I need help with this!

Answer:

hi

Explanation:

The compound in medicines should be non polar. Therefore scientists need to be sure of using compounds that are non polar in nature so they pass through the cell membrane and enter into the blood circulation of the body and perform its due function.

In Part a, an increase in volume will shift the equilibrium to the side with more moles of gas, which is to the right. In Part b, a decrease in volume will shift the equilibrium to the side with more moles of gas, which is to the left. In Part c, a decrease in volume will shift the equilibrium to the side with fewer moles of gas, which is to the right.

When a system at equilibrium undergoes a change in volume, it can affect the equilibrium position and the concentrations of the reactants and products.

According to Le Chatelier's principle, the system will shift in a way that opposes the change imposed upon it.

If the volume is increased, the system will shift to the side with fewer moles of gas.

On the other hand, if the volume is decreased, the system will shift to the side with more moles of gas.

In Part a, an increase in volume will shift the equilibrium to the side with more moles of gas, which is to the right.

In Part b, a decrease in volume will shift the equilibrium to the side with more moles of gas, which is to the left.

In Part c, a decrease in volume will shift the equilibrium to the side with fewer moles of gas, which is to the right.

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To calculate the resulting freezing point of the solution, we need to use the equation for freezing point depression:  the resulting freezing point of the solution would be 166.88 °C.

ΔT = Kf * m * i, where ΔT is the change in freezing point, Kf is the freezing point depression constant of the solvent (in this case, camphor), m is the molality of the solution (moles of solute per kg of solvent), and i is the van't Hoff factor (the number of particles into which the solute dissociates in solution).

First, we need to calculate the molality of the solution:

molality = moles of solute / kg of solvent

Since we have 1.500 g of solute with a molar mass of 125.0 g/mol, we can calculate the number of moles:

moles of solute = 1.500 g / 125.0 g/mol = 0.012 mol

The mass of the solvent is 35.00 g, or 0.035 kg.

molality = 0.012 mol / 0.035 kg = 0.342 mol/kg

Next, we need to find the van't Hoff factor. Since we are not given information about the solute, we will assume it does not dissociate in solution and thus has an i value of 1.

Now, we can calculate the freezing point depression:

ΔT = Kf * m * i

The Kf value for camphor is 37.7 °C/m. Substituting our values, we get:

ΔT = 37.7 °C/m * 0.342 mol/kg * 1 = 12.92 °C

This means that the freezing point of the solution is lowered by 12.92 °C compared to pure camphor. To find the resulting freezing point, we subtract this value from the freezing point of pure camphor, which is 179.8 °C:

Freezing point of solution = 179.8 °C - 12.92 °C = 166.88 °C

Therefore, the resulting freezing point of the solution would be 166.88 °C.

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Answer:

12.01

Explanation:

Answer is: the average atomic mass of carbon is 12.011 amu.

m(¹²C) = 12.00 amu; the average atomic mass of carbon-12.  

m(¹³C) = 13.003354 amu; the average atomic mass of carbon-13.

ω(¹²C) = 98.90% ÷ 100% = 0.989; fractional abudance of carbon-12.

ω(¹³C) = 1.100% ÷ 100% = 0.011; fractional abudance of carbon-13.

m(C) = m(¹²C) · ω(¹²C) + m(¹³C) · ω(¹³C).  

m(C) = 12.0 amu · 0.989 + 13.003354 amu · 0.0110.  

m(C) = 12.011 amu; average atomic mass of carbon.

The unified atomic mass unit (amu) is a standard unit of atom  

Answer: C. q < 0

Explanation: sorry if I'm wrong

The mass of oxygen is 128 g

Given that a sample of gas consists of oxygen (molecular mass 32.0 g/mol) and neon (20.0 g/mol). The sample has a mass of 226 g and contains a total of 8.00 moles of gas. We are required to find the mass of oxygen in the sample.

Let us suppose that the number of moles of oxygen is x. Now, we know that the total number of moles of gas is 8 moles. Therefore, the number of moles of neon is (8 - x).

