A B C D 1) NaNH2 2) MeI 3) 9-BBN 4) H2O2, NaOH 1) Br2 2) Excess NaNH2 3) H2O 1) Br2 2) Excess NaNH2 3) H2O 4) H2SO4, H2O, HgSO4 1) NaNH2 2) EtI 3) Na, NH3 (l) E F G H 1) 9-BBN 2) H2O2, NaOH 1) Excess NaNH2 2) H2O 3) Br2 (1 eq), CCl4 1) Excess NaNH2 2) H2O 3) NaNH2 4) MeI 5) Na, NH3 (l) 1) NaNH2 2) 3) H2, Pt

Answers

Answer 1

Answer:

The question has some part missing which I have added in the attachment.

Explanation:

The reactions given occur under certain reagents, which we are told to pick for each of the reaction synthesis. Conventionally, halogens have the ability to undergo addition reactions with hydrocarbons by breaking down the double or triple bond in them to a single bond, this usually occur by electron donation and electron acceptor.

The attachment shows the reactions and the necessary reagents required for each

A B C D 1) NaNH2 2) MeI 3) 9-BBN 4) H2O2, NaOH1) Br2 2) Excess NaNH2 3) H2O1) Br2 2) Excess NaNH2 3)

Related Questions

The vapor pressure of benzene (C6H6) is 73.0 mm Hg at 25 °C. What is the vapor pressure of a solution consisting of 179 g of benzene and 0.217 mol of a nonvolatile nonelectrolyte?

Answers

Answer:

Vapor pressure of solution is 66.7 mmHg

Explanation:

Colligative property about vapor pressure lowering. That's we must use to solve this problem.

Formula is: ΔP = P° . Xm

P° is Vapor pressure of pure solvent

ΔP = P° - P' (vapor pressure of solution)

Xm = mole fraction of solute (mol of solute / total moles)

Let's determine the total moles, firstly.

Total moles = moles of solute + moles of solvent

Moles of solute → 0.217 mol

Moles of solvent → 179 g / molar mass of benzene

179 g / 78 g/mol = 2.29  mol

2.29 mol + 0.217 mol = 2.507 moles

Xm for solute = 0.217 mol / 2.507 mol = 0.0865

Let's replace the data in the formula:

73 mmHg - P' = 73 mmHg  . 0.0865

P' = - (73 mmHg  . 0.0865  - 73mmHg)

P' = 66.7 mmHg

For your research project, your group is planning to treat embryos with acetaminophen at a concentration of 50 ug/L plus varying concentrations of dextromethorphan between O and 100 uM. You have stock solutions of 500 ug/L acetaminophen and 1 mM dextromethorphan to work with. If you wanted to treat some embryos with final concentrations of 50 ug/L acetaminophen and 20 uM dextromethorphan in a total volume of 10 ml, how would you make up 9 ml of solution containing both acetaminophen and dextromethorphan, to which you could add 1 ml of embryo water containing the embryos? A. Add 0.5ml of 500 ug/1 acetaminophen and 0.5 ml of 1 mM dextromethorphan to 8 ml of water B. Add 1 ml of 500 ug/l acetaminophen and 0.2 ml of 1 mM dextromethorphan to 8.8 ml of water. C. Add 0.1 ml of 500 ug/l acetaminophen and 0.5 ml of 1 mM dextromethorphan to 8.4 ml of water. D. Add 1 ml of 500 ug/I acetaminophen and 0.2 ml of 1 mM dextromethorphan to 7.8 ml of water. E. Add 1 ml of 500 ug/I acetaminophen and 2 ml of 1 mM dextromethorphan to 6 ml of water.

Answers

Answer:

D

Explanation:

When a dilution is made, a volume of the stock solution is collected and then is mixed to the solvent. The total amount of the solute (number of moles or mass), must be equal in the volume of the sample and at the final volume, because of the Lavoiser's law (the matter can't be created nor destructed).

The mass or the number of moles is the concentration (C) multiplied by the volume (V), so, if 1 is the sample of the stock solution, and 2 the diluted solution:

C1*V1 = C2*V2

The final volume of the solution is 10 mL. So, let's identify the volume needed for each stock solution.

Acetaminophen

C1 = 500 ug/L

C2 = 50 ug/L

500*V1 = 50*10

V1 = 1 mL

Dextromethorphan

C1 = 1 mM = 1000 uM

C2 = 20 uM

1000*V1 = 20*10

V1 = 0.2 mL

So, the volume of water needed is the total less the volume of the stocks solutions less the volume of the embryo water:

V = 10 - 1 - 1 - 0.2 = 7.8 mL

Thus, to to the solution, it's necessary to add at 1 mL of the embryo water 1 ml of 500 ug/I acetaminophen and 0.2 ml of 1 mM dextromethorphan, and 7.8 ml of water.

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooling system holds 5.60 gal, what is the boiling point of the solution? For the calculation, assume that at normal filling conditions, the densities of engine coolant and water are 1.11 g/mL and 0.998 g/mL respectively. Also, assume that the engine coolant is pure ethylene glycol ( HOCH 2 CH 2 OH ) , which is non‑ionizing and non‑volatile, and that the pressure remains constant at 1.00 atm. The boiling‑point elevation constant for water will also be needed.

Answers

Answer:

109.09°C

Explanation:

Given that:

the capacity of the cooling car system = 5.6 gal

volume of solute = volume of the water; since a 50/50 blend of engine coolant and water (by volume) is used.

∴ [tex]\frac{5.60}{2}gallons = 2.80 gallons[/tex]

Afterwards, the mass of the solute and the mass of the water can be determined as shown below:

mass of solute = [tex](M__1}) = Density*Volume[/tex]

                          [tex]= 1.1g/mL *2.80*\frac{3785.41mL}{1gallon}[/tex]

                         [tex]= 11659.06grams[/tex]

On the other hand; the mass of water = [tex](M__2})= Density*Volume[/tex]

                         [tex]= 0.998g/mL *2.80*\frac{3785.41mL}{1gallon}[/tex]

                        [tex]= 10577.95 grams[/tex]

Molarity = [tex]\frac{massof solute*1000}{molarmassof solute*massofwater}[/tex]

              =  [tex]\frac{11659.06*1000}{62.07*10577.95}[/tex]

              = 17.757 m

              ≅ 17.76 m

∴  the boiling point of the solution is calculated using the  boiling‑point elevation constant for water and the Molarity.

