An oxygen molecule is adsorbed on a patch of surface (see sketch at right). This patch is known to contain adsorption sites. The molecule has enough energy to move from site to site, so it could be on any one of them. Suppose part of the surface becomes inaccessible, so that only adsorption sites are now available for the molecule. Calculate the change in entropy. Round your answer to significant digits, and be sure it has the correct unit symbol.

Answer:

D.

the father parent shares his genetic information with the mother during sexual intercourse( the sperm and egg)

Units of the SI System

There are seven base units in the SI system:

the kilogram (kg), for mass

the second (s), for time

the kelvin (K), for temperature

the ampere (A), for electric current

the mole (mol), for the amount of a substance

the candela (cd), for luminous intensity

the meter (m), for distance

Answer: Ammonia Ice

Explanation: Like Jupiter, Saturn boasts layers of clouds. The upper layers of clouds are made up of ammonia ice. Traveling toward the core, clouds of water ice form, with bands of ammonium hydrosulfide ice intermixed. The lower layers of Saturn see higher temperatures and pressures.

Answer:

It is 98 degrees C.

Explanation:

Answer: I believe it's C

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Can you please make my answer brainly

Answer:

the pasta will soften

Explanation:

the pasta will boil and will soften up as a result of it being regular pasta

Answer:

(8.3×40)+(8.3×71)

921.3grames

Answer:

[tex]\Large \boxed{\sf 921.3 \ g}[/tex]

Explanation:

Use formula

[tex]\displaystyle moles=\frac{mass}{M_r}[/tex]

[tex]M_r=40+35.5 \times 2 = 111[/tex]

Solve for mass

[tex]mass=8.3 \times 111=921.3[/tex]

the answer is 375 joules

Answer:

120.366 g/mol

Explanation:

Answer:

Electrostatic smoke precipitators work by forcing dirty flue gas (the gas escaping from a smokestack) past two electrodes (electrical terminals), which take the form of metal wires, bars, or plates inside a pipe or smokestack

Explanation:

This question is incomplete, the complete question is;

The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.20 m² at 500C.

Assume a diffusion coefficient of 1.0 × 10⁻⁸ m²/s, that the concentrations at the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained

Answer:

the number of kilograms of hydrogen that pass per hour through is 2.592 × 10⁻³ kg

Explanation:

Given the data in the question;

thickness of sheet t = 5 mm

Area A = 0.20 m²

Temperature T = 500°C

diffusion coefficient ∝ = 1.0 × 10⁻⁸ m²/s

concentration high pressure side C[tex]_A[/tex] = 2.4

concentration low pressure side C[tex]_B[/tex] = 0.6 kg

from the question, we calculate the concentration gradient

dc/dx = (C[tex]_A[/tex] -  C[tex]_B[/tex])/dt

so we substitute

dc/dx = (2.4 - 0.6) / ( - 5 × 10⁻³ )

dc/dx = -360

now, mass of hydrogen per hour that diffuse through a pd sheet

M = -∝AT(dc/dx)

where time t is 1 hour ( 3600 sec )

we substitute

M = -(1.0 × 10⁻⁸) × 0.20 × 3600 (-360)

M = 0.002592

M = 2.592 × 10⁻³ kg per one hour

Therefore, the number of kilograms of hydrogen that pass per hour through is 2.592 × 10⁻³ kg

The number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.20 m² at 500°C is 2.592 × 10⁻³ kg.

Purification: It refers to the process of removing impurities or contaminants from a substance in order to make it cleaner, purer, or more refined. It also involves the elimination or reduction of unwanted substances or components that may be present in the original material.

