Blood alcohol concentration (BAC) is a measure of alcohol in your body, expressed as grams of alcohol per 100ml of blood. For a BAC of 0.05, every 100ml of your blood contains 0.05 grams of alcohol. Your BAC can be measured through your breath, blood, and urine. Drivers with an open licence have to keep under a BAC of 0.05.

Scientists estimate the mass of blood in a human body to be approximately 8 percent of body weight. If we assume that 17 % of the alcohol that a person drinks from beer goes into the bloodstream, how many litres of beer can an 71 kg person drink to get a BAC of 0.05? The alcohol content of beer is 6 % volume, and the blood density is 1025kg/m3, the alcohol density is 789 kg/m3.

Answers

Answer 1

Answer:

A 71 kg person will get a BAC of 0.05 when drinking 0.3442 L of beer

Explanation:

The mass of blood in a human body is approximately 8%, so if a person weighs 71 kg, the mass of blood would be:

71 kg * 8/100 = 5.68 kg of blood.

Using blood density, we can calculate the volume that 5.68 kg of blood occupies:

5.68 kg * [tex]\frac{1m^{3}}{1025kg}[/tex] = 0.0055415 m³

We convert m³ into mL, keeping the unit that we want to convert to in the numerator; and the unit that we want to convert in the denominator:

[tex]0.0055415m^{3}*\frac{1000L}{1m^{3}} *\frac{1000mL}{1L}=5541.5mL[/tex]

Now we calculate the amount of alcohol that would be needed in the bloodstream to get a BAC of 0.05:

[tex]\frac{5541.5mL}{100mL}*0.05g[/tex] = 2.77 g of alcohol are needed in the bloodstream in order to have a BAC of 0.05

The amount of alcohol that needs to be ingested is higher than 2.77 g, due to the fact that only 17% of the alcohol goes into the bloodstream, so:

2.77 g [tex]*\frac{100}{17} =[/tex] 16.29 g of alcohol need to be ingested

Then we use the alcohol concentration of beer to calculate the volume of beer needed, using the alcohol density. First we convert the alcohol density to g/L, making sure the units that we want to convert cancel each other:

[tex]789\frac{kg}{m^{3}}*\frac{1000g}{1kg} *\frac{1m^{3}}{1000L}  =789g/L[/tex]

Now we use the density to calculate the litres of alcohol needed, keeping in mind that 16.29 g of alcohol are needed:

[tex]16.29g*\frac{1L}{789g}=[/tex] 0.02065 L of alcohol are needed.

Finally we calculate the litres of beer needed, keeping in mind the concentration of alcohol in beer:

[tex]0.02065L_{alcohol}*\frac{100L_{beer}}{6L_{alcohol}} =[/tex]0.3442 L of beer are needed.

Answer 2
Final answer:

To have a Blood Alcohol Concentration (BAC) of 0.05, a 71 kg person would need to consume approximately 271.57 liters of beer with a 6% alcohol content, assuming that 17% of the alcohol ends up in the bloodstream.

Explanation:

First, let's calculate the total mass of blood in the body. The total weight of the body is 71 kg and blood comprises about 8 percent of body weight. Therefore, the mass of blood = 0.08 * 71 = 5.68 kg. Because the blood density is 1025 kg/m³, this implies that the volume of the blood is 5.68/1025 = 0.00554 m³, or 5.54 liters. When the BAC is 0.05, this means that there are 0.05 grams of alcohol in every 100 ml of blood, or 0.05 grams of alcohol per 0.1 liter of blood. Therefore, to have a BAC of 0.05 in 5.54 liters, there needs to be 0.05 * 5.54 * 10 = 2.77 grams of alcohol in the blood. We know that 17 percent of the alcohol that a person drinks goes into their bloodstream. Therefore, the total mass of beer consumed to get this amount of alcohol is = 2.77 / 0.17 = 16.294 grams of pure alcohol. Beer has an alcohol equivalent to 6 % of its volume, so we find that the total volume of beer consumed = 16.294 / 0.06 = 271.57 liters. Because alcohol density is 789 kg/m³, this would be approximately 0.271 m^3 or 271.57 liters of beer.

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Related Questions

A liquid mixture contains 1 kg of water, 1.9 lb of ethanol, and 4.6 lb of methyl acetate. What is the weight fraction of ethanol in this mixture? Report your answer to the nearest hundredths place.

Answers

Answer: The weight fraction of ethanol in the mixture is

Explanation:

We are given:

Mass of water = 1 kg = 2.205 lb   (Conversion factor:  1 kg = 2.205 lb)

Mass of ethanol = 1.9 lb

Mass of ethyl acetate = 4.6 lb

Mass of mixture = [2.205 + 1.9 + 4.6] = 8.705 lb

To calculate the percentage composition of ethanol in mixture, we use the equation:

[tex]\%\text{ composition of ethanol}=\frac{\text{Mass of ethanol}}{\text{Mass of mixture}}\times 100[/tex]

Mass of mixture = 8.705 lb

Mass of ethanol = 1.9 lb

Putting values in above equation, we get:

[tex]\%\text{ composition of ethanol}=\frac{1.9lb}{8.705lb}\times 100=21.8\%[/tex]

Hence, the weight fraction of ethanol in the mixture is 21.8 %

A consistent explanation of known observations is called A) an experiment. B) a prediction. C) a hypothesis. theory

Answers

Final answer:

A theory is a well-substantiated explanation of observed phenomena that has undergone rigorous testing and validation through the scientific method.

