Suppose the reaction: A + 2B → AB2 occurs by the following mechanism:
Step 1: 2A + B→ A2B slow
Step 2: A2B + B→ 2AB fast
Step 3: AB + B→ AB2 fast
Overall: A + 2B → AB2
The Rate Law Expression must be Rate = _____.

This question is incomplete, here´s the complete question.  

How does the percentage of monounsaturated and polyunsaturated fatty acids in olive oil compare to that of canola oil? Match the words in the left column to the appropriate blanks in the sentences on the right.

Olive oil has about ____ monounsaturated fats, while canola oil has about ___.

Olive oil has about ___ polyunsaturated fats, while canola oil has about ___.

6%

10%

30%

84%

5%

65%

Answer:

Olive oil has about 84% monounsaturated fats, while canola oil has about 65%.

Olive oil has about 5% polyunsaturated fats, while canola oil has about 30%.

Explanation:

Olive and canola oil are the major sources of monounsaturated fatty acids. Although vegetable oils usually have high concentrations of polyunsaturated fatty acids and less monounsaturated fats, olive and canola oils have comparatively less polyunsaturated fatty acids, and more monounsaturated fatty acids.

Answer:

Explanation:

As,

Law of reflection states that , angle of incidence is always equal to angle of reflection.

Answer:

Frequency of spinal trait in the population = 0.11

Explanation:

Given:

Total spineless trait = 600

Number of spinal trait = 80

Find:

Frequency of spinal trait in the population

Computation:

Total population = Total spineless trait + Number of spinal trait

Total population = 600 + 80

Total population = 680

Frequency of spinal trait in the population = Number of spinal trait / Total population

Frequency of spinal trait in the population = 80 / 680

Frequency of spinal trait in the population = 0.11

To determine the volume of N2(g) produced by the complete decomposition of 1 mole of nitroglycerin (C3H5N3O9), we need to consider the balanced chemical equation for the decomposition reaction.

The balanced equation for the decomposition of nitroglycerin is as follows:

4 C3H5N3O9(s) → 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g) From the balanced equation, we can see that for every 4 moles of nitroglycerin, 6 moles of N2(g) are produced. Since we are considering the decomposition of 1 mole of nitroglycerin, we can use this ratio to determine the moles of N2(g) produced, which is 6/4 = 1.5 moles of N2(g). Now, at STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 liters. Therefore, 1.5 moles of N2(g) would occupy approximately 33.6 liters

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To determine the number of atoms of titanium in 0.820 mole of each compound, we need to use Avogadro's number, which is 6.022 x 10²³ atoms/mol.

1. Ilmenite, FeTiO3:

2. Titanium(IV) chloride, TiCl4:

Thus, the number of titanium atoms in 0.820 mole of ilmenite is 4.917 x 10²³ atoms, and the number of titanium atoms in 0.820 mole of titanium(IV) chloride is 4.917 x 10²³ atoms.

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Answer:

redox and AlF2

Explanation:

Answer:

Explanation:

Proteins have two main secondary structures: alpha helices, which are spirals formed by hydrogen bonds between amino acids, and beta pleated sheets, [ which are formed by a bend in the amino acid with alternating hydrogen bonding between amino acids.

Answer:

Explanation:

The density of a substance can be found by using the formula

From the question

mass = 54 g

volume = 20 cm³

Substitute the values into the above formula and solve for the density

That's

We have the final answer as

Hope this helps you

Answer:

why

xy=gh

Explanation:

common science

The cube is among the five Platonic solids. 0.56g/cm³ is the density if a 2.5cm cube of bone weighs 8.75g.

A cube is indeed a three-dimensional physical block having six square faces, facets, or sides, three of which meet at each vertex. It is a hexagon when viewed from a corner, and its net is commonly portrayed as a cross.

The cube is among the five Platonic solids and the only regular hexahedron. It has six faces, twelve edges, and eight vertices. In addition, the cube is a square parallelepiped, a equilateral cuboid, a right rhombohedron, and a 3-zonohedron.

density= mass/ volume

volume of cube =2.5×2.5×2.5=15.62cm³

density=  8.75/ 15.62=0.56g/cm³

Therefore, 0.56g/cm³  is the density if a 2.5cm cube of bone weighs 8.75g.

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Answer:

D

Explanation:

The new molarity of the solution is 0.5M.

The molarity of a solution can be calculated by using the following expression:

C1V1 = C2V2

Where;

According to this question, a chemist dilutes 2.0 L of a 1.5 M solution with water until the final volume is 6.0 L.

1.5 × 2 = 6 × C2

3 = 6C2

C2 = 3/6 = 0.5M

Therefore, the new molarity of the solution is 0.5M.

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Answer:

This question lacks options, the options are;

A. Absorption of heat

B. A chemical change

C. A physical change

D. Formation of a new compound

The answer is C. A physical change

Explanation:

A physical change is a type of change that does not involve the chemical composition of the substance(s) involved. Physical changes affect the form but not the chemical content of a substance. Examples of physical changes are freezing, change of state, boiling etc.

