What type of intermolecular forces occur between 2 oil molecules?

Answer:

The answer is agriculture.

Explanation:

Answer: Agriculture

Explanation:

I got it right on my exam

Answer:

Nitrogen cycle

Explanation:

Answer: The Anode reaction is [tex]Fe(s)\rightarrow Fe^{2+}(aq)+2e^-[/tex]

Explanation:

Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.

Oxidation reaction is defined as the reaction in which a substance looses its electrons.Oxidation occurs at anode.

Anode : [tex]Fe(s)\rightarrow Fe^{2+}(aq)+2e^-[/tex]

Reduction reaction is defined as the reaction in which a substance gains electrons. Reduction occurs at cathode.

Cathode : [tex]Cd^{2+}(aq)+2e^-\rightarrow Cd(s)[/tex]

Answer:

1.75 g HCl

Explanation:

First we convert 2.00 g of PCl₅ into moles, using its molar mass:

Then we convert PCl₅ moles into HCl moles, using the stoichiometric coefficients of the equation:

Finally we convert HCl moles into grams, using its molar mass:

Answer: Kilograms

Explanation: Milliliters and Kilometers are the wrong unit of measurement, and Milligrams are too small. Kilograms would be most appropriate to measure the weight of a heavy animal.

Answer:

OOH a lot of points! too bad you scammed me :) its what you get i gueess

Explanation:

Answer:

The number of bees increased as the heat increased.

Explanation:

The line is on a steady increase.

Answer:

The correct answer is -

a) Sodium acetate: least Ammonia: _______ Sodium ethoxide: most

b) Sodium acetate: most Sodium chloroacetate: _______ Sodium fluoroacetate: least

c)  Sodium acetate: least Sodium methoxide: most Sodium phenoxide:

Explanation:

A) In this case the basicity can be found by the stability of the base, Sodium acetate is the most stable base due to resonance stabilization of acetate anion. Sodium acetate is the least basic as the lone pair of electrons involved in the resonance here. An ethoxide ion has more electrons to donate an electron pair easily.

Thus, the correct answer is - Sodium acetate is the least basic while sodium ethoxide is the most basic.

B) Fluroacetate gives a very strong acid and the acetate gives the least strong acid among all three as described in case of a), thus, the most basic is acetate and the least is fluoroacetate.

]C) comparing the acidity of the acids formed by the addition of a proton to the anions is an easy and correct way to find the basicity of compounds.

Acetate ion adds a proton ⇒ acetic acid

methoxide adds a proton ⇒ methanol

phenoxide ion adds proton ⇒ phenol

As it is known that the stronger the acid, the weaker is the conjugate base formed by the loss of the proton.

acetic acid is the strongest acid because it can easily give up its proton to form acetate. Methanol is the weakest among all three and hence methoxide anion is the strongest base.

Thus, the correct answer is - sodium acetate is the least basic while sodium methoxide is the most basic.

Answer:

The correct answer is - 1 and 3 are true.

Explanation:

Hydrogen bonds require a lot of energy to be broken as these bonds are relatively strong intermolecular forces which makes the boiling point of liquid like water very high, as it requires a lot of heat to break the intermolecular bonds holding the water molecules together.

An increase in the surface area of the liquid also increases the rate of evaporation as it takes place among the molecules of the liquid on the surface.

Answer:

i)  B

ii) D

Explanation:

Bond length is determined by the size of the atoms involved and the bond order

A) C-I

B)  H-I

answer : H-I has the shortest bond length because H has an electronegativity value of 2.2  while C has an electronegativity value of 2.5 hence the bond between H-I is greater than C - I due the electronegativity difference between H-I is greater as well.

C) H-Cl

D) H-I

answer :  H-Cl has the shortest bond length due the electronegativity difference between H-CI is greater as well.

Answer:

The correct answer is - 4-nitroaniline.

Explanation:

It is given that all three solid compounds salicylic acid + 4-nitroaniline + naphthalene are equal in the ratio in the mixture and then 1 gram of this mixture is dissolved in the diethyl ether and run a drop of the solution on TLC plate. This plate shows three spots.