[tex]\Delta T_{boiling} = k_{boiling}M[/tex]

where,

[tex]k_{boiling}[/tex] = 0.512 °C/m

[tex]\Delta T_{boiling}[/tex] =  100°C + 17.56 × 0.512

              = 109.09 °C

Determine the electric field (magnitude and direction) at the point A (8.00 nm, 6.00 nm) caused by a particle located at the origin and carrying a charge of 7.00 μC .

Answers

Answer:

E1 = 9.83 x [tex]10^{20}[/tex] [tex]NC^{-1}[/tex]

E2 = 1.748 x [tex]10^{21}[/tex] [tex]NC^{-1}[/tex]

Explanation:

E1 = k Q/r2 = 8.99 x [tex]10^{9}[/tex] x 7 x [tex]10^{-6}[/tex] / 8 x [tex]10^{-9}[/tex] x 8 x [tex]10^{-9}[/tex] = 9.83 x [tex]10^{20}[/tex] [tex]NC^{-1}[/tex]

E2 = k Q/r2 = 8.99 x [tex]10^{9}[/tex] x 7 x  [tex]10^{-6}[/tex] / 6 x  [tex]10^{-9}[/tex] x 6 x [tex]10^{-9}[/tex] = 1.748 x [tex]10^{21}[/tex] [tex]NC^{-1}[/tex]

The direction of the electric field will be from E1 to E2...

A solution of NaCl ( aq ) is added slowly to a solution of lead nitrate, Pb ( NO 3 ) 2 ( aq ) , until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 12.11 g CaCl2 ( s ) is obtained from 200.0 mL of the original solution.

Answers

Answer:

0.218 M of Pb(NO3)2

Explanation:

Equation of the reaction

Pb(NO3)2(aq) + 2NaCl(aq) --> PbCl2(s) + 2NaNO3(aq)

1 mole of Pb(NO3)2 reacts to precipitate 1 mole of PbCl2

Molar mass of PbCl2 = 207 + (35.5*2)

= 278 g/mol

Number of moles of PbCl2 precipitated = mass/molar mass

= 12.11/278

= 0.04356 mol

Since 0.04356 moles of PbCl2 was precipitated, therefore by stoichiometry; 0.04356 moles of Pb(NO3)2 reacted.

Molarity is defined as the number of moles of solute in 1 liter of solution.

Molarity = number of moles/volumes

= 0.04356/0.2

= 0.218 M

A 3.3×10-2 mg sample of a protein is dissolved in water to make 0.25 mL of solution. The osmotic pressure of the solution is 0.56 torr at 25°C. What is the molar mass of the protein?

Answers

Answer: 4376.6g/mol

Explanation:Please see attachment for explanation

The molar mass of the protein is 4376.6 g/mol.

What is molar mass?

The molar mass of a chemical compound is defined as the mass of a sample divided by the amount of substance in that sample, measured in moles.

Solution:

Given, the mass of protein is 3.3×10-2 mg

Mass converted into gram = 3.3×10-2 mg = M 0.000033 g

Therefore, xg of the protein dissolve in 1 L

xg pf protein =[tex]\bold{\dfrac{0.000033}{0.00025} = 0.132\; g/L}[/tex]

Given, the osmotic pressure of 0.56 torr.

[tex]\bold{\pi = 0.56\;torr =\dfrac{0.56}{760} = 0.00737\; atm p}[/tex]

Thus, the volume of the solution is 0.007 L

Given, Temperature = 25 °C

Converting into kelvin = 25+273 = 298 k.

R = 0.082 atm.

Now, by the formula  [tex]\bold{M_2= \dfrac{W_2RT}{\pi V}}[/tex]

[tex]\bold{M_2= \dfrac{0.132\times0.082\times298}{0.00737\times0.007} =4376.6 \;g/mol}[/tex]

Thus, the molar mass of protein is 4376.6 g/mol.

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Consider the market for peanut butter. If there is a decrease in the price of deli turkey slices (a
substitute in consumption for peanut butter) along with a decrease in the price of peanut brittle
(a substitute in production for peanut butter), the
A) equilibrium quantity of peanut definitely decreases.
B) equilibrium quantity of peanut butter definitely increases.
C) equilibrium price of peanut butter definitely falls.
D) equilibrium price of peanut butter definitely rises.
E) equilibrium price of peanut butter might rise or fall

Answers

Answer:

The correct answer is option C) "equilibrium price of peanut butter definitely falls".

Explanation:

The equilibrium price of a product is the market price established at a point where the quantity of products is equal to the quantity demanded by the consumers. In this case, the market of peanut butter is facing the decrease in the price of two substitutes of peanut butter: deli turkey slices and peanut brittle. As a result, the equilibrium price of peanut butter definitely falls since the quantity demanded by the consumers will certainly fall.

Final answer:

A decrease in the price of deli turkey slices and peanut brittle affects the demand and supply for peanut butter differently, making the final impact on the market for peanut butter ambiguous without further information.

Explanation:

The question considers the effects of changes in the prices of substitute goods in consumption (deli turkey slices) and production (peanut brittle) on the market for peanut butter. A decrease in the price of deli turkey slices, a substitute in consumption, would lead to consumers substituting away from peanut butter towards deli turkey slices, thereby decreasing the demand for peanut butter. Conversely, a decrease in the price of peanut brittle, a substitute in production, might signal that manufacturers could switch to producing more peanut brittle as it becomes more profitable, potentially decreasing the supply of peanut butter if resources are diverted. Consequently, the equilibrium quantity of peanut butter might increase or decrease, depending on the relative magnitudes of the shifts in supply and demand, while the equilibrium price might rise or fall for similar reasons. Therefore, without additional information on the extents of these changes, the final impact on the equilibrium price and quantity of peanut butter cannot be definitively determined, making option E the correct answer.