In this question, a liquid is cooled during an investigation causing it to solidify. The cooling of this liquid represents a PHYSICAL CHANGE because the liquid state of the substance is only changed to a solid state but does not involve changing the chemical composition of the substance.

A liquid is cooled during an investigation physical change causing it to solidify.

Physical change occurs whenever there is difference in the appearance of any object observed.

In the question it is given that during an investigation a liquid is cooled down which results in the solidification of that liquid. During that change state of material changes from liquid state to the solid state, which is an example of a physical change.

Hence, in the question physical change occurs.

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Answer:

Electrons accelerated to high velocities travel in straight lines through an empty cathode ray tube and strike the glass wall of the tube, causing excited atoms to fluoresce or glow.

Explanation:

1. I noticed that the soil particles settled out of the water with the larger, heavier particles settling out first and on the bottom, and each layer has smaller and lighter particles in it.

2. Sand settled on the bottom because they are heavier and have more mass than the smaller particles such as silt and clay.

3. Yes, this is a good way to determine particle size distribution. The method used in the experiment allows for the separation of soil particles based on their size and weight, which is a good indicator of their distribution in the soil sample.

4. The results in this experiment may look different when done in different locations because the particle size distribution of soil can vary depending on the location, climate, and geology of the area. Soils in different locations may have different compositions, including different amounts of sand, silt, and clay, which can affect the results of the experiment. Additionally, the sedimentation rate of the particles can also be influenced by factors such as the temperature, the presence of organic matter, and the presence of clay minerals, which can further affect the results.

The mass of oxygen gas produced, given that 15.8 g of potassium permanganate is heated until no more oxygen gas is given off is 1.6 g

From the question given, the following data were obtained:

The mass of oxygen gas produced from the reaction can be obtained as follow:

Mass of potassium permanganate = Mass of remaining substance + mass of oxygen

Inputting the given parameters, we have:

15.8 = 14.2 + mass of oxygen

Collect like terms,

Mass of oxygen = 15.8 - 14.2

= 1.6 g

Thus, we can conclude that the mass of oxygen gas produced from the reaction is 1.6 g

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Answer:Oxygen has a molar mass of 16.00 g/mol. One mole of CO2 contains 32.00 grams of oxygen. One mole equals 22.4L at standard temperature and pressure. 0.60L CO2 equals 0.0268 mol at STP.

Explanation:

The total cooling performed by the coil is 29168.21 BTU/Hr and the sensible cooling performed by the coil is 1209.93 BTU/Hr.

Inlet Condition: TDB, in = 84 °F and TWB, in = 72 "F

Outlet Condition: TDB,

out = 55 F and TDP out = 50 'F

Flow Rate: CFM of air = 3000

Standard Density of Air = 0.075 lbm/ft³

Specific Heat of Air = 0.24 BTU/lbm-'F.

The total cooling performed by the coil can be calculated using the following formula;

Sensible Cooling = CFM x Density x Specific Heat x (TDB, in - TDB, out)

Latent Cooling = CFM x Density x (HFG x (W, in - W, out)

Total Cooling = Sensible Cooling + Latent Cooling

First, we need to calculate the density of air at the inlet condition and outlet condition using the formula:

Density = 0.075 x (460 + TDB) / (460 + TWB)  at inlet and outlet condition

Density at inlet condition = 0.075 x (460 + 84) / (460 + 72)Density at inlet condition = 0.0666 lbm/ft³

Density at outlet condition = 0.075 x (460 + 55) / (460 + 50)Density at outlet condition = 0.068 lbm/ft³

The sensible cooling performed by the coil is;

Sensible Cooling = CFM x Density x Specific Heat x (TDB, in - TDB, out)

Sensible Cooling = 3000 x 0.0666 x 0.24 x (84 - 55)

Sensible Cooling = 1209.93 BTU/Hr

The total cooling performed by the coil is;

Total Cooling = Sensible Cooling + Latent Cooling

Here, Latent Cooling = CFM x Density x (HFG x (W, in - W, out))

At the inlet condition; W, in = (0.62198 x PWS) / (PB - PWS)W,

in = (0.62198 x 0.6237) / (14.433 - 0.6237)W,

in = 0.0427

At the outlet condition; W, out = (0.62198 x PWS) / (PB - PWS)W,

out = (0.62198 x 0.315) / (14.266 - 0.315)W,

out = 0.0237

HFG at average of inlet and outlet air temperature = 1074 BTU/lbm

Latent Cooling = CFM x Density x (HFG x (W, in - W, out))

Latent Cooling = 3000 x 0.0666 x (1074 x (0.0427 - 0.0237))

Latent Cooling = 27958.28 BTU/Hr

Therefore, Total Cooling = Sensible Cooling + Latent Cooling

Total Cooling = 1209.93 + 27958.28

Total Cooling = 29168.21 BTU/Hr

Therefore, the total cooling performed by the coil is 29168.21 BTU/Hr and the sensible cooling performed by the coil is 1209.93 BTU/Hr.