Explanation:

A consistent explanation of known observations is called a theory. In the realm of science, a theory is more than just a simple guess; it is a well-substantiated explanation that is grounded in a significant body of evidence. This evidence consists of a series of facts and observations that, when considered collectively, give a reliable account of a part of the natural world. To arrive at a theory, the scientific method is typically employed, involving systematic observations, forming hypotheses, conducting experiments, and refining those hypotheses based on experimental results. When a hypothesis withstands rigorous testing and is capable of explaining a large and diverse range of phenomena, it can evolve into a theory. Scientific theories are valued because they offer comprehensive explanations that are testable and can be modified in the face of new evidence.

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Topic 1 Homework Homework – Due in 13 hours T2HW Question 16 - Challenge Homework – Unanswered The distance from San Francisco to Los Angeles is approximately 385 miles. You and your friends decide to cycle from San Francisco to Los Angeles. If the distance between the cities is about 385 miles and your doctor tells you that you need to drink 1 L of water for every 1 km that you cycle, how many Lof water will each cyclist need to drink on the journey? Enter your answer as a number using 3 significant figures without units. Do not enter the word "liters" as part of your answer. 1609 m = 1.0 mi Numeric Answer Unanswered

Answers

Answer:

Each cyclist will need to drink 619 l

Explanation:

Hi there!!

First, let´s convert the miles to kilometers:

If 1.0 mi = 1609 m, then 385 mi will be:

385 mi · (1609 m/ 1.0 mi) ·  (1 km/ 1000 m) = 619 km

Now, if each cyclist need to drink one liter water per kilometer ( I think that´s a lot of water!), for the entire journey each cyclist will need to drink:

619 km · 1 l/km = 619 l

Then, each cyclist will need to drink 619 l.

ObIel WiLll unt COl.. USSMS A certain chemical reaction releases 31.2 kJ/g of heat for each gram of reactant consumed. How can you calculate what mass of reactant will produce 1080. J of heat? Set the math up. But don't do any of it. Just leave your answer as a math expression. Also, be sure your answer includes all the correct unit symbols. mass X ? Explanation Check ducabon ARghtsserved T of Ue Phacy The I'm esc

Answers

Answer:

The expression to calculate the mass of the reactant is [tex]m = \frac{1.080kJ}{31.2kJ/g}[/tex]

Explanation:

The amount of heat released is equal to the amount of heat released per gram of reactant times the mass of the reactant. To keep to coherence between units we need to transform 1,080 J to kJ. We do so with proportions:

[tex]1,080J.\frac{1kJ}{10^{3}J } =1.080kJ[/tex]

Then,

[tex]1.080kJ=31.2\frac{kJ}{g} .m\\m = \frac{1.080kJ}{31.2kJ/g}[/tex]

Answer:

1. The expression is: [tex]m=\frac{E}{\Delta _rH}[/tex]

2. The computed mass is: [tex]m=0.0346g[/tex]

Explanation:

Hello,

In this case, we know the so called enthalpy of reaction whose symbol and value is shown below:

[tex]\Delta _rH=31.2\frac{kJ}{g}[/tex]

In addition, we know that the energy released by the involved reactant is:

[tex]E=1080 J[/tex]

Therefore, the expression to compute the required mass, based on the given units is:

[tex]m=\frac{E}{\Delta _rH}[/tex]

Finally, the computed mass turns out:

[tex]m=\frac{1080J*\frac{1kJ}{1000J} }{31.2\frac{kJ}{g}} \\m=0.0346g[/tex]

Best regards.

A certain insulation has a thermal conductivity of 0.2 W/m °C What thickness is necessary to effect a temperature drop of 400°C for a heat flux of 200 W/m2?

Answers

Answer:

It is necessary 1 meter of thickness.

Explanation:

You need to know the equation (P/A)=(k*T)/(L), where P/A is the heat flux, k is the thermal conductivity, T the temperature drop, and L the unknown thickness. So if rearrange the variables you will have that L = (k*T)/(P/A), and substituting the terms L = (0.2*400)/(200) = 1 meter

Final answer:

The necessary thickness of insulation to achieve a temperature drop of 400°C with a heat flux of 200 W/m^2, given a thermal conductivity of 0.2 W/m°C, is 0.4 meters.

Explanation:

The necessary thickness of insulation for a specific temperature drop and heat flux. By using the equation of thermal conduction Q = (kAΔT)/d, where Q is the heat flux, k is the thermal conductivity, A is the area, ΔT is the temperature difference, and d is the thickness, we can solve for the thickness d.

To start, we know that k (thermal conductivity) is 0.2 W/m°C, the temperature drop ΔT is 400°C, and the heat flux Q is 200 W/m^2. Plugging these values into the equation gives us d = (kAΔT)/Q. Since the area A is not specified, it will cancel out in this case, allowing us to solve for d directly. Rearranging the equation yields d = (kΔT)/Q, which equals (0.2 W/m°C * 400°C)/200 W/m^2, resulting in d = 0.4 m. Therefore, the thickness required for this insulation is 0.4 meters.