In a concentrated solution there is ____.a. no solvent c. a small amount of soluteb. a large amount of solute d. no solute

Answers

Answer:a large amount of solute

Explanation:

A solution is composed of a solute and a solvent. If the amount of solute is greater than that of solvent, the solution is concentrated. A concentrated solution contains quite a large amount of solute while a dilute solution contains less amount of solute. This is the difference between diluted and concentrated solutions.

Answer:

Answer:a large amount of solute

Explanation:

A solution is composed of a solute and a solvent. If the amount of solute is greater than that of solvent, the solution is concentrated. A concentrated solution contains quite a large amount of solute while a dilute solution contains less amount of solute. This is the difference between diluted and concentrated solutions.

Explanation:

Predict which member of each pair will be more acidic. Explain your answers. (a) methanol or tert-butyl alcohol (b) 2-chloropropan-1-ol

Answers

Answer:

A. Methanol

B. 2-chloropropan-1-ol

C. 2,2-dichloroethanol

D. 2,2-difluoropropan-1-ol

Explanation:

Primary alcohols are stronger acids than secondary alcohols which are stronger than tertiary alcohols.

This trend is so because of the stability of the alkoxide ion formed(stabilising the base, increases the acidity). A more stabilised alkoxide ion is a weaker conjugate base (dissociation of an acid in water).

By electronic factors, When there are alkyl groups donating electrons, the density of electrons on th O- will increase a d thereby make it less stable.

By stearic factors, More alkyl group bonded to the -OH would mean the bulkier the alkoxide ion which would be harder to stabilise.

Down the group of the periodic table, basicity (metallic character) decreases as we go from F– to Cl– to Br– to I– because that negative charge is being spread out over a larger volume that is electronegativity decreases down the group.

Electronegative atoms give rise to inductive effect and a decrease in indutive effects leads to a decrease in acidity. Therefore an Increasing distance from the -OH group lsads to a decrease in acidity.

From above,

A. Methanol

B. 2-chloropropan-1-ol

C. 2,2-dichloroethanol

D. 2,2-difluoropropan-1-ol

Final answer:

Methanol is more acidic than tert-butyl alcohol because it forms a more stable alkoxide ion on deprotonation. Likewise, 2-chloropropan-1-ol is more acidic than propan-1-ol due to the stabilizing effect of its chlorine atom on the alkoxide ion.

Explanation:

The acidity of an alcohol is determined by the stability of the resulting alkoxide ion on deprotonation. In other words, an alcohol that can form a more stable alkoxide ion will be more acidic.

(a) Between methanol and tert-butyl alcohol, methanol would be more acidic. This is because the alkoxide ion formed when methanol loses a proton is more stable. In tert-butyl alcohol, the large, bulky tert-butyl group hinders the solvation of the alkoxide ion, making it less stable and the alcohol less acidic.

(b) 2-chloropropan-1-ol would be more acidic than propan-1-ol because the electron-withdrawing chlorine atom stabilizes the alkoxide ion, making the alcohol more acidic.

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) If 23 g of FeCl2 reacts with 41 grams of Na3PO4, what is the limiting reagent? How much NaCl can be formed?

Answers

Answer: The limiting reagent is iron (II) chloride and the mass of sodium chloride formed is 21.2 grams

Explanation:

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]      .....(1)

For iron (II) chloride:

Given mass of iron (II) chloride = 23 g

Molar mass of iron (II) chloride = 126.8 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of iron (II) chloride}=\frac{23g}{126.8g/mol}=0.181mol[/tex]

For sodium phosphate:

Given mass of sodium phosphate = 41 g

Molar mass of sodium phosphate = 164 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of sodium phosphate}=\frac{41g}{164g/mol}=0.25mol[/tex]

The chemical equation for the reaction of iron (II) chloride and sodium phosphate follows:

[tex]3FeCl_2+2Na_3PO_4\rightarrow 6NaCl+Fe_3(PO_4)_2[/tex]

By Stoichiometry of the reaction:

3 moles of iron (II) chloride reacts with 2 moles of sodium phosphate

So, 0.181 moles of iron (II) chloride will react with = [tex]\frac{2}{3}\times 0.181=0.1206mol[/tex] of sodium phosphate

As, given amount of sodium phosphate  is more than the required amount. So, it is considered as an excess reagent.

Thus, iron (II) chloride is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of iron (II) chloride produces 6 mole of sodium chloride.

So, 0.181 moles of iron (II) chloride will produce = [tex]\frac{6}{3}\times 0.181=0.362moles[/tex] of sodium chloride.

Now, calculating the mass of sodium chloride from equation 1, we get:

Molar mass of sodium chloride = 58.5 g/mol

Moles of sodium chloride = 0.362 moles

Putting values in equation 1, we get:

[tex]0.362mol=\frac{\text{Mass of sodium chloride}}{58.5g/mol}\\\\\text{Mass of sodium chloride}=(0.362mol\times 58.5g/mol)=21.2g[/tex]

Hence, the limiting reagent is iron (II) chloride and the mass of sodium chloride formed is 21.2 grams

Final answer:

To determine the limiting reagent between FeCl2 and Na3PO4, the masses of the reactants are converted to moles and compared according to the stoichiometry of the balanced chemical equation. The reactant that produces the least amount of product is the limiting reagent, and using this information, the mass of NaCl produced can be calculated.

Explanation:

The student is asking which reactant is the limiting reagent in the reaction between FeCl2 and Na3PO4, and the amount of NaCl that can be formed as a result. To solve this, we must first balance the chemical equation and then convert the masses of the reactants to moles. After that, we use stoichiometry to compare the mole ratios and identify the limiting reactant. Lastly, we calculate the mass of NaCl produced by the reaction, based on the limiting reactant's moles.

Let's balance the equation: FeCl2 + Na3PO4 → Fe3(PO4)2 + NaCl. We don't have the balanced equation here, but typically, you'd see NaCl being produced alongside iron(III) phosphate. To compute the moles of each reactant, we use their molar masses (FeCl2 = 126.75 g/mol, Na3PO4 = 163.94 g/mol) and determine the mole ratio according to the balanced equation. The smaller mole ratio indicates the limiting reagent. Using the stoichiometry of the balanced reaction, we then calculate the amount of NaCl that can be formed.