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Answer:

protons

Atomic number and protons

After the discovery of protons , scientists realised that the atomic number of an element is the same as the number of protons in its nucleus . In the modern periodic table, the elements are arranged according to their atomic number - not their relative atomic mass .

Answer:

\(\boxed {\boxed {\sf 142.2 \ K}}\)

Explanation:

This problem asks us to find the temperature change necessary to make the volume change. We use Charles's Law which states that temperature is directly proportional to the volume of a gas. The formula is:

\(\frac{V_1}{T_1}=\frac{V_2}{T_2}\)

We know the balloon originally had a volume of 6.5 liters and a temperature of 280 Kelvin. Then, the temperature was cooled so the new volume is 3.3 liters. However, the exact new temperature is unknown.

Substitute all known values into the formula.

\(\frac {6.5 \ L}{280 \ K}=\frac{3.3 \ L}{T_2}\)

Now, solve the new temperature (T₂). First, cross multiply. Multiply the 1st numerator by the 2nd denominator. Then, multiply the 1st denominator by the 2nd numerator.

\(280 \ K * 3.3 \ L = 6.5 \ L * T_2\)

Multiply the left side.

\(924 \ K *L=6.5 \ L *T_2\)

We must isolate the variable. Currently, it is being multiplied by 6.5 liters. The inverse of multiplication is division. Divide both sides by 6.5 L.

\(\frac { 924 \ K*L}{6.5 \ L}= \frac{ 6.5 \ L *T_2}{ 6.5 \ L}\)

\(\frac { 924 \ K*L}{6.5 \ L}= T_2\)

The units of liters (L) cancel.

\(\frac { 924 \ K}{6.5 }= T_2\)

\(142.153846 \ K= T_2\)

Let's round to the nearest tenth place. The 5 in the hundredth place tells us to round the 1 up to a 2.

\(142.2 \ K= T_2\)

The temperature must be cooled to approximately 142.2 Kelvin.

Explanation:

To find the freezing point of the solution, we can use the freezing point depression equation:

ΔT = Kᵣ x m

Where ΔT is the change in freezing point, Kᵣ is the freezing point depression constant of benzene, and m is the molality of the solution.

Substituting the values from the problem, we get:

ΔT = 5.12 °C/m x 2.8 m

ΔT = 14.34 °C

Since ΔT = Tᵢ - T, where Tᵢ is the freezing point of the solvent (benzene) and T is the freezing point of the solution, we can rearrange the equation to solve for T:

T = Tᵢ - ΔT

T = 5.50 °C - 14.34 °C

T = -8.84 °C

Therefore, the freezing point of the solution is -8.84 °C.

Answer:The freezing point of the solution is -8.84 °C.

Explanation:

Mass of NaCl (MM = 58.5 g/mol)  present in 100 µL of the PBS solution is B. 5.85 µg.

To calculate the mass of NaCl in 100 µL of a 0.1 mM NaCl solution, we need to first convert mM (millimoles per liter) to µM (micromoles per liter) by multiplying by 1000.

0.1 mM NaCl = 100 µM NaCl

Next, we can calculate the number of moles of NaCl in 100 µL of the solution using the formula:

moles = concentration (in µM) x volume (in liters)

volume = 100 µL = 0.0001 L

moles = 100 µM x 0.0001 L = 0.00001 moles

Finally, we can calculate the mass of NaCl using its molar mass:

mass = moles x molar mass

mass = 0.00001 moles x 58.5 g/mol = 0.585 µg

Therefore, the answer is B. 5.85 µg.

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Answer:

La materia está constituida por átomos, que a su vez forman moléculas. Las moléculas constituyen la mínima parte en la que se puede fragmentar una sustancia para que conserve sus propiedades. Todo lo anterior permite que la materia se pueda presentar en estado sólido, líquido y gaseoso.

Explanation:

The partial pressure of oxygen at sea level in dry air is approximately 159 mmHg (millimeters of mercury). This value represents the fraction of atmospheric pressure that is due to oxygen.

When the outside air is almost saturated with water vapor at 37 °C and the water vapor partial pressure is 45 mmHg, the partial pressure of oxygen is reduced. The partial pressure of oxygen in this scenario depends on the atmospheric pressure at that location. Assuming a typical atmospheric pressure of 760 mmHg at sea level, and subtracting the water vapor partial pressure of 45 mmHg, the partial pressure of oxygen would be approximately 715 mmHg.

In the alveoli of the lungs, where there is saturation with water vapor (partial pressure of 47 mmHg) and elevated carbon dioxide levels (partial pressure of 45 mmHg), the partial pressure of oxygen is reduced. The normal partial pressure of oxygen in the alveoli is around 104 mmHg. However, with the elevated carbon dioxide levels, the partial pressure of oxygen will be lower. Subtracting the carbon dioxide partial pressure of 45 mmHg, the partial pressure of oxygen in the alveoli would be approximately 59 mmHg.

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