The salicylic acid and naphthalene would stay dissolved in the diethyl ether solution due to the 4-nitroaniline could be extracted by adding aqueous acid and involve in the aqueous layer and thus spot of 4-nitroaniline would be with largest Rf value.

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is [tex]-22Jmole^-^1k^-^1[/tex]

(b) The normal boiling point of water (J·K−1·mol−1) is [tex]-109Jmole^-^1K^-^1[/tex]

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is  109J/mole

Explanation:

Lets calculate

(a) - General equation -

      [tex](\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p[/tex] = [tex]-5_m(\beta )+5_m(\alpha )[/tex] =  [tex]-\frac{\Delta H}{T}[/tex]

 [tex]\alpha ,\beta[/tex] → phases

ΔH → enthalpy of transition

T → temperature transition

 [tex](\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p[/tex] =[tex]= -\frac{\Delta_fH}{T_f}[/tex]

            = [tex]\frac{-6.008kJ/mole}{273.15K}[/tex] ( [tex]\Delta_fH[/tex] is the enthalpy of fusion of water)

           = [tex]-22Jmole^-^1k^-^1[/tex]

(b) [tex](\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}[/tex]

                                  = [tex]\frac{40.656kJ/mole}{373.15K}[/tex] ([tex]\Delta_v_a_p_o_u_rH[/tex] is the enthalpy of vaporization)

                               = [tex]-109Jmole^-^1K^-^1[/tex]

(c) [tex]\Delta\mu =\Delta\mu(l)-\Delta\mu(s)[/tex] =[tex]-S_m\DeltaT[/tex]

[tex][\mu(l-5[/tex]°[tex]C)-\mu(l,0[/tex]°[tex]C)][/tex] =  [tex][\mu(s-5[/tex]°[tex]C)-\mu(s,0[/tex]°[tex]C)][/tex][tex]=-S_m[/tex]ΔT

[tex]\mu(l,-5[/tex]°[tex]C)-\mu(s,-5[/tex]°[tex]C)=-Sm\DeltaT [\mu(l,0[/tex]

[tex]\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)[/tex]

     = 109J/mole

Answer:

1. Electron-deficient molecules

2. Odd electron molecules

3. Expanded valence shell molecules

Explanation:

Answer:

B-100

Explanation:

ppm is an unit of concentration that could be defined as the mass in mg of solute (In this case, NaCl) per kg of solution.

Now, a solution of NaCl that is 0.01% by mass, contains 0.01g of NaCl in 100g of solution.

To solve this question, we must convert the mass of NaCl to mg and the mass of solution to kg:

Mass NaCl:

0.01g * (1000mg / 1g) = 10mg

Mass Solution:

100g * (1kg / 1000g) = 0.10kg

The ppm are:

10mg / 0.10kg =

100ppm

Right answer is:

They can be replenished fast.

Answer:

4.48 L O2

Explanation:

At STP, a mole of any gas contains 22.4 liters. Therefore, we simply have to multiply the amount of moles by 22.4

0.2mol O2 ( 22.4 L) = 4.48 L O2

Answer:

Cr 6+ & SO4 2-

Explanation:

Sulfate is a polyatomic ion that has a charge of 2-. There are three of them in the chemical formula so it equates to a 6- total charge. Thus, chromium must have a 6+ charge to give the compound a neutral charge.

Answer:

c and d

Explanation:

mitochondria and vacoule

Answer:

It is actually D and A

Explanation:

I took this assignment before on edge and the guy who said C and D is wrong

Answer:

iron and chlorine

Explanation:

Answer:

Life in the Han Dynasty (206 BCE - 220 CE) differed significantly from today in family, government, social structure, religion, and trade. For example, the Han Dynasty emphasized a patriarchal family structure, where the eldest male held authority, and filial piety was highly valued. In contrast, contemporary societies embrace more egalitarian family dynamics with shared decision-making.

The government system of the Han Dynasty relied on a centralized bureaucracy and emphasized meritocracy, while modern societies often adopted democratic systems. Socially, the Han Dynasty followed a hierarchical model influenced by Confucian principles, whereas contemporary societies strive for greater equality and social mobility.