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How many grams of Aldol product can be produced from the complete reaction of 0.2 grams of vanillin with an excess of acetone in the presence of aqueous base? Enter only the number with two significant figures.

Answers

Answer:

There is 0.25 grams of C11H12O3 produced

Explanation:

Step1: Data given

vanillin = C8H8O3

Mass of vanillin = 0.2 grams

Molar mass of vanillin = 152.15 g/mol

Acetone =  58.08 g/mol

Step 2: The balanced equation

C8H8O3 + C3H6O → H2O + C11H12O3

Step 3: Calculate moles of C8H8O3

Moles C8H8O3 = mass / molar mass

Moles C8H8O3 = 0.2 grams / 152.15 g/mol

Moles C8H8O3 = 0.0013 moles

Step 4: Calculate moles of C11H12O3

For 1 mol vanillin we need 1 mol acetone to produce 1 mol C11H12O3

Step 5: Calculate mass of C11H12O3

Mass C11H12O3 = moles * molar mass

Mass C11H12O3 = 0.0013 moles * 192.21 g/mol

Mass C11H12O3 = 0.25 grams

There is 0.25 grams of C11H12O3 produced

A vertical piston-cylinder device contains water and is being heated. During the heating process, 70 kJ of heat is transferred to the water, and heat losses from the side walls amount to 8 kJ. The piston rises as a result of evaporation, changing the volume by 0.5 m3. When the atmospheric pressure is 1 atm and the mass of the piston is 50 kg and its cross-sectional area is 0.05 m2, determine the change in the energy of the water for this process.

Answers

The change in the energy of the water for this process is calculated as 11.3375 kJ, considering the heat transferred and work done.

To determine the change in energy of the water within the piston-cylinder setup, we need to account for the heat transferred and the work done. The given values are:

Heat transferred to water: 70 kJHeat losses from the side walls: 8 kJVolume change: 0.5 m³Atmospheric pressure: 1 atm (101.325 kPa)Mass of the piston: 50 kgCross-sectional area of the piston: 0.05 m²

The net heat energy (Q) added to the system is:

[tex]Q_{net[/tex] = 70 kJ - 8 kJ = 62 kJ

For the work done by the system, we use:

W = P × ΔV

where W is work, P is pressure (101.325 kPa), and ΔV is the volume change (0.5 m³). Converting kPa to kJ by recognizing 1 kPa×m³ = 1 kJ, we get:

W = 101.325 kPa × 0.5 m³ = 50.6625 kJ

Using the first law of thermodynamics, the change in internal energy (ΔU) is:

ΔU = Qnet - W

Substituting the values, we get:

ΔU = 62 kJ - 50.6625 kJ = 11.3375 kJ

Therefore, the change in the energy of the water for this process is 11.3375 kJ.

A metallic object holds a charge of −4.8 × 10−6 C. What total number of electrons does this represent? (e = 1.6 × 10−19 C is the magnitude of the electronic charge.)

Answers

Answer:

[tex]n=3.0\times 10^{13}[/tex]

Explanation:

Charge on 1 electron = [tex]-1.6\times 10^{-19}\ C[/tex]

The expression for charge is:-

[tex]Charge=n\times q_e[/tex]

Given that:- Charge = [tex]-4.8\times 10^{-6}\ C[/tex]

[tex]-4.8\times 10^{-6}=n\times (-1.6\times 10^{-19})[/tex]

[tex]n=\frac{4.8\times 10^{-6}}{1.6\times 10^{-19}}=3.0\times 10^{13}[/tex]

Total number of electrons, n = [tex]3.0\times 10^{13}[/tex]

The total number of electrons the charge represent is 3.0 × 10¹³

Calculating number of electrons

From the question, we are to calculate the total number of electrons

Using the formula,

Q = ne

Where Q is the total charge

n is the number of electrons

and e is the charge of an electron

From the given information,

q = - 4.8 × 10⁻⁶ C

(NOTE: The negative sign indicates the type of charge)

e = 1.6 × 10⁻¹⁹ C

Then,

4.8 × 10⁻⁶ = n × 1.6 × 10⁻¹⁹

n = [tex]\frac{4.8 \times 10^{-6} }{1.6 \times 10^{-19} }[/tex]

n = 3.0 × 10¹³

Hence, the total number of electrons the charge represent is 3.0 × 10¹³

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How does the aufbau principle, in connection with the periodic law, lead to the format of the periodic table?

Answers

Final answer:

The Aufbau principle leads to the format of the periodic table by arranging elements based on their electron configurations and the number of valence electrons they have.

Explanation:

The Aufbau principle, in connection with the periodic law, leads to the format of the periodic table by arranging elements based on their electron configurations and the number of valence electrons they have. The periodic table is organized in a way that groups elements with similar properties in the same vertical columns, also known as groups. This arrangement follows the filling of subshells with electrons according to the Aufbau principle.

A molecular solid coexists with its liquid phase at its melting point. The solid-liquid mixture is heated, but the temperature does not change while the solid is melting.
The best explanation for this phenomenon is that the heat absorbed by the mixture:

A. is lost to the surroundings very quickly
B. is used in overcoming the intermolecular attractions in the solid
C. is used in breaking the bonds within the molecules of the solid
D. causes the nonbonding electrons in the molecules to move to lower energy levels

Answers

The best explanation for this phenomenon is that the heat absorbed by the mixture is lost to the surroundings very quickly. Hence option B is correct.

What is mixture?

Mixture is defined as a substance consisting of two or more unrelated chemicals that are not chemically linked. The two categories of mixtures are heterogeneous and homogeneous mixtures. While heterogeneous mixtures include distinct components, homogenous mixtures appear consistent throughout.

Although the mixture of solid and liquid is heated, the temperature remains constant as the solid melts. The best explanation for this phenomena is that the heat that the mixture absorbs is swiftly transferred to the environment and used to overcome the intermolecular forces.

Thus, the best explanation for this phenomenon is that the heat absorbed by the mixture is lost to the surroundings very quickly. Hence option B is correct.

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Final answer:

The heat absorbed by the solid-liquid mixture at its melting point is used in overcoming the intermolecular attractions in the solid, allowing it to transition into the liquid phase without an increase in temperature.