According to the question, given data is:

Thickness of sheet t = 5 mm

Area A = 0.20 m²

Temperature T = 500°C

Diffusion coefficient ∝ = 1.0 × 10⁻⁸ m²/s

Concentration high pressure side [tex]C_A[/tex] = 2.4

Concentration low pressure side[tex]C_B[/tex]= 0.6 kg

from the question, we calculate the concentration gradient,

[tex]dc/dx = (C _A- C_B)/dt[/tex]

so, we substitute the values,

[tex]dc/dx[/tex] = (2.4 - 0.6)/ ( - 5 × 10⁻³)

[tex]dc/dx = -360[/tex]

Now, mass of hydrogen per hour that diffuse through a Palladium sheet,

M = -∝[tex]AT(dc/dx)[/tex]

where time t is 1 hour (3600 sec)

We substitute,

M = -(1.0 × 10⁻⁸) × 0.20 × 3600 (-360)

M = 0.002592

M = 2.592 × 10⁻³ kg per hour.

Therefore, the number of kilograms of hydrogen that pass per hour through a palladium sheet is approximately 2.592 × 10⁻³ kg.

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This question is incomplete, the complete question is;

The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.20 m² at 500C.

Assume a diffusion coefficient of 1.0 × 10⁻⁸ m²/s, that the concentrations at the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.

The given question is incomplete. The complete question is:

Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen. A plant has 88.0 g of carbon dioxide and 64.0 g of water available for photosynthesis. Determine the mass of glucose (C6H1206) produced

Answer: 60.0 g of glucose

Explanation:

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]

a) moles of [tex]CO_2[/tex]

[tex]\text{Number of moles}=\frac{88.0g}{44g/mol}=2.0moles[/tex]

b) moles of [tex]H_2O[/tex]

[tex]\text{Number of moles}=\frac{64.0g}{18g/mol}=3.5moles[/tex]

[tex]6CO_2+6H_2O\rightarrow C_{6}H_{12}O_6+6O_2[/tex]

According to stoichiometry :

6 moles of [tex]CO_2[/tex] require = 6 moles of [tex]H_2O[/tex]

Thus 2.0 moles of [tex]CO_2[/tex] require=[tex]\frac{6}{6}\times 2.0=2.0moles[/tex]  of [tex]H_2O[/tex]

Thus [tex]CO_2[/tex] is the limiting reagent as it limits the formation of product.

As 6 moles of [tex]CO_2[/tex] give =  1 moles of glucose

Thus 2.0 moles of [tex]CO_2[/tex] give =[tex]\frac{1}{6}\times 2.0=0.33moles[/tex]  of glucose

Mass of glucose = [tex]moles\times {\text {Molar mass}}=0.33moles\times 180g/mol=60g[/tex]

Thus 60.0 g of glucose will be produced from 88.0 g of carbon dioxide and 64.0 g of water  

In green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen. A plant has 88.0 g of carbon dioxide and 64.0 g of water available for photosynthesis produces 60.0 g of glucose.

To calculate the moles, we use the equation:

Mole fraction is a unit of concentration, described to be same to the range of moles of a element divided via way of means of the entire range of moles of a solution.

Thus it is well explained.

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Answer:

Br>Ge>Kr

Explanation:

Electron affinity is the energy change that results from adding an electron to a gaseous atom(Khan Academy). It is also defined as "the amount of energy released when an electron is attached to a neutral atom or molecule in the gaseous state to form a negative ion" (Wikipedia).

Electron affinity increases from left to right in the periodic table. However, the noble gases possess a complete outermost electron shell.

Hence, electron affinity can be raked from the most negative to the least negative for the three atoms as follows; Br>Ge>Kr

Answer:

3.45E24

Explanation:

calculator I'm not that great at math but I'm good at science but that isn't really science but it's aix between it tho

Answer:

The pistil and the stamen.

Explanation:

Please Mark Me Branliest.

The pistil and the stamen are the specialized  structures increase the probability of successful plant reproduction

Reproduction in Flowering Plants occur both asexually and sexually, flowering plants or angiosperms, use the sexual mode of reproduction.

Reproduction in Flowering Plants occurs in the flower as  both male and the female gametes are present in it, various parts of flower involve in the process of reproduction.

There are four important steps in the reproduction of flowering plants. such as Pollination, Gametogenesis, Fertilization

The flower is a specialized structure where reproductive part of the plant, containing the male and female reproductive organs which facilitate fertilization.

