has 8 protons (derived from the atomic number) and 8 neutrons (derived from subtracting the amount of protons from the atomic mass).

We can craft an equation to show the relationship between these variables.

M - N = P, where M = Mass, N = Neutrons, and P = Protons

This equation can be rearranged to show the relationship between the neutrons and protons leading to the atomic mass. Simply add N to both sides of the equation.

M = N + P

This shows that atomic mass is equivalent to the sum of protons and neutrons in an atom's nucleus.

Van der Waals forces hold molecules together by inducing temporary dipoles that attract each other. That is option C

Van Der Waals forces are example of those intermolecular forces which are weaker than ionic and covalent bonds that exists between molecules.

Van Der Waals forces was postulated by a Dutch physicist known as Van Der Waals. He postulated the existence of weak, short-range forces of attraction, which are independent of normal bonding forces, between non-polar molecules. He came to this conclusion after studying the  of real gases at low temperatures and high pressures that:

This is how weak Van Der Waals forces are set up. Therefore, option C is CORRECT

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Lakes, oceans, glaciers, clouds, etc. It categorizes all forms of water on earth.

hydro = water

Answer:

Lakes, streams, ground water, polar ice caps, glaciers, water vapor, and rivers!

Explanation:

The hydrosphere is made up of all the water on Earth. So anything that is water, like oceans, can be found in the hydrosphere:)

Answer:

See figure 1

Explanation:

We have to remember that in the isomer structures we have to change the structure but we have to maintain the same formula, in this case [tex]C_4H_1_1N[/tex].

In the formula, we have 1 nitrogen atom. Therefore we will have as a main functional group the amine group.

In the amines, we have different types of amines. Depending on the number of carbons bonded to the "N" atom. In the primary amines, we have only 1 C-H. In the secondary amines, we have two C-N bonds and in the tertiary amines, we have three C-N bonds.

With this in mind, we can have:

-) Primary amines:

1) n-butyl amine

2) sec-butyl amine including 2 optical isomers

3) isobutyl amine

4) tert-butyl amine

-) Secondary amines:

5) N-methyl n-propyl amine

6) N-methyl isopropyl amine

7) N, N-diethyl amine

-) Tertiary amines:

8) N-ethyl N, N-dimethyl amine

See figure 1

I hope it helps!

Answer:

1. This reaction is: B. Endothermic.

2. When the temperature is increased the equilibrium constant, K: A. Increases.

3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.

Explanation:

Hello,

In this case, considering the images, we can state that the red color at high temperature is due to the presence of nitrogen dioxide (product) and the lower coloring is due to the presence of dinitrogen tetroxide (reactant) at low temperature.

With the aforementioned, we can conclude that the chemical reaction:

[tex]N_2O_4(g) \rightleftharpoons 2 NO_2(g)[/tex]

Is endothermic since high temperatures favor the formation of the product and the low temperatures favor the consumption of the the reactant. thereby:

1. This reaction is: B. Endothermic.

2. When the temperature is increased the equilibrium constant, K: A. Increases. In this particular case, since the dinitrogen tetroxide has 1 molecule and nitrogen dioxide two molecules in the chemical reaction, the entropy change should be positive, therefore, by increasing the T, the Gibbs free energy of reaction becomes more negative:

[tex]G=H-TS[/tex]

As Gibbs free energy becomes more negative, the equilibrium constant becomes bigger given their relationship:

[tex]K=exp(-\frac{\Delta G}{RT} )[/tex]

3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.

Regards.

Izopropanol doesn't form by Hydration of propene.

Acetone, propanone, or dimethyl ketone, exists as an organic compound with the formula (CH₃)₂CO. It stands for the simplest and smallest ketone. Reduction of ketones can be accomplished by hydrogenation (or) utilizing Grignard Reagent. Let us now decrease propanone to propan-2-ol by hydrogenation. The reduction of propane in the existence of catalyst platinum along with hydrogen provides the product propan-2-ol.

The correct answer is option A.

Izopropanol doesn't form by Hydration of propene.

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Answer:

water molecules

Explanation:

Redox reactions are carried out under acidic or basic conditions as the case may be.

If the reaction is carried out in an acid medium, then we must balance the hydrogen ions on the lefthand side of the reaction equation with water molecules on the righthand side of the reaction equation.

For instance, the equation for reduction of MnO4^- under acidic condition is shown below;

MnO4^-(aq) + 5e + 8H^+(aq) --------> Mn^2+(aq) + 4H2O(l)

Answer:

AgI I the limiting reactant.

Explanation:

The balanced equation for the reaction is given below:

2AgI + HgI2 → Ag2HgI4

Next, we shall determine masses AgI and HgI2 that reacted from the balanced equation.

This is illustrated below:

Molar mass of Agl = 108 + 127 = 235 g/mol

Mass of AgI from the balanced equation = 2 x 235 = 470 g

Molar mass of HgI2 = 201 + (2x127) = 455 g/mol

Mass of HgI2 from the balanced equation = 1 x 455 = 455 g

From the balanced equation above,

470 g of AgI reacted with 455 g of HgI2.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above,

470 g of AgI reacted with 455 g of HgI2.

Therefore, 2 g of AgI will react with

= (2 x 455)/470 = 1.94 g of HgI2.

From the calculations made above, we can see that only 1.94 g out of 3.50 g of HgI2 given is needed to react completely with 2 g of AgI.

Therefore, AgI I the limiting reactant.

Answer:

А It is hard because its ions are held together by strong electrostatic attractions.

B It is malleable because its atoms can easily move past one another without disrupting the bonding.  

Explanation:

These are correct explanations of the properties of magnesium.

C is wrong. Mg is a good conductor of electricity and it has metallic bonds.