Explanation:

When a solid coexists with its liquid phase at its melting point, the heat absorbed by the mixture is used in overcoming the intermolecular attractions in the solid. As a solid is heated, the particles vibrate more rapidly and gain enough energy to overcome the attractive forces within the solid. This allows the solid to transition into the liquid phase without an increase in temperature. The heat added during melting is not lost to the surroundings quickly or used in breaking the bonds within the molecules of the solid, but rather in overcoming the intermolecular attractions in the solid.

What is a pseudo–noble gas configuration? Give an example of one ion from Group 3A(13) that has it.

Answers

Explanation:

Pseudo-noble gas configuration which can also be called pseudo inert configuration is when elements have fully filled d-orbital, along with s- and p- orbitals. Examples are ions of elements, this is because they lose their valence electrons to be fully filled in order to have a stable octet.

Example of Group 3A(13) ion is Aluminium ion; Al3+. Al3+ takes up the configuration of Neon when it loses its 3 valence electrons.

Electronic configuration of Al and Al3+

Al = [Ne] 3s2 3p1

Al3+ = 1s2 2s2 2p6

Final answer:

A pseudo-noble gas configuration occurs when an ion or atom has an electron configuration that resembles that of a noble gas. An example of an ion from Group 3A(13) that has a pseudo-noble gas configuration is aluminium (Al3+).

Explanation:

A pseudo-noble gas configuration refers to the electron configuration of an ion or atom that resembles that of a noble gas. This occurs when an ion gains or loses electrons to acquire the same electron configuration as a noble gas. An example of an ion from Group 3A(13) that has a pseudo-noble gas configuration is aluminium (Al3+).

Aluminum has an electron configuration of 1s2 2s2 2p6 3s2 3p1 in its ground state. When it loses its three valence electrons, its configuration becomes 1s2 2s2 2p6, which is the same as that of the noble gas, neon (Ne). Therefore, aluminium has a pseudo-noble gas configuration.

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Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. What mass of each substance is present after 21.75 g of calcium nitrate and 22.66 g of ammonium fluoride react completely

Answers

g calcium nitrate = 21.75 g, g ammonium fluoride = 22.66 g

g calcium fluoride = 10.34 g, g dinitrogen monoxide = 5.82 g, and

g water = 2.38 g.

The balanced chemical equation for the reaction is:

[tex]\rm 3 Ca(NO_3)2 + 6 NH_4F \rightarrow 6 NH_4NO_3 + CaF_2 + N_2O + 3 H_2O[/tex]

Given the molar masses:

[tex]\rm Ca(NO_3)2[/tex] = 164.09 g/mol

[tex]\rm NH_4F[/tex] = 37.04 g/mol

[tex]\rm CaF_2[/tex] = 78.08 g/mol

[tex]\rm N_2O[/tex] = 44.02 g/mol

[tex]\rm H_2O[/tex] = 18.02 g/mol

First, calculate the moles of each reactant:

Moles of [tex]\rm Ca(NO_3)2[/tex] = 21.75 g / 164.09 g/mol ≈ 0.1323 mol

Moles of [tex]\rm NH_4F[/tex] = 22.66 g / 37.04 g/mol ≈ 0.6113 mol

Based on the balanced equation, the limiting reactant is [tex]\rm Ca(NO_3)2[/tex], which reacts with 0.1323 moles.

Calculate the masses of the products:

g calcium fluoride = 0.1323 mol * 78.08 g/mol ≈ 10.34 g

g dinitrogen monoxide = 0.1323 mol * 44.02 g/mol ≈ 5.82 g

g water = 0.1323 mol * 18.02 g/mol ≈ 2.38 g

Since [tex]\rm NH_4F[/tex] is in excess, some of it remains unreacted:

Unreacted [tex]\rm NH_4F[/tex] = (0.6113 mol - 0.1323 mol) * 37.04 g/mol ≈ 21.90 g

In summary:

g calcium nitrate = 21.75 g

g ammonium fluoride = 22.66 g

g calcium fluoride = 10.34 g

g dinitrogen monoxide = 5.82 g

g water = 2.38 g

The reaction consumes calcium nitrate and ammonium fluoride to produce calcium fluoride, dinitrogen monoxide, and water vapour.

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How is it possible for a protein to change over 70% of its amino acids and still fold in the same way?

Answers

The complete question is

Comparison of a homeodomain protein from yeast and Drosophila shows that only 17 of its 60 amino acids are identical. How is it possible for a protein to change over 70% of its amino acids and still fold in the same way?

Answer:

Many different strings of the amino acid can give rise to the identical protein folds. The amino acid differences between the Drosophila and homeodomain proteins from yeast are the functional proteins which do not change their structure and function.

There is a several amount of folding required which can cause change in the protein structure.

There is a basic functional proteins common in both the organism which do not change its structure and function.

Fluoridation is the process of adding fluorine compounds to drinking water to help fight tooth decay. A concentration of 1 ppm of fluorine is sufficient for the purpose (1 ppm means one part per million, or 1 g of fluorine per 1 million g of water). The compound normally chosen for fluoridation is sodium fluoride, which is also added to some toothpastes. Calculate the quantity of sodium fluoride in kilograms needed per year for a city of 50,000 people if the daily consumption of water per person is 115.0 gallons. (Sodium fluoride is 45.0 percent fluorine by mass. 1 gallon = 3.79 L; 1 ton=2000lb; 1 lb= 453.6 g; density of water =1.0 g/mL)

Answers

Answer:

The quantity of sodium fluoride is 17,676,14 kilograms is needed per year for a city of 50,000 people.