Pressure drop in packed column ..... a tray column
a) More than
b) Less than
c) Equal to
d) No relation to

Answers

Answer:

The answer is b)

Explanation:

In packed column the mass transfer area is higher and the packing can be random, that could provoque an increase in the pressure, because obstruction of the mass flux, While in a tray column the sieves can have specific porosity and the flux of mass can be facilitated, thus the pressure drop.  

When a compound is dissolved in hot ethanol during a recrystallization, what is changing on a molecular level? The intermolecular forces between the solute and solvent The covalent bonds of the solute The compound slowly melts into the hot solvent The covalent bonds between the solute and solvent

Answers

Answer:

The intermolecular forces between the solute and solvent.

Explanation:

When you are heating a solvent, the intermolecular forces are reduced because the distances between molecules are large. Thus, in a solution where solvent is hot the intermolecular forces between solute and solvent are lower than those solutions where solvent is in room temperature.

The covalent bonds do not change because this mean a chemical reaction that doesn't occur in a solution.

Usually solid solutes melts in a higher temperature than boiling point in solvents. Thus, a compound normally doesn't melt in a hot solvent.

I hope it helps!

Final answer:

When a compound is dissolved in hot ethanol during recrystallization, the intermolecular forces between the solute and solvent are changing on a molecular level. The process of solution formation involves the formation of new intermolecular forces between the solute and solvent molecules, allowing the compound to dissolve in the solvent.

Explanation:

Intermolecular forces are attractive or repulsive forces between molecules. They include London dispersion forces, dipole-dipole interactions, and hydrogen bonding. These forces influence physical properties like boiling points and solubility in molecular substances. The intermolecular forces between the solute and solvent are changing on a molecular level when a compound is dissolved in hot ethanol during recrystallization.

When the compound is dissolved in the hot solvent, the solute-solvent interactions which is also known as solvatio then naturally takes place. In addition to this, these interactions involve the formation of new intermolecular forces between the solute and solvent molecules, which are nearly as strong as the intermolecular forces within the solute and solvent alone. This favorable solution formation process allows the compound to dissolve in the solvent.

Calculate the mass of manganese in potassium permanganate.

Answers

Answer:

The percentage mass of Manganese is 34.76%

Explanation:

The formula of potassium permanganate is KMnO4

Its molar mass is 158.034 g/mol.

The molar mass of Mn is 54.938 g/mol.

Percentage mass of Mn is:

[tex]Mn=\frac{54.938}{158.034}100[/tex]

Mn=34.76%

List and discuss two major sources of errors in numerical methods

Answers

Answer:

There are three major sources of errors in numerical method, namely human error, truncation error and round off error, but here we have to discuss only two sources-

Human error: This type of error occurs when the experiment is being performed but student isn't attentive during the experiment, or noted down the wrong observation, this error is very difficult to detect.There are rules for significant figures and rounding off, but if we round off the values incorrectly it will lead to error in final answer.

Answer:

sdfsdfsdf

Explanation:dsf

Air enters a diffuser witha velocity of 400 m/s, a pressure of 1 bar and temperature of 25 C. It exits with a temperature of 100 C. What is the exit velocity of the air? Assume there are no heat losses or change in potential energy Data:= 0.718 kJ/kg.°C. MW = 28.9 g/mol

Answers

Answer:

Exit velocity of air is 96.43 m/s.

Explanation:

Given that

[tex]V_1=400\ m/s[/tex]

[tex]T_1=25C[/tex]

[tex]T_2=100C[/tex]

For air

[tex]C_p=1.005\ KJ/kg.K[/tex]

Now from first law of thermodynamics for open system at steady state

[tex]h_1+\dfrac{V_1^2}{2000}+Q=h_2+\dfrac{V_2^2}{2000}+w[/tex]

For diffuser

[tex]h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}[/tex]

We know that

[tex]h=C_pT[/tex]

[tex]h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2000}[/tex]

[tex]1.005\times 25+\dfrac{400^2}{2000}=1.005\times 100+\dfrac{V_2^2}{2}[/tex]

[tex]V_2=96.43\ m/s[/tex]

So the exit velocity of air is 96.43 m/s.

Use the References to access imp A sample of hydrogen gas has a mass of 48.3 grams. How many H, molecules are there in the sample? molecules Submit Answer & Next

Answers

Answer: The number of hydrogen molecules in the given amount is [tex]1.454\times 10^{25}[/tex]

Explanation:

To calculate the number of moles, we use the equation:

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

Given mass of hydrogen gas = 48.3 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of hydrogen gas}=\frac{48.3g}{2g/mol}=24.15mol[/tex]

According to mole concept:

1 mole of a compound contains [tex]6.022\time 10^{23}[/tex] number of molecules

So, 24.15 moles of hydrogen gas will contain = [tex]24.15\times 6.022\times 10^{23}=1.454\times 10^{25}[/tex] number of molecules.

Hence, the number of hydrogen molecules in the given amount is [tex]1.454\times 10^{25}[/tex]

2. Sitting on a bench top are several samples: lithium metal (d = 0.53 g/mL), gold (d = 19.3 g/mL), aluminum (d = 2.70 g/mL), and lead (d = 11.3 g/mL). If all of the samples have the same mass, which one occupies the largest volume? Why?