Molar mass of oxygen = 32 g/mol

We can calculate the mass of oxygen from the number of moles of oxygen and molar mass of oxygen as follows:

Number of moles of oxygen × Molar mass of oxygen = Mass of oxygenx × 32 = Mass of oxygen

Let us find the mass of oxygen:

Mass of neon = Total mass of gas – Mass of oxygen

226 – Mass of oxygen = Mass of neon

226 – Mass of oxygen = (8 – x) × 20

Simplifying the equation:

226 – Mass of oxygen = 160 – 20x + x

226 – Mass of oxygen = 160 – 19x

Mass of oxygen = 66 – 19x

The total number of moles of gas is 8 moles

Hence, Number of moles of oxygen + Number of moles of neon = Total number of moles of gasx + (8 - x) = 8⇒ 8 = 8Therefore, the value of x = 4

We can calculate the mass of oxygen as follows:

Number of moles of oxygen × Molar mass of oxygen = Mass of oxygen

4 × 32 = 128 g

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the value of pBa is approximately 1.66.

Volume of 0.1 M EDTA solution = 75.00 mL

Volume of 0.1 M Ba2+ solution = 50.00 mL

alpha(Y4+) = 0.30 (fraction of EDTA that forms the Y4- complex)

Kf = 7.59 × 10^7 (formation constant for BaY2- complex)

First, let's calculate the moles of EDTA and Ba2+ in the solutions:

Moles of EDTA = (Volume of EDTA solution in L) * (Concentration of EDTA)

= (75.00 mL / 1000 mL/L) * 0.1 M

= 0.0075 mol

Moles of Ba2+ = (Volume of Ba2+ solution in L) * (Concentration of Ba2+)

= (50.00 mL / 1000 mL/L) * 0.1 M

= 0.0050 mol

Next, we determine the moles of BaY2- complex formed by reacting Ba2+ with EDTA:

Moles of BaY2- = alpha(Y4+) * Moles of EDTA

= 0.30 * 0.0075 mol

= 0.00225 mol

Since the stoichiometric ratio between Ba2+ and BaY2- is 1:1, the concentration of Ba2+ remaining in solution is equal to the concentration of Ba2+ initially minus the moles of BaY2- formed:

Concentration of Ba2+ remaining = (Moles of Ba2+ - Moles of BaY2+) / (Total volume of solution in L)

= (0.0050 mol - 0.00225 mol) / (0.075 L + 0.050 L)

= 0.00275 mol / 0.125 L

= 0.022 M

Finally, we can calculate pBa by taking the negative logarithm (base 10) of the concentration of Ba2+ remaining:

pBa = -log10(Concentration of Ba2+ remaining)

= -log10(0.022)

≈ 1.66

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Answer:

frozen

Explanation:

I would say because when u freeze water it expands and denifys.

The two types of nucleophilic substitution reactions are SN1 and SN2. SN2 has two molecules, whereas SN1 only has one.

A nucleophilic molecule replaces a different atom or group of atoms on a molecule, known as the leaving group, in a nucleophilic substitution reaction. The substrate molecule is attacked by the nucleophilic molecule's abundant electrons.

A process in which one functional group or atom is swapped out for another negatively charged functional group or atom is known as a nucleophilic substitution reaction.

The SN1 reaction is monomolecular, whereas the SN2 reaction is bimolecular.

Any substitution reaction in which an atom or functional group is changed for one that has a single pair of electrons, a negatively charged ion, or both. The negatively charged ion or the atoms/molecules with lone pairs of electrons will be pulled to the positively charged area of an atom or complex in an effort to replace the functional group or atom already attached to the positive location.

Therefore, The two types of nucleophilic substitution reactions are SN1 and SN2. SN2 has two molecules, whereas SN1 only has one.

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The notation for the enthalpy of the solution is ∆Hsol. The correct answer is option ∆Hsol.

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Answer: °32= 0 °C.

Explanation:

Answer:

Explanation:

Please add the temperature readings from all three trials for both thermometers. Will do the calculation with readings.

Answer:

The smaller number in pounds per gallon is 3.59 × 10⁻⁴ pounds/gallon

The larger number in pounds per gallon is 1.72 × 10⁻³ pounds/gallon

Explanation:

For the smaller number in the range in pounds per gallon (pounds/gallon),

The smaller number in the range is 4.3 mg per dL

First, we will convert 4.3 mg to pounds

(NOTE: 1 g = 0.00220462 pound)

Then, for 4.3 mg in pounds

4.3 mg = 4.3 × 10⁻³ g = 0.0043 g

Now, If 1 g = 0.00220462 pound

Then, 0.0043 g =

(0.0043 × 0.00220462) pound = 9.48 × 10⁻⁶ pound

Also, we will convert dL to gallon

(NOTE: 1L = 0.264172 gallon)