Explanation:

Population of the city = 50,000

Volume of water consumed by an individual in day= 115.0 gallons

Volume of water consumed by 50,000 people in day: V

V = 115.0 gallons × 50,000 = 5,750,000 gallons

1 gallon = 3.79 L

[tex]V=5,750,000 gallons=5,750,000\times 3.79 L = 21,792,500 L[/tex]

[tex]V=21,792,500 L=21,792,500,000 mL[/tex]

( 1L = 1000 mL)

Mass of water = m

Density of water = d = 1.0 g/mL

[tex]m=d\times V=1.0 g/mL\times 21,792,500,000 mL=21,792,500,000 g[/tex]

Concentration of Fluorine in water = 1 ppm = 1 gram of F /1 million grams of water

Then mass of fluorine present in 21,792,500,000 g of water:

[tex]\frac{1}{10^6}\times 21,792,500,000 g=21,792.5 g[/tex]

Mass of sodium fluoride with 21,792.5 g of F = M

Percentage of fluorine in NaF = 45.0 %

[tex]45\%=\frac{21,792.5 g}{M}\times 100[/tex]

M = 48,427.78 g = 48.427 kg ( 1g = 0.001 kg)

48.427 kg of NaF should be added to water in day for city population.

Amount of NaF needed per year for city with 50,000 population :

(1 year = 365 days)

[tex]48.427 kg\times 365=17,676.14 kg[/tex]

The quantity of sodium fluoride is 17,676,14 kilograms is needed per year for a city of 50,000 people.

Final answer:

Requiring a fluorine concentration of 1 ppm in a city's water supply for 50,000 people, each consuming 115 gallons of water per day, would necessitate 17,677 kg of sodium fluoride per annum.

Explanation:

First, we'd need to calculate the total amount of water consumed by the entire city in a year. This would be 50,000 people x 115 gallons/person/day x 365 days/year = 2,098,750,000 gallons/year. After converting this to liters (1 gallon = 3.79 L), the total water consumption is 7,954,625,000 L/year or 7,954,625,000,000 g/year (since 1L of water = 1g).

For a fluorine concentration of 1 ppm (1 g of fluorine/1,000,000 g of water), the yearly requirement of fluorine is 7,954,625 g of fluorine. Sodium fluoride is 45.0 percent fluorine by mass, so to get this amount of fluorine, we will need 7,954,625 g / 0.45 = 17,676,944 g or 17,677 kg of sodium fluoride per year.

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B. Steve selects an amino acid that you wrote for part A and dissolves 0.1 moles in 1 liter of water. He adjusts the pH to 7.2, but then absentmindedly adds 0.03 more moles of HCl to the solution. Is the solution still a good buffer

Answers

Answer:

i asked to where my uncle shoes he told my  no

Explanation:

2. Mrs. Roberts, in a diabetic coma, has just been admitted to Noble Hospital. Her blood pH indicates that she is in severe acidosis, and measures are quickly instituted to bring her blood pH back within normal limits. (a) Define pH and note the normal pH of blood. (b) Why is severe acidosis a problem?

Answers

Answer:

Explanation:

a) The pH can be define as the measurement of the hydrogen or hydroxide ion concentration in a solution. The pH for blood is 7.4. In severe acidosis condition the pH of blood is 7.35 or lower.

b) Severe acidosis is a condition which is caused due to the overproduction of acid and building up of acid in the blood. This occurs due to the excessive loss of bicarbonate from the blood or buildup of carbon dioxide  in the blood. This leads to poor functioning of the lungs and depressed breathing.

The severe acidosis is critical condition because because it can adversely affect the cell membranes, muscle contraction, and function of the kidneys and neural activity of the body.  

Final answer:

pH is a scale used to measure how acidic or alkaline a substance is, with normal blood pH being between 7.35 and 7.45. Severe acidosis, often due to diabetic ketoacidosis, disrupts normal bodily functions, impairs oxygen transport, and can lead to life-threatening symptoms like dehydration, lethargy, and coma if untreated.

Explanation:

pH is a measure of the acidity or alkalinity of a solution, where a pH below 7 is acidic, and a pH above 7 is alkaline. The normal pH of blood is tightly regulated between 7.35 and 7.45. Severe acidosis is a condition where the blood pH falls significantly below this range.

Severe acidosis is problematic because it can disrupt biological processes. Specifically, it affects the ability of hemoglobin to transport oxygen and may lead to symptoms like labored breathing, dehydration, lethargy, and loss of appetite. The condition can be life-threatening, as it may lead to a coma if not promptly treated.

Diabetic ketoacidosis is a common cause of acidosis in diabetic individuals, occurring when excessive ketone bodies in the blood lower the pH well below normal levels, often leading to a value around 6.9, thus disrupting the acid-base balance and necessitating medical intervention.

In a first order decomposition in which the rate constant is 0.0808 sec-1, how long will it take (in minutes) until 0.358 mol/L of the compound is left, if there was 0.52 mol/L at the start? (give answer to 3 decimal places)?

Answers

Answer:

t = 4.62 sec        

Explanation:

For every first order reaction the rate constant K is given as

           [tex]k =(\frac{2.303}{t} )log\frac{[A_{o} ]}{[A]}[/tex]

        [tex][A_{o} ] = initial concentration = 0.52\frac{mol}{L}[/tex]

       [tex][A] =final concentration = 0.358 \frac{mol}{L}[/tex]

         [tex]K = 0.0808 sec^{-1}[/tex]

      [tex]t = (\frac{2.303}{K} ) log (\frac{[A_{o} ]}{[A]} )[/tex]

         [tex]= (\frac{2.303}{0.0808} )log (\frac{0.52}{0.358} )[/tex]

        t = 4.62 sec

       

Answer:0.0771mins

Explanation:The first order rate law eqn =Ca=Caoe^-kt

Ca=final mass remaining

Cap=initial mass

K=rate or decay constant

t=time

e=exponential

ca=0.358

cao=0.52

K=0.0808/sec

Substituting,we have

0.358=0.52e^-0.0808t

0.358/0.52=e^-0.0808t

0.688=e^-0.0808t

Taking naturaing logarithm of both sides(ln of both sides)

ln(0.688)=-0.0808t

-0.3739=-0.0808t

t=0.3739/0.0808

t=4.628secs

In mins,4.628/60=0.0771mins.

Suppose a 0.049 M aqueous solution of sulfuric acid ( H 2 SO 4 ) is prepared. Calculate the equilibrium molarity of SO 4 − 2. You'll find information on the properties of sulfuric acid in the ALEKS Data resource. Round your answer to 2 significant digits.