Answers

Answer:

The sample of lithium occupies the largest volume.

Explanation:

Given the densities for the four elements, we have the expression [tex]d=\frac{m}{V}[/tex] that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

[tex]d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL[/tex]

[tex]d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL[/tex]

[tex]d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL[/tex]

[tex]d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL[/tex]

The problem says that all the samples have the same mass, so:

[tex]m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m[/tex]

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

[tex]V_{lithium}=\frac{m}{d_{lithium}}[/tex]

[tex]V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m[/tex]

[tex]V_{lithium}=1.88\frac{mL}{g}*m[/tex]

[tex]V_{gold}=\frac{m}{d_{gold}}[/tex]

[tex]V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m[/tex]

[tex]V_{gold}=5.18*10^{-2}\frac{mL}{g}*m[/tex]

[tex]V_{aluminum}=\frac{m}{d_{aluminum}}[/tex]

[tex]V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m[/tex]

[tex]V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m[/tex]

[tex]V_{lead}=\frac{m}{d_{lead}}[/tex]

[tex]V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m[/tex]

[tex]V_{lead}=8.85*10^{-2}\frac{mL}{g}*m[/tex]

If we assume m = 1g, we find that:

[tex]V_{lithium}=1.88mL[/tex]

[tex]V_{gold}=5.18*10^{-2}mL[/tex]

[tex]V_{aluminum}=3.70*10^{-1}mL[/tex]

[tex]V_{lead}=8.85*10^{-2}mL[/tex]

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.

A 1.28-kg sample of water at 10.0 °C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg at 215 °C into it. After the sizzling subsides, what is the final equilibrium temperature? (Make the reasonable assumptions that any steam produced condenses into liquid water during the process of equilibration and that the evaporation and condensation don’t affect the outcome, as we’ll see in the next section.)

Answers

Answer:

[tex]T_{2}=16,97^{\circ}C[/tex]

Explanation:

The specific heats of water and steel are  

[tex]Cp_{w}=4.186 \frac{KJ}{Kg^{\circ}C}[/tex]

[tex]Cp_{s}=0.49 \frac{KJ}{Kg^{\circ}C}[/tex]

Assuming that the water and steel are into an adiabatic calorimeter (there's no heat transferred to the enviroment), the temperature of both is identical when the system gets to the equilibrium [tex]T_{2}_{w}= T_{2}_{s}[/tex]  

An energy balance can be written as

[tex] m_{w}\times Cp_{w}\times (T_{2}- T_{1})_{w}= -m_{s}\times Cp_{s}\times (T_{2}- T_{1})_{s} [/tex]  

Replacing

[tex] 1.28Kg\times 4.186\frac{KJ}{Kg^{\circ}C}\times (T_{2}-10^{\circ}C)= -0.385Kg\times 0.49 \frac{KJ}{Kg^{\circ}C} \times (T_{2}-215^{\circ}C)[/tex]

Then, the temperature [tex]T_{2}=16,97^{\circ}C[/tex]

The equilibrium temperature given that You drop a piece of steel at 215 °C into the water at 10.0 °C in the calorimeter is 16 °C

How to calculate the equilibrium temperature?

The equilibrium temperature given that You drop a piece of steel at 215 °C into the water at 10.0 °C can be calculated as follow:

Mass of steel (M) = 0.385 KgTemperature of steel (T) = 215 °CSpecific heat capacity of steel (C) = 420 J/kg°CMass of water (Mᵥᵥ) = 1.28 KgTemperature of water (Tᵥᵥ) = 10.0 °CSpecific heat capacity of water (Cᵥᵥ) = 4184 J/Kg°CEquilibrium temperature of steel and water mixture (Tₑ) =?

[tex]MC(T - T_e) = M_wC_w(T_e - T_w)\\\\0.385\ \times\ 420(215 - T_e) = 1.28\ \times\ 4184(T_e - 10)[/tex]

[tex]34765.5 - 161.7T_e = 5355.52T_e - 53555.2\\\\34765.5\ +\ 53555.2 = 5355.52T_e\ +\ 161.7T_e\\\\88320.7 = 5517.22T_e\\\\T_e = \frac{88320.7}{5517.22} \\\\T_e = 16\ \textdegree C[/tex]

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For the following pair, indicate which element has the lower first ionization energy: Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer.

1. Given the elements Cl and Ge ,_________ has the smaller first ionization energy.
2. Given the elements Te and Se , __________ has the smaller first ionization energy.
3. Given the elements Ba and Ti , __________ has the smaller first ionization energy.
4. Given the elements Cu and Ag ,___________ has the smaller first ionization energy.

to fill in the blank:

(A) Ag
(B) Ge
(C) Cl
(D) Ti
(E) Cu
(F) Ba
(G) Se
(H) Te

Answers

Answer:

For 1: The correct answer is Ge.

For 2: The correct answer is Te.

For 3: The correct answer is Ba.

For 4: The correct answer is Ag.

Explanation:

Ionization energy is defined as the energy required to remove an electron from the outermost shell of an isolated gaseous atom. It is represented as [tex]E_i[/tex]

[tex]X(g)\rightarrow X^+(g)+1e^-;E_i[/tex]

Ionization energy increases as we move from left to right in a period. This happens because the atomic radius of an element decreases moving across a period, which increases the effective attraction between the negatively charged electrons and positively-charged nucleus. Hence, the removal of electron from the outermost shell becomes difficult and requires more energy.