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[tex]\huge\underline\mathbb\pink{ANSWER\::}\\\\[/tex]

_BaS + _PtF2 → _BaF2 + _Pts

Barium Sulfide + Platinum(II) Fluoride = Barium Fluoride + Platinum(II) Sulfide.

=> 1 BaS + 1 PtF2 → 1 BaF2 + 1 PtS

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Answer:

i think answer is B negative effects and benefits

Answer:

The balanced equation is given below:

SiO2 + 4HF —> SiF4 + 2H2O

The coefficients are 1, 4, 1, 2

Explanation:

The equation for the reaction is given below:

SiO2 + HF —> SiF4 + H2O

The above equation can be balance by as follow:

There are 4 atoms of F on the right side and 1 atom on the left side. It can be balance by putting 4 in front of HF as shown below:

SiO2 + 4HF —> SiF4 + H2O

Therefore are 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of H2O as shown below:

SiO2 + 4HF —> SiF4 + 2H2O

Now the equation is balanced.

Explanation:

A balanced equation obey the law of conservation of mass. According to this law, the mass can neither be created nor be destroyed. Here the balanced equation for the given reaction is 2KClO₃ → 2KCl + 3O₂.

A chemical equation in which the number of atoms of reactants and products are equal on both sides of the equation is defined as the balanced chemical equation. The coefficients are the numbers that are used to balance a chemical equation.

The substances which appear on the left hand side of the equation are called the reactants whereas those which appear on the right hand side are called the products. The coefficients are essential to balance a equation.

The given equation can be balanced as follows:

2KClO₃ → 2KCl + 3O₂

Here the coefficients are 2, 2 and 3.

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Answer:

huh

Explanation:

I like ribs too eat lamfoaoa

HCO3-

Sodium bicarbonate (IUPAC name: sodium hydrogen carbonate), is commonly known as baking soda or bicarbonate of soda, is a chemical compound with the formula NaHCO3. It is a salt composed of a sodium cation (Na+) and a bicarbonate anion (HCO3−). Sodium bicarbonate is a white solid that is crystalline, but often appears as a fine powder.

Answer:

B

Explanation:

Answer:

B

Explanation:

Edge 2022

Answer:

partB: n=2N205

partc:17.20

Answer: [tex]BaCl_2[/tex] is the limiting reagent and [tex]Al_2(SO_4)_3[/tex] is the excess reagent.

Explanation:

To calculate the moles :

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

[tex]\text{Moles of aluminium sulphate}=\frac{150g}{342.15g/mol}=0.438moles[/tex]

[tex]\text{Moles of barium chloride}=\frac{200g}{208.23g/mol}=0.960moles[/tex]

The balanced chemical reaction is:

[tex]Al_2(SO_4)_3+3BaCl_2\rightarrow 2AlCl_3+3BaSO_4[/tex]  

According to stoichiometry :

3 moles of [tex]BaCl_2[/tex] require = 1 mole of [tex]Al_2(SO_4)_3[/tex]

Thus 0.960 moles of [tex]BaCl_2[/tex] will require=[tex]\frac{1}{3}\times 0.960=0.320moles[/tex]  of [tex]Al_2(SO_4)_3[/tex]

Thus [tex]BaCl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]Al_2(SO_4)_3[/tex] is the excess reagent as it is left.

hmm.......iphone ¯\_(ツ)_/¯

Answer:

A Brønsted-Lowry acid is any species that can donate a proton (H +) to another molecule. A Brønsted-Lowry base is any species that can accept a proton from another molecule. In short, a Brønsted-Lowry acid is a proton donor (PD), while a Brønsted-Lowry base is a proton acceptor (PA).

Answer:

hello

Bromate is a bromine oxoanion and a monovalent inorganic anion. It is a conjugate base of a bromic acid.

Explanation:

thanks

A scientist that carelessly pours cultures of living microbes down the drain represents a type of bacterial pollution.

Bacterial pollution refers to the uncontrolled proliferation of bacteria in the surrounding environment and ecosystems.

In conclusion, a scientist that carelessly pours cultures of living microbes down the drain represents a type of bacterial pollution.

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