Answers

Final answer:

The equilibrium molarity of SO4 2- in a 0.049 M aqueous solution of H2SO4 will be approximately 0.049 M after the initial dissociation, as sulfuric acid is a strong acid and dissociates completely in its first step, and the second dissociation is generally weaker.

Explanation:

The question involves the dissociation of sulfuric acid (H2SO4) in water, which occurs in two steps. The first dissociation is strong, with the equation H2SO4 (aq) → 2H+ (aq) + SO42- (aq). Because sulfuric acid is a strong acid, the initial dissociation is essentially complete, and the equilibrium molarity of SO42- will be equal to the initial molarity of H2SO4, barring any further reactions.

Given a 0.049 M solution of H2SO4, after the first dissociation, we have 0.049 M of SO42-. The second dissociation of HSO4- to form SO42- is weak. However, without the acid dissociation constant (Ka) value or any other provided equilibrium concentrations, one cannot calculate the additional contribution of SO42- from the second dissociation. Therefore the most straightforward answer, assuming the second dissociation's contribution is negligible compared to the first, is that the equilibrium molarity of SO42- due to the first dissociation is approximately 0.049 M.

The flame test for sodium is based on the intense yelloworange emission at 589 nm; the test for potassium is based on its emission at 404 nm. When both elements are present, the Na⁺ emission is so strong that the K⁺ emission can’t be seen, except by looking through a cobalt-glass filter. (a) What are the colors of these Na⁺ and K⁺ emissions? (b) What does the cobalt-glass filter do? (c) Why are the oxidizing agents in fireworks made of KClO₄ or KClO₃, rather than the corresponding sodium salts?

Answers

Answer:

a) The flame test of Na, at 589nm will show a golden yellow color when the sodium ion is emitted. These is due to the electropositive nature of the alkali metals.

While the flame test for K at 404nm will ignite a violet color when the Potassium ion is emitted. This properties are due to the high electropositive nature of the group1 elements which also indicate their strong reducing agent.

b) The cobalt glass filter act as hindrance during the flame test to seperate or filter the golden yellow color caused as a result of the presence of the sodium, as it makes the violet color to be more visible.

c) These is due to the Oxidizing ability of KClO₄ or KClO₃ compared to the salts of sodium. Also is the low solubility of the two salts and their solubility constant (Ksp) compared to sodium salts.

Explanation:

a) The flame test of Na, at 589nm will show a golden yellow color when the sodium ion is emitted. These is due to the electropositivity nature of the alkali metals.

While the flame test for K at 404nm will ignite a violet color when the Potassium ion is emitted. This properties are due to the high electropositivity nature of the group1 elements which also indicate their strong reducing agent.

b) The cobalt glass filter act as hindrance during the flame test to seperate or filter the golden yellow color caused as a result of the presence of the sodium, as it makes the violet color to be more visible.

c) These is due to the Oxidizing ability of KClO₄ or KClO₃ compared to the salts of sodium. Also is the low solubility of the two salts and their solubility constant (Ksp) compared to sodium salts.

Ammonium carbamate (NH2COONH4) is a salt of carbamic acid that is found in the blood and urine of mammals. At 250.°C, Kc = 1.58 × 10−8 for the following equilibrium:
NH2COONH4(s) ⇌ 2 NH3(g) + CO2(g)
If 11.51 g of NH2COONH4 is put into a 0.500−L evacuated container, what is the total pressure at equilibrium? atm

Answers

Answer:

The total pressure at equilibrium is 0.07503 atm

Explanation:

The partial pressure of the product at equilibrium will be calculated as follows;

Kp = Kc[RT]³

given;

equilibrium constant Kc = 1.58 X 10⁻⁸

gas constant R = 0.0821 L.atm/mol.K

temperature T = (250 +273) = 523 k

Kp = 1.58 X 10⁻⁸ *(0.0821)³ *(523)³ = 1.251 X 10⁻³

NH₂COONH₄(s) ⇌ 2NH₃(g) + CO₂

NH₂COONH₄(s): Kp = 0, since it is in solid state

2NH₃(g) + CO₂: Kp = 1.251 X 10⁻³

I.C.E Analysis on the product

          2NH₃(g)       CO₂

I  :      0                  0

C :      2x                x

E :     (2x-0)          (x-0)

At equilibrium, E: (2x-0)(x-0) = 1.251 X 10⁻³

(2x)(x) = 1.251 X 10⁻³

2x² = 1.251 X 10⁻³

x²   = (1.251 X 10⁻³)/2

x²  = 6.255 X 10⁻⁴

x  = √(6.255 X 10⁻⁴)

x  = 0.02501 atm

Partial pressure of 2NH₃(g)  = 2x = 2(0.02501 atm) = 0.05002 atm

Partial pressure of  CO₂ = x = 0.02501 atm

Total pressure = P(NH₃(g)) +P(CO₂)

Total pressure = 0.05002 atm + 0.02501 atm = 0.07503 atm

Therefore, the total pressure at equilibrium is 0.07503 atm

From data provided, the total pressure at equilibrium is 0.203 atm.

What is the total pressure at equilibrium?

The total pressure, Ptotal at equilibrium is calculated from the equation of the reaction given below:

NH2COONH4(s) ⇌ 2 NH3(g) + CO2(g)

From the equation of the reaction, If x moles of ammonium carbamate decomposes, it will produce 2x moles of NH3(g) and x moles of CO2(g).

Ammonium carbamate is a solid, and so it does not appear in the expression for Kc.

Kc = 1.58 × 10^-8

Therefore:

Kc = [NH3(g)]^2[CO2(g)]

Kc = (2x)^2(x) = 4x^3

1.58 × 10-8 = 4x^3

Thus, x = 0.00158 M

Hence:

[NH3(g)] = 2 × 0.00158 = 0.00316 M

[CO2(g)] = 0.00158 M

From the ideal gas equation:

PV = nRTP = nRT/V

where

R = 0.08206 L.atm/K.molT = 250°C = 523 K

Also, concentration is given by:

c = n/V

Therefore, P = cRT.

Substituting and calculating for Ptotal:

Ptotal = (0.00316 M) × RT + (0.00158 M) × RT

= ((0.00316 + 0.00158) M) × (0.08206) × ((523) K)

Ptotal = 0.203 atm

Therefore, the total pressure at equilibrium is 0.203 atm.