Ionization energy decreases on moving from top to bottom in a group. This happens because the number of shells increases as we move down the group. The electrons get added in the new shell. So, the shielding of outermost electrons from the inner ones is more which decreases the attraction between the electrons and the nucleus. Hence, the removal of electron from the outermost shell becomes easy and requires less energy.

For the given options:

Option 1:

Chlorine is the 17th element of the periodic table belonging to Period 3 and Group 17.

Germanium is the 32nd element of the periodic table belonging to Period 4 and Group 14.

Hence, germanium will have smaller first ionization energy.

Option 2:

Tellurium is the 52nd element of the periodic table belonging to Period 5 and Group 16.

Selenium is the 34th element of the periodic table belonging to Period 4 and Group 16.

Hence, tellurium will have smaller first ionization energy.

Option 3:

Barium is the 56th element of the periodic table belonging to Period 6 and Group 2.

Titanium is the 22nd element of the periodic table belonging to Period 4 and Group 4.

Hence, barium will have smaller first ionization energy.

Option 4:

Copper is the 29th element of the periodic table belonging to Period 4 and Group 11.

Silver is the 47th element of the periodic table belonging to Period 5 and Group 11.

Hence, silver will have smaller first ionization energy.

Ionization energy increases across the period but decreases down the group.

First ionization energy of an element refers to the energy required to remove an electron from the atom. Ionization energy is a periodic trend that increases across the period but decreases down the group.

The following are true regarding the statements in the question;

Given the elements Cl and Ge, Ge has the smaller first ionization energyGiven the elements Te and Se, Te has the smaller first ionization energy.Given the elements Ba and Ti, Ba has the smaller first ionization energy. Given the elements Cu and Ag, Ag has the smaller first ionization energy.

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Ethyl acetate can be prepared by an SN2 reaction. Draw the alkylbromide and nucleophile used in the reaction. Remember to include formal charges (do not include counterions).

Answers

Answer:

Few important points related to [tex]S_N2[/tex] reaction:

1. [tex]S_N2[/tex] is a one-step reaction that follows second order kinetics.

2. In [tex]S_N2[/tex] reaction, a transition state is formed in situ.

3. Strong nucleophiles like [tex]OH^- \ or\  CN^-[/tex] are used in case of bi-molecular nucleophilic substitution reaction.

Ethyl acetate can be prepared by a second-order nucleophilic substitution reaction between acetic acid and ethyl bromide.  

The reaction between acetic acid and ethyl bromide is drawn below:

Final answer:

In the SN2 reaction to create ethyl acetate, the nucleophile acetate ion attacks the alkyl bromide, leading to the formation of ethyl acetate and bromide ions.

Explanation:

Ethyl acetate can be prepared by an SN2 reaction. In this context, an alkylbromide and nucleophile initiate the reaction. Ethyl bromide (CH3CH2Br) would be the alkylbromide used in the reaction. The nucleophile, in this case, would be acetate ion (CH3COO-).

Ethyl bromide is a good leaving group and acetate ion is a strong nucleophile, which comes from acetic acid (ethanoic acid). In SN2 reactions, the nucleophile attacks the substrate, and the leaving group (Br- in this case) leaves, leading to a reverse configuration. In this reaction, the nucleophile (acetate ion) will attack the alkyl bromide, resulting in the formation of ethyl acetate, CH3COOCH2CH3, and bromide ions, which is a good leaving group.

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If a substance increases the OH- concentration of pure water by 10^4. what is the pH of the solution?

Answers

Answer:

pH = 11

Explanation:

The concentration of OH⁻ in pure water is 10⁻⁷ M. If a substance increases OH⁻ concentration by 10⁴, the new concentration will be:

[OH⁻] = 10⁴ x 10⁻⁷ M = 10⁻³M

We can calculate pOH using it's definition:

pOH = -log[OH⁻] = -log (10⁻³) = 3

Then, we can find out pH using the following relation:

pH + pOH = 14

pH = 14 - pOH = 14 -3 = 11

Since pH = 11 is higher than 7, we can confirm that the substance is a base.



As the volume of confined gas decreases at the constant temperature, the pressure exerted by the gas________________.
Fluctuates
decreases
Increases
stays the same

Answers

As the volume of the confined gas decreases at the constant temperature, the pressure exerted by the gas increases.

As the volume of confined gas decreases at the constant temperature, the pressure exerted by the gas increases .  The correct option is C.

What is the Boyle's law?

Boyle–Mariotte law is a gas law, shows the relation between pressure and volume. With the increase in the volume, the pressure decreases.

The pressure exerted by the mass, is inversely proportional to the volume of the gas.

Thus, the correct option is C, increases.

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What type of chemical bond would form between an atom of carbon (C) and an atom of nitrogen (N). Explain specifically why this type of bond would form.

Answers

Answer:

Due to the low difference of electronegativity in carbon and nitrogen, these will form a covalent bond, and this covalent union of C-N is one of the most common bonds in the organic chemistry and biological systems. This type of bonds can be found in amines, amides, imines, etc. Also, a single atom of C and a single atom of N can form a cyanide, that is a triple covalent bond between the atoms.