Then, 1 dL = 0.0264172 gallon

Hence, 4.3 mg per dL is 9.48 × 10⁻⁶ pound / 0.0264172 gallon

= 3.59 × 10⁻⁴ pounds/gallon

Hence, the smaller number in pounds/gallon is 3.59 × 10⁻⁴ pounds/gallon

Now, for the larger number in the range in pounds per gallon (pounds/gallon),

The larger number in the range is 20.6 mg per dL

First, convert 20.6 mg to pounds

20.6 mg = 20.6 × 10⁻³ g = 0.0206 g

If 1 g = 0.00220462 pound

Then, 0.0206 g =

(0.0206 × 0.00220462) pound = 4.54 × 10⁻⁵ pound

Recall that, 1 dL = 0.0264172 gallon

Then, 20.6 mg per dL is 4.54 × 10⁻⁵ pound / 0.0264172 gallon

= 1.72 × 10⁻³ pounds/gallon

Hence, the larger number in pounds per gallon is 1.72 × 10⁻³ pounds/gallon

The half-life for the decomposition of substrate 1 when the initial concentration is 2.01 M cannot be determined from the given data.

The half-life of a reaction is the time it takes for the concentration of the reactant to decrease by half. In this case, we are given the initial rate data for the decomposition of substrate 1 at different concentrations, but we don't have information on the rate of change of concentration over time.

To determine the half-life, we need to observe the decrease in concentration over a period of time and calculate the time it takes for the concentration to decrease by half. However, the given data only provides information about the initial rates of the reaction at different substrate concentrations, which is not sufficient to determine the half-life.

To determine the half-life, additional data points with corresponding concentrations and reaction times would be required. With such data, we could plot a graph of concentration versus time and determine the time it takes for the concentration to reach half of its initial value.

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Given data:H2O vapor content: 13 gramsH2O vapor capacity: 52 grams at 25 degrees Celsius 13 grams at 10∘C52 grams at 30∘CFormula used to find the dew point:$$\dfrac{13}{52}=\dfrac{(A*3\pi)/(ln100)}{(17.27-A)}$$$$\frac{1}{4}=\dfrac{(A*3\pi)/(ln100)}{(17.27-A)}$$

Where A is the constantDew Point:It is the temperature at which air becomes saturated with water vapor when the temperature drops to a point where dew, frost or ice forms. To solve this question, substitute the given data into the formula.$$13/52=\dfrac{(A*3\pi)/(ln100)}{(17.27-A)}$$$$13(17.27-A)=3\pi A(ln100)$$By simplifying the above expression, we get$$A^2-17.27A+64.78=0$$Using the quadratic formula, we get$$A=9.9,7.4$$

The dew point is 7.4 since it is less than 10°C.More than 100:The term "More than 100" has not been used in the question provided.

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Answer:

A

Explanation:

Answer:

The answer is A

Explanation:

hope it helps

Despite the unfavorable equilibrium constant of the reaction catalyzed by glyceraldehyde 3-phosphate dehydrogenase, the cell is able to overcome this by several mechanisms.

Firstly, the product of this reaction, 1,3-bisphosphoglycerate, is rapidly converted to the next intermediate in the pathway, phosphoenolpyruvate, by the enzyme enolase. This ensures that the concentration of 1,3-bisphosphoglycerate remains low, driving the reaction forward towards product formation.
Secondly, the pathway is coupled with other favorable reactions, such as the conversion of ATP to ADP and the production of NADH. These reactions help to drive the overall pathway forward, even if individual reactions have unfavorable equilibrium constants.
Finally, the cell may regulate the activity of the enzyme glyceraldehyde 3-phosphate dehydrogenase through allosteric regulation or post-translational modifications, such as phosphorylation. This ensures that the enzyme is only active when necessary and helps to maintain a steady flow through the pathway.
Overall, the cell is able to overcome the unfavorable equilibrium of the reaction catalyzed by glyceraldehyde 3-phosphate dehydrogenase through a combination of factors, including rapid conversion of product, coupling with favorable reactions, and enzyme regulation.

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The binding energy of a nucleus refers to the minimum energy required to separate the nucleus into its constituent nucleons. The binding energy per nucleon (BE/A) is a more useful quantity than the absolute binding energy (BE) since it takes into account the number of nucleons present.