D is wrong. Mg has no molecules. It has no intermolecular forces.

Answer:

.217, .311, and .472, respectively.

Explanation:

The total number of moles of gas is 3.50 + 5.00 + 7.60 = 16.10 (to preserve significant digits).

X of helium=3.50/16.10 = .217

X of krypton=5.00/16.10 = .311

X of neon=7.60/16.10 = .472

Explanation:

Introducción:

La designación de los compuestos orgánicos puede hacerse utulizando alguno de los

siguientes sistemas:

a) Mediante nombres triviales o comunes, que expresen alguna propiedad característica

(sabor, color, acción fisiológica, etc.) o hagan referencia a la materia de la cual se extrajo el

compuesto.

b) Mediante nombres racionales que proporcionen una idea de su constitucion química y

destaquen sus analogías estructurales.

La necesidad de una nomenclatura sistemática, que expresara en forma clara, conforme

a normas precisas, el nombre y la estructura de los compuestos orgánicos, ha sido motivo de

preocupación permanente y observada a través de los numerosos congresos internacionales

que, al efecto, se han realizado en diversas oportunidades.

Las bases del actual sistema de nomenclatura fueron establecidas por una comisión que

se reunió en Ginebra en 1892. Posteriormente, fue perfeccionado y ampliado por el Comité de

Nomenclatura de la Unión Internacional de Química Pura y Aplicada, por lo que se conoce

como sistema I.U.P.A.C. (International Union of Pure and Applied Chemistry).

En las reglas aprobadas se ha tratado de introducir los menores cambios posibles a la

terminología universalmente adoptada. El sistema tiene la necesaria flexibilidad como para

adaptar la forma precisa de las palabras, de las terminaciones, etc. a las características de

distintos idiomas.

El nombre de los hidrocarburos consta de tres partes: a) la raíz, que indica el esqueleto

carbonado; b) la terminación o sufijo, que indica el grado de saturación, y c) el prefijo que

diferencia las distintas estructuras isoméricas (distintas estructuras construidas con exactamente

los mismos átomos).

Ej.: CH3-CH2-CH2-CH2-CH3 pentano (tambien llamado n-pentano)

penta: raíz que señala el número de átomos de carbono que componen la cadena principal

del compuesto.

-ano: sufijo que indica que el hidrocarburo es saturado

Answer:

Each chlorine atom shares a pair of electrons with the sulfur atom. ... Each chlorine atom shares all its valence electrons with the sulfur atom

Answer:

Explanation:

For pH of a buffer solution , the formula is

pH = pKa + log [ Base ] / [ conjugate acid ]

=   pKa + log [ NH₃ ] / [ NH₄⁺ ]

Ka = Kw / Kb

Kb for NH₄OH = 1.8 x 10⁻⁵

Ka = 10⁻¹⁴ / 1.8 x 10⁻⁵

= 5.6 x 10⁻¹⁰

pH = - log ( 5.6 x 10⁻¹⁰ ) + log 0.2 / 0.2

= 10 - log 5.6

= 9.25

Effect of addition of HCl

H⁺ of HCl will react with NH₃ to produce NH₄⁺

25 mL of .1 HCl = 2.5 mM of HCl

25 mL of .1 NH₄⁺ = 2.5 mM of NH₄⁺

65 mL of .2 M NH₃ = 13 mM of NH₃

65 mL of .2 M NH₄⁺ = 13 mM of NH₄⁺

NH₃ + H⁺ = NH₄⁺

NH₄⁺ formed = 2.5 + 13  mM

15.5 mM of NH₄⁺

NH₃ = 13 mM

Concentration of NH₃ = 13 / 90

Concentration of NH₄⁺ = 15.5 / 90

pH of final buffer mixture

= 9.6 + log 13 / 15.5

= 9.25 - .076

= 9.174

The pH value  is mathematically given as

pH= -6.332.

Question Parameters:

the pH of a 0.20 M NH3/0.20 M NH4Cl buffer

the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.

Generally, the equation for the Chemical Reaction  is mathematically given as

HCl + NH3 --> NH4^+ + Cl^-

Therefore

pH= pka + log(13/14).

pH= -6.3 + log 0.93.

pH= -6.3+ (-0.032).

pH= -6.332.

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Answer:

The total photons required for this radiation = 5.1938 × 10²⁸ photons

Explanation:

Given that:

A microwave oven heats by radiating food with microwave radiation, which is absorbed by the food and converted to heat.

If the radiation wavelength is 12.5 cm,

density of water = 1g/cm³

volume of the container = 0.250 L = 250 cm³

density = mass/volume

mass of the water = density × volume

mass of the water =  1g/cm³  × 250 cm³

mass of the water = 250 g

specific heat capacity of water = 4.182 J/g°C

The change in temperature was from 20.0° C to 99° C

ΔT =( 99 -20.0)° C

ΔT = 79.0° C

The heat absorbed in the process is calculated by using the formula,

q = mcΔT

q = 250 g × 4.182 J/g°C ×  79.0° C

q = 82594.5 Joules

Recall that the radiation wavelength λ = 12.5 cm = 0.125 m

The amount of energy of one photon of the radiation wavelength is determined by using the formula:

E = hv  

since v = c/λ

E = hc/λ

where;

h = Planck's constant = 6.626 × 10⁻³⁴ J.s

c = velocity of light = 3.0 × 10⁸ m/s

∴

E = (6.626 × 10⁻³⁴ J.s × 3.0 × 10⁸ m/s)/ 0.125 m

E = 1.59024⁻²⁴ Joules

The total photons required for this radiation = total heat energy/energy of radiation

The total photons required for this radiation = 82594.5 Joules/1.59024⁻²⁴ Joules

The total photons required for this radiation = 5.1938 × 10²⁸ photons