Conversion of flow quantities by the continuity equation Water flows through a 3-inch, schedule 40 steel pipe (inside diameter- 3.068 inches) with an average velocity of 4 ft/s. Compute: (a) mass flow rate in lb/h, (b) volumetric flow rate in ft /h, and (c) mass velocity (mass flux) in lb/h-ft.

Answers

Answer:

a) m  = 2398256.64 Ib/hr

b) Q = 38433.6 ft^3/hr

c) J = = 898560 lb/h-ft2

Explanation:

a) To get the mass flow rate in lb/h we are going to use this formula:

m = ρνA

note: we have to change some units to reach to the final units

when m is the mass flow rate which we need to calculate

ρ is the density of water in lb/ft3 = 62.4 lbs/ft3 it is a constant numbers you can get it from anywhere

ν is the velocity of the water = 4 ft/s we need to change it to ft/h unit so  

when 1 ft/s = 3600 ft/hr (note : 3600 is a result of converting hr to sec 60min*60sec)

so  the velocity of the water = 4 ft/s * 3600 ft/hr= 14400 ft/hr

now , we need to get the area of the pipe in feet also when:

A = π*diameter * length

diameter = 3.068 inches

when 1 inch = 0.0833333333 feet so,

diameter in feet = 3.068 * 0.0833333333 =.300 0.2557ft

and length = 40 * 0.0833333333  = 3.333 ft

so the area of the pipe in feet^2 = 3.14*0.255*3.33 =2.669 ft^2

by substitution in the mass flow rate:

m = 62.4 lbs/ft3 * 14400 ft/hr *2.669 ft^2 = 2398256.64 Ib/hr

b) now to calculate volumetric flow rate in ft /h  we are going to use this formula:

Q = AV

when Q is  the volume flow rate

A is the cross-sectional area filled by water =2.669 ft^2

V is the average velocity of the water = 14400 ft/hr

so , by substitution:

Q = 2.669  ft^2* 14400 ft/hr= 38433.6 ft^3/hr

C) To calculate mass velocity (mass flux) in lb/h-ft  we are going to use this formula:

J = m / A

when J is the mass flux

and m is the mass flow rate which we calculated above = 2398256.64 Ib/hr

and A is the cross section area of the pipe which we calculated above and = 2.669 ft^2

so, by substitution:

J = 2398256.64 Ib/hr /  2.669 ft^2

 = 898560 lb/h-ft2

We know that one mole of Oxygen gas contains two moles of O atoms (remember that O is a diatomic element in nature and is written O2). If we have a sample of O2 that contains 1.0000 x 10-20 mol, we will have (blank)
atoms of O in the sample.

Answers

Answer:

12.4 × 10∧3 atoms

Explanation:

Given data:

moles of oxygen molecule= 1.0000 x 10-20 mol

atoms =?

Solution:

32 g O2 = 1 mol = 6.02 × 10∧23

1.0000 x 10∧-20 mol × 6.02 × 10∧23 × 2 = 12.4 × 10∧3 atoms

Vanessa pours 52 kilograms of table sugar into 100 mL of coffee. Some of the sugar dissolves, but most of it does not dissolve. The coffee solution is... a....unsaturated with sugar. Ob...saturated with sugar.

Answers

b ) saturated with sugar
Final answer:

The coffee solution is unsaturated with sugar.

Explanation:

A sugar and water solution is created when sugar dissolves in water. In this case, Vanessa pours 52 kilograms of table sugar, which is the solute, into 100 mL of coffee, which is the solvent. Since most of the sugar does not dissolve, the coffee solution can be considered unsaturated with sugar.

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A weak acid. What is the pH of a 0.1 M solution of acetic acid (pKa = 4.75)?
(Hint: Let x be the concentration of H+ ions released from acetic acid when it dissociates. The solutions to a quadratic equation of the form ax^2 + bx + c = 0 are x = (-b +- squareroot (b^2- 4ac)/2a.

Answers

Answer:

pH of acetic acid solution is 2.88

Explanation:

[tex]pK_{a}=4.75[/tex]

or, [tex]-log(K_{a})=4.75[/tex]

or, [tex]K_{a}=10^{-4.75}=1.78\times 10^{-5}[/tex]

We have to construct an ICE table to determine concentration of [tex]H^{+}[/tex] and corresponding pH. Initial concentration of acetic acid is 0.1 M.

[tex]CH_{3}COOH\rightleftharpoons CH_{3}COO^{-}+H^{+}[/tex]

I(M):  0.1                            0                    0

C(M): -x                              +x                 +x

E(M): 0.1-x                         x                    x

So, [tex]\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}=K_{a}[/tex]

or, [tex]\frac{x^{2}}{0.1-x}=1.78\times 10^{-5}[/tex]

or, [tex]x^{2}+(1.78\times 10^{-5}\times x)-(1.78\times 10^{-6})=0[/tex]

So, [tex]x=\frac{-(1.78\times 10^{-5})+\sqrt{(1.78\times 10^{-5})^{2}+(4\times 1\times 1.78\times 10^{-6})}}{2\times 1}[/tex](M)

so, [tex]x=1.33\times 10^{-3}M[/tex]

Hence [tex]pH=-log[H^{+}]=-log(1.33\times 10^{-3})=2.88[/tex]

Water flows at the rate of 0.5 kg/s in a 2.5-cm-diameter tube having a length of 4 m. A constant heat flux is imposed at the tube wall so that the tube wall temperature is always 50°C higher than the water temperature. Calculate the heat transfer rate and estimate the temperature rise in the water as it exits the pipe if the water enters at 5°C. The water is pressurized so that boiling cannot occur. (Assume h 3000 W/m2 oC). Note the interplay between heat transfer and thermo!!