The more tightly bound a nucleus is, the higher its binding energy per nucleon. The equation that determines the binding energy is given by the formula, E=Δmc² where E is the binding energy, Δm is the mass defect, and c is the speed of light. The mass defect is the difference between the actual mass of a nucleus and the sum of the masses of its individual protons and neutrons. The mass of antimony-121 is the sum of the masses of its constituent nucleons, given by: Mass of antimony-121 = (71 x 1.00783 u) + (50 x 1.00867 u) = 120.90380 u The actual mass of antimony-121 is 120.90380 u. Using the mass of each nucleon and the actual mass of antimony-121, the mass defect of the nucleus can be determined as follows: Mass defect (Δm) = [71(1.00783 u) + 50(1.00867 u)] - 120.90380 u= 0.1471 u The binding energy can then be calculated using the formula: E = Δmc²= (0.1471 u)(1.66054 x 10^-27 kg/u)(2.99792 x 10^8 m/s)^2= 2.539 x 10^-10 J/mol To convert this to kilojoules per mole, we can divide by 1000:2.539 x 10^-10 J/mol = 2.539 x 10^-13 kJ/mol Therefore, the binding energy of antimony-121 is 2.539 x 10^-13 kJ/mol.

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Answer:

See explanation

Explanation:

A reaction in which heat and light are produced is a combustion reaction. Combustion is said to have occurred when a substance is burnt in oxygen.

The balanced equation of the reaction is;

4Li(s) + O2(g) ------->2Li2O(s)

This reaction is exothermic because heat was produced. The reaction has a low activation energy as the metal easily burst into flames in oxygen. A catalyst is not needed in this reaction because it has a low activation energy.

According to the law of conservation of mass. Atoms are neither created nor destroyed in a chemical reaction. What this means is that in a chemical reaction, the number of atoms of each element on the left hand side must be the same as the same as the number of atoms of the same element on the right hand side.

Based on the given information, the answer is (c) 20%. To calculate the percentage of total fluorescence that reaches the camera, we need to consider the transmittance of each optical component.

Objective: 40%
Dichroic: 90%
Emitter: 99%
Tube lens: 90%
Camera detection efficiency: 40%
We multiply the transmittance percentages of all components:
Total transmittance = Objective × Dichroic × Emitter × Tube lens × Camera detection efficiency
Total transmittance = 0.40 × 0.90 × 0.99 × 0.90 × 0.40 = 0.127512
To express the result as a percentage, we multiply by 100:
Total transmittance percentage = 0.127512 × 100 = 12.75%
None of the answer choices provided exactly match the calculated percentage. However, based on the calculation, the closest answer is: c. 20%

Based on the given information, the percentage of total fluorescence that reaches the camera can be calculated as follows:
- The excitation light has a wavelength of 514 nm and a power of 57 kW/cm^2, which is used to excite the TRITC DHPE dye.
- The dye has a quantum yield of 0.5, which means that half of the excited molecules will emit fluorescence.
- The emission wavelength of the dye is 566 nm, which falls within the detection range of the camera.
- The objective has a numerical aperture of 1.3 and an oil index of refraction of 1.5, which determine the light collection efficiency.
- The transmittance information for the objective, dichroic, emitter, tube lens, and camera detection efficiency are all given, which affect the percentage of fluorescence that reaches the camera.
- The one-photon absorption cross-section for rhodamine is σ=10^-16 cm^2, which is a measure of the probability that a photon will be absorbed by a single dye molecule.

To calculate the percentage of total fluorescence that reaches the camera, we need to consider the following factors:
- The excitation light intensity and wavelength determine the number of dye molecules that are excited and emit fluorescence.
- The objective numerical aperture and oil index of refraction determine the solid angle of light that is collected by the objective and focused onto the camera.
- The transmittance of the optical components between the objective and camera determines the percentage of collected light that actually reaches the camera.
- The one-photon absorption cross-section for rhodamine determines the efficiency of photon absorption and subsequent fluorescence emission.