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Consider two acids: CH3CO2H (acetic acid, pKa = 4.8) and C6H5CO2H (benzoic acid, pKa = 4.2). Which acid is the stronger acid? Select the single best answer.
a. benzoic acid
b. acetic acid

Answers

Answer:

Benzoic acid

Explanation:

The strength of an acid is principally a measure of its dissociative capabilities in aqueous solutions. While strong acids dissociate completely in solution, weak acids dissociates only partially.

The relative strength of an acid can be obtained from its pKa value. The pKa value is the negative logarithm of the concentration of the Ka value.

Stronger acids have a pKa value usually negative. This is a pointer to the fact that the lower the pKa value, the stronger the strength of the acid in question.

Relatively therefore, Benzoic acid is stronger than acetic acid because it has a lesser value for pKa

Final answer:

The strength of an acid is determined by its pKa value, with lower values indicating stronger acids. Therefore, between CH3CO2H (acetic acid, pKa = 4.8) and C6H5CO2H (benzoic acid, pKa = 4.2), benzoic acid is the stronger acid.

Explanation:

When comparing acids, the strength of an acid is determined by its pKa value. The lower the pKa value, the stronger the acid is. In this case, the two acids that are being compared are CH3CO2H (acetic acid, pKa = 4.8) and C6H5CO2H (benzoic acid, pKa = 4.2). Therefore, given that the pKa of benzoic acid is lower than that of acetic acid, we can conclude that benzoic acid is the stronger acid. To summarize, the best answer to the question 'Which acid is the stronger acid, CH3CO2H (acetic acid) or C6H5CO2H (benzoic acid)?' is option a. benzoic acid.

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A gas mixture contains 3.0 mol of hydrogen (H2) and 7.3 mol of nitrogen (N2). The total pressure of the mixture is 304 kPa. What is the mole fraction and partial pressure of H2?

Answers

Answer:

Mole fraction H₂ = 0.29

Partial pressure of H₂ → 88.5 kPa

Explanation:

You need to know this relation to solve this:

Moles of a gas / Total moles = Partial pressure of the gas / Total pressure

Total moles = 3 mol + 7.3 mol → 10.3 moles

Mole fraction H₂ → 3 moles / 10.3 moles = 0.29

Mole fraction = Partial pressure of the gas / Total pressure

0.29 . 304 kPa = Partial pressure of H₂ → 88.5 kPa

Final answer:

The mole fraction of H₂ in the gas mixture is 0.291 and the partial pressure of H₂ is 88.28 kPa.

Explanation:

In order to find the mole fraction of H₂, we need to first calculate the total number of moles in the mixture. The total moles will be the sum of the moles of hydrogen and nitrogen: 3.0 mol + 7.3 mol = 10.3 mol.

The mole fraction of H₂ is calculated by dividing the moles of H₂ by the total moles in the mixture: 3.0 mol / 10.3 mol = 0.291. This means that the mole fraction of H₂ is 0.291.

The partial pressure of H₂ can be calculated using Dalton's Law of Partial Pressures. The total pressure of the mixture is given as 304 kPa, which is equal to the sum of the partial pressures of the gases: PH₂ + PN₂ = 304 kPa. We want to find the partial pressure of H₂, so we can rearrange the equation: PH₂ = 304 kPa - PN₂.

Now we need to find the partial pressure of N₂. The mole fraction of N₂ can be calculated as 1 - mole fraction of H₂: XN₂ = 1 - 0.291 = 0.709. We can then use this mole fraction and the total pressure to find the partial pressure of N₂: PN₂ = XN₂ * total pressure = 0.709 * 304 kPa = 215.72 kPa. With the partial pressure of N₂ known, we can plug it back into the rearranged equation for PH₂: PH₂ = 304 kPa - 215.72 kPa = 88.28 kPa.

If particles have wavelike motion, why don’t we observe that motion in the macroscopic world?

Answers

Answer:

The usual 'particles' that we see in the macroscopic world (let's call them objects), are big and massive. This usually means that they have high characteristic frequencies => low wavelengths.Explanation: umm on google i found

A small portion of a crystal lattice is sketched below. What is the name of the unit cell of this lattice? Your answer must be a word, a very short phrase, or a standard abbreviation. Spelling counts!

Answers

Final answer:

The unit cell described is called the simple cubic unit cell, which contains one atom total due to each corner atom being shared by eight unit cells.

Explanation:

The name of the unit cell of the crystal lattice you've described is the simple cubic unit cell, also known as primitive cubic unit cell. In a simple cubic lattice, the unit cell that repeats itself in all directions to form the entire lattice is a cube with atoms at its corners. These atoms effectively 'touch' each other, and each corner atom is shared among eight unit cells; hence, each unit cell contains one-eighth of an atom at each of its eight corners, totaling one atom per unit cell.

Calculate the distance olive oil (a lipid) could move in a membrane in 15 seconds assuming the diffusion coefficient is 1 μm2/s. Use the equation where S is distance traveled, t is time, and D is the diffusion coefficient.S = (4Dt)^1/2

Answers

Answer:

The answer according to the given equation is S = 0.00077 cm

Explanation:

According to this equation

S = (4Dt)^1/2

S = (4* 1^e-8 * 15)^0.5

S = 0.00077 cm

According to the Approximation equation for diffusion time

t ≅ S^2 / 2D

S = 0.00055 cm  

Final answer:

Using the equation S = (4Dt)^1/2, the average distance olive oil can move in a membrane in 15 seconds is calculated to be 7.75 micrometers.

Explanation:

To calculate the distance that olive oil, a lipid, could move in a membrane in 15 seconds with a diffusion coefficient of 1 μm2/s, we use the equation S = (4Dt)1/2. Plugging in the values, we get S = (4 × 1 μm2/s × 15 s)1/2. We perform the calculation as follows:

Calculate the product of 4, the diffusion coefficient (D), and time (t): 4 × 1 μm2/s × 15 s = 60 μm2Take the square root of 60 μm2 to find the distance: √60 μm2 = 7.75 μm

The average distance that olive oil can move in the membrane in 15 seconds is 7.75 micrometers.

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