Answers

Answer:

Q = 47.1 kW

Texit = 25.93ºC

Explanation:

The heat transfer is occurring by convection, which means that the heat is flowing from different materials (tube for water). So, by Newton Cooling Law:

Q = h x A x ΔT

The area of the tube will be :

A = πdL

Where d is the diameter and L is the length.

d = 2.5 cm = 0.025 m

A = 3.14x0.025x4 = 0.314 m²

Then

Q = 3000x0.314x50

Q = 47100 W = 47.1 kW

By the heat equation for water, knowing that the specific heat (c) of water is 4.5 kJ/ºC

Q = mcΔT

47.11 = 0.5x4.5x(Texits - Tenters)

47.1 = 2.25x(Texits - 5ºC)

Texits - 5ºC = 20.93 ºC

Texits = 25.93ºC

47.1 kW is the rate of heat transfer and 25.93ºC is the temperature of the water at exit.

How we calculate the heat transfer?

We can calculate the rate of heat transfer of the water by using the Newton's cooling law as:

Q = h x A x ΔT, where

h = heat transfer coefficient = 3000 W/m².°C (given)

A = surface area of heat transfer

ΔT = change in temperature = 50°C (given)

First we calculate the area by using the below formula:

A = πdL, where

d = diameter of tube = 2.5cm = 0.025m

L = length of tube = 4m

A = 3.14 x 0.025 x 4 = 0.314 m²

Therefore, rate of heat transfer is:

Q = 3000 x 0.314 x 50 = 47100 W = 47.1 kW

Now we calculate the rise in temperature by using the below formula:

Q = mcΔT

47.11 = 0.5 x 4.5 x (Texits - Tenters)

47.1 = 2.25 x (Texits - 5ºC)

Texits - 5ºC = 20.93 ºC

Texits = 25.93ºC

Hence, 47.1 kW is the rate of heat transfer and 25.93ºC is the temperature of exit water.

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How many atoms are found in 4.20 g of Magnesium?

Answers

Answer:

There are 1.041×10²³ atoms in 4.20g of Magnesium.

Explanation:

To find the amount of atoms in 4.20 g of Magnesium we need de molar mass of Mg: 24.305 g/mol

According to Avogadro number there are 6.022×10²³ particles in 1 mol, so the number of atoms of Mg is:

[tex]4.20 g Mg*\frac{1 molMg}{24.305gMg} *\frac{6.022*10^{23}atoms }{1 mol Mg} = 1.041*10^{23}atoms Mg[/tex]

Answer: The number of atoms found in given amount of magnesium is [tex]1.042\times 10^{23}[/tex]

Explanation:

To calculate the number of moles, we use the equation:

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

Given mass of magnesium = 4.20 g

Molar mass of magnesium = 24.31 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of magnesium}=\frac{4.20g}{24.31g/mol}=0.173mol[/tex]

According to mole concept:

1 mole of an element contains [tex]6.022\times 10^{23}[/tex] number of atoms

So, 0.173 moles of an element contains = [tex]0.173\times 6.022\times 10^{23}=1.042\times 10^{23}[/tex] number of atoms

Hence, the number of atoms found in given amount of magnesium is [tex]1.042\times 10^{23}[/tex]

When 0.491 grams of a protein were dissolved in 44 mL of benzene at 24.4 degrees C, the osmotic pressure was found to be 50.9 torr. Calculate the molar mass of the protein.

Answers

Answer:

4057.85 g/mol

Explanation:

Hello, the numerical procedure is shown in the attached file.

- In this case, since we don't have the density of the protein, we must assume that the volume of the solution is solely given by the benzene's volume, in order to obtain the moles of the solute (protein).

-Van't Hoff factor is assumed to be one.

Best regards.

Final answer:

To calculate the molar mass of the protein, use the formula M = (RT) / (V * P), where M is the molar mass, R is the ideal gas constant, T is the temperature, V is the volume, and P is the osmotic pressure.

Explanation:

To calculate the molar mass of the protein, we can use the formula:

M = (RT) / (V * P)

Where M is the molar mass, R is the ideal gas constant (0.0821 L.atm/mol.K), T is the temperature in Kelvin (24.4 + 273 = 297.4 K), V is the volume in liters (44 mL / 1000 = 0.044 L), and P is the osmotic pressure in atm (50.9 torr / 760 = 0.067 atm).

Substituting the values into the formula:

M = (0.0821 * 297.4) / (0.044 * 0.067) = 25567.14 g/mol

Therefore, the molar mass of the protein is approximately 25567.14 g/mol.

What is the chemical formula for Iron(III) hydroxide? Hydroxide is OH O a. Fe(OH)2 O b. FeOH Fe2OH O c. O d. Fe(OH)3 O e. Fe2(OH)3

Answers

Final answer:

The correct chemical formula for Iron(III) hydroxide is Fe(OH)3, which is formed by combining one iron ion with a +3 charge (Fe3+) with three hydroxide ions (OH-).