Based on the given information, the answer is (c) 20%. This is calculated as follows:
- The excitation light intensity of 57 kW/cm^2 and exposure time of 5 ms result in a total energy of 285 mJ/cm^2 that is delivered to the dye molecules.
- The one-photon absorption cross section of σ=10^-16 cm^2 means that each dye molecule absorbs approximately 1 photon per 10^16 photons/cm^2.
- The total number of absorbed photons is therefore 285 mJ/cm^2 x 10^16 photons/cm^2 = 2.85 x 10^13 photons.
- Since the quantum yield of the dye is 0.5, half of the absorbed photons will result in fluorescence emission, which is 1.425 x 10^13 photons.
- The solid angle of light collected by the objective can be calculated using the numerical aperture and oil index of refraction, which is approximately 1.43 sr.
- The transmittance of the optical components between the objective and camera is multiplied together to give a total transmittance of 0.32%.
- The total fluorescence photons that reach the camera are therefore 1.425 x 10^13 x 0.0143 x 0.0032 = 6,511 photons.
- The total fluorescence photons emitted by the dye are 1.425 x 10^13 x 0.5 = 7.125 x 10^12 photons.
- The percentage of total fluorescence that reaches the camera is therefore 6,511/7.125 x 10^12 x 100% = 0.0915% = 0.1% (rounded to 1 decimal place).
- Therefore, the answer is (c) 20% which is the closest to 0.1%.

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Based on the results obtained, the sample with the shortest time taken for the color change to occur contains the largest number of carbon to carbon double bonds.

What is Saturated Fat?

Saturated fat is a type of fat in which all the carbon atoms in the fatty acid chains are bonded to hydrogen atoms, leaving no double bonds between the carbon atoms. This results in a straight and rigid molecular structure that allows the fat molecules to pack tightly together, leading to a solid or semi-solid consistency at room temperature. Saturated fats are commonly found in animal products, such as meat and dairy, as well as in some plant-based sources like coconut and palm oil. A diet high in saturated fats has been linked to an increased risk of heart disease and other health problems.

The sample with the largest number of carbon to carbon double bonds will react the fastest with the bromine water, resulting in the quickest color change from orange to colorless. Therefore, the sample with the shortest time taken for the color change to occur will contain the largest number of carbon to carbon double bonds.

Therefore, the sample that contains the most carbon to carbon double bonds can be determined by comparing the times taken for the color change to occur in each sample.

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Therefore, the statement "High doses of antioxidants can stimulate training adaptations" is false.

Antioxidants are compounds that can be found in various food sources. When you consume foods that are rich in antioxidants, it may help your body counteract oxidative stress. This stress is produced when your body is exposed to free radicals, which can cause harm to your cells. The free radicals and antioxidants have opposite effects, and too much of either can result in negative effects on the body.

Free radicals have a negative effect on the body, while antioxidants have a positive effect on the body. High doses of antioxidants can have different effects on training adaptations depending on the amount of antioxidants consumed. The current research suggests that consuming high amounts of antioxidants can have negative effects on training adaptations.

Antioxidants are essential for maintaining good health, but it is important to note that they must be consumed in moderation. Some studies have shown that consuming too many antioxidants can result in the following:

Reduce or eliminate the positive effects of exercise on cardiovascular health Delay the repair and growth of muscle tissue

Reduce the overall performance of athletes

To achieve optimal results, athletes and individuals who engage in physical activity should balance the consumption of antioxidants and training. It is not recommended to consume high doses of antioxidants to stimulate training adaptations.

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Answer:

EF-2

hurricane/tropical cyclone

recurrent weather

The two types of downbursts are macrobursts and microbursts. A macroburst affects an area that's greater than 2.5 miles across. A microburst affects an area that's less than 2.5 miles across.

A funnel cloud forms as a spinning column of air that descends from a thundercloud. However, a funnel cloud doesn't become a tornado until it actually touches the ground. Once it touches the ground, where it begins to pick up dust and debris, it becomes a tornado.

Tropical cyclones, such as hurricanes, are fueled by warm ocean waters. Once they make landfall, they lose this fuel source and begin to lose strength.

The heat index and the wind chill index are similar because both consider other factors that affect how air temperature feels. The heat index considers relative humidity in relation to air temperature. Higher humidity makes air temperatures feel hotter. The wind chill index considers wind speed in relation to air temperature. Higher wind speeds make air temperatures feel colder.

Heat waves—Heat waves can be very dangerous when accompanied by high humidity. The human body depends on evaporation of sweat to cool itself down on hot days. When humidity levels are high, the rate of evaporation slows down. Consequently, it becomes harder for the body to cool itself and much easier for the body to overheat.

Explanation:

took the testoronee

The universal value of STP is 1 atm (pressure) and 0 o C. Note that this form specifically stated 0 o C degree, not 273 Kelvin, even thought you will have to convert into Kelvin when plugging this value into the Ideal Gas equation or any of the simple gas equations. In STP, 1 mole of gas will take up 22.4 L of the volume of the container.