Religion in the Han Dynasty combined Confucianism, Taoism, and Buddhism, whereas modern societies exhibit diverse religious beliefs. Lastly, trade in the Han Dynasty thrived along the Silk Road, while modern trade was globally interconnected and facilitated by technological advancements. These differences highlight the evolution of society over time.

Explanation:

1. The standard cell potential for this reaction is 0.14 V.

2. The standard cell potential for this reaction is 0.46 V.

3. The reduction potential for Hg2+(aq) + 2e^- → Hg(l) is 0.79 V.

1. The half-reactions are:

Oxidation: Sn2+(aq) → Sn(s) + 2e^-

Reduction: Pt2+(aq) + 2e^- → Pt(s)

To balance the charges, we multiply the oxidation half-reaction by 2:

2Sn2+(aq) → 2Sn(s) + 4e^-

Now, we can combine the half-reactions to form the overall cell reaction:

2Sn2+(aq) + Pt2+(aq) → 2Sn(s) + Pt(s)

The cell notation for this reaction is:

Sn(s) | Sn2+(aq) || Pt2+(aq) | Pt(s)

To calculate the standard cell potential (E°), we can look up the reduction potentials for each half-reaction. The reduction potential for Pt2+(aq) + 2e^- → Pt(s) is typically listed as 0.00 V. The reduction potential for Sn2+(aq) + 2e^- → Sn(s) is -0.14 V. The standard cell potential is the sum of the reduction potentials:

E° = E°(reduction) - E°(oxidation)

E° = 0.00 V - (-0.14 V) = 0.14 V

2. The half-reactions are:

Oxidation: Co2+(aq) → Co(s) + 2e^-

Reduction: Cr3+(aq) + 3e^- → Cr(s)

To balance the charges, we multiply the reduction half-reaction by 2:

2Cr3+(aq) + 6e^- → 2Cr(s)

Now, we can combine the half-reactions to form the overall cell reaction:

Co2+(aq) + 2Cr3+(aq) + 6e^- → Co(s) + 2Cr(s)

The cell notation for this reaction is:

Co(s) | Co2+(aq) || Cr3+(aq) | Cr(s)

To calculate the standard cell potential (E°), we look up the reduction potentials for each half-reaction. The reduction potential for Co2+(aq) + 2e^- → Co(s) is typically listed as -0.28 V. The reduction potential for Cr3+(aq) + 3e^- → Cr(s) is -0.74 V. The standard cell potential is the sum of the reduction potentials:

E° = E°(reduction) - E°(oxidation)

E° = -0.28 V - (-0.74 V) = 0.46 V

3. The half-reactions are:

Oxidation: Cr2+(aq) → Cr(s) + 2e^-

Reduction: Hg2+(aq) + 2e^- → Hg(l)

The balanced overall cell reaction is:

Cr2+(aq) + 2Hg2+(aq) + 4e^- → Cr(s) + 2Hg(l)

The cell notation for this reaction is:

Hg(l) | Hg2+(aq) || Cr2+(aq) | Cr(s)

To calculate the standard cell potential (E°), we look up the reduction potentials for each half-reaction. The reduction potential for Cr2+(aq) + 2e^- → Cr(s) is typically listed as -0.91 V.

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Answer:

Linoleic acid is omega-6 fatty acid, Oleic acid is omega-9 fatty acid and Palmitic acid  is omega-7 fatty acid

Explanation:

Linoleic acid is omega-6 fatty acid which is not naturally produced by the body while Oleic acid is omega-9 fatty acid which occurs naturally in various animal and vegetable fats and oils.  

As compared to oleic acid, Linoleic acid is lightweight and thin.  

Palmitic acid  is omega-7 monounsaturated fatty acid which occurs naturally and is found in plant and marine sources.  

Palmitic acid is lighter than both the oleic acid and Linoleic acid

Answer:

A)Renewable Energy Sources

Explanation:

Answer:

A. 4.0 mol

Explanation:

Divide 16 g by the molar mass of He, which is 4.0 g/mol.

16 g / 4.0 g/mol = 4.0 mol

Answer:

0.29 atm

Explanation: add both partial