Explanation:

The chemical formula for Iron(III) hydroxide is Fe(OH)3. This compound consists of an iron ion with a +3 charge (Fe3+) and three hydroxide ions (OH-). The number in parentheses after iron indicates the oxidation state of iron, which in this case is +3. The formula reflects the stoichiometry required to balance the charges, resulting in one Fe3+ for every three OH- ions to form the neutral compound. Therefore, the correct answer is Fe(OH)3.

The correct answer is d. Fe(OH)₃.

To determine the correct chemical formula for Iron(III) hydroxide, we need to understand the charge on the Iron(III) ion and the hydroxide ion. Iron(III) (Fe³⁺) has a +3 charge, and the hydroxide ion (OH⁻) has a -1 charge. To balance the charges, we need three hydroxide ions to balance the +3 charge of one Iron(III) ion.The chemical equation is as follows: Fe³⁺ (aq) + 3OH⁻ (aq) → Fe(OH)₃(s).Thus, the correct formula for Iron(III) hydroxide is Fe(OH)₃. The option that correctly represents this formula is d. Fe(OH)₃.

What is an isochoric process? (1 mark) b) Can heat be exchanged in an isochoric process? (1 mark) c) A 100L container holding an ideal gas at an initial pressure of 10MPa is raised to a pressure of 15MPa. How much work is done?

Answers

Explanation:

(a)  An isochoric process is defined as the process in which volume of a system remains constant.

For example, an air conditioner follows an isochoric process.

(b) As pressure and temperature can change in an isochoric process but volume will remain the same.

Hence, heat can also be exchanged in an isochoric process. Whereas in an adiabatic process heat remains 0. Therefore, it means the in adiabatic process there occurs no exchange of heat between the system and its surrounding.

(c) It is given that V = 100 L = 0.1 [tex]m^{3}[/tex]   (as 1 L = 0.001 [tex]m^{3}[/tex])

               [tex]P_{1}[/tex] = [tex]10 \times 10^{6} Pa[/tex]

               [tex]P_{2}[/tex] = [tex]15 \times 10^{6} Pa[/tex]

Relation between work, pressure and volume is as follows.

                     W = [tex]\Delta PV[/tex]

                         = PdV + VdP

                          = 0 + 0.1 [tex]m^{3} (15 - 10) \times 10^{6}[/tex]

(Since V = constant so, dV = 0)

                          = [tex]5 \times 10^{5} J[/tex]

Thus, we can conclude that work done is [tex]5 \times 10^{5} J[/tex].

2.4 Bromium has two naturally occurring isotopes: 79Br, with an atomic weight of 78.918 amu, and 81Br, with an atomic weight of 80.916 amu. If the average atomic weight for Br is 79.903 amu, calculate the fraction-of-occurrences of these two isotopes.

Answers

Answer:

49.3% 81Br and 50.7% 79Br or [tex]\frac{493}{1000}[/tex] 81Br and [tex]\frac{507}{1000}[/tex] 79BR

Step-by-step explanation:

To get the fraction-of-occurrences of the isotopes we must write the following equation. x is the isotopic abundance of 81Br, we can use 1 - x to get the isotopic abundance of 79Br.

(78.918)(1 - x) + (80.916)(x) = 79.903

78.918 - 78.918x + 80.916x = 79.903

1.998x = 0.985

x = 0.493

0.493 × 100 = 49.3% 81Br

100 - 49.3 = 50.7% 79Br

49.3% 81Br and 50.7% 79Br or [tex]\frac{493}{1000}[/tex] 81Br and [tex]\frac{507}{1000}[/tex] 79BR

What volume of concentrated (10.2 M) HCl would be required to prepare 1.11 x 104 mL of 1.5 M HC1? Enter your answer in scientific notation. Be sure to answer all parts. x 10 (select) L

Answers

Answer:

The required volume is 1.6 x 10³mL.

Explanation:

When we want to prepare a dilute solution from a concentrated one, we can use the dilution rule to find out the required volume to dilute. This rule states:

C₁ . V₁ = C₂ . V₂

where,

C₁ and V₁ are the concentration and volume of the concentrated solution

C₂ and V₂ are the concentration and volume of the dilute solution

In this case, we want to find out V₁:

C₁ . V₁ = C₂ . V₂

[tex]V_{1} = \frac{C_{2}.V_{2}}{C_{1}} = \frac{1.5M \times1.11.10^{4}mL }{10.2M} =1.6\times10^{3} mL[/tex]

Alkynes are hydrocarbons that have.. O a.... at least one triple bond between carbon atoms O b. None of the above. O c. ...all single bonds between carbon atoms. O d....at least one double bond between carbon atoms

Answers

Answer: Option (a) is the correct answer.

Explanation:

Alkynes are the hydrocarbons, that is, they contain only atoms of carbon and hydrogen. Their general chemical formula is [tex]C_{n}H_{2n-2}[/tex], where n is a whole number.

A compound that contains a double or triple bond is known as an unsaturated compound. An alkyne contains a triple bond and it is also an unsaturated compound.

Thus, we can conclude that alkynes are hydrocarbons that have at least one triple bond between carbon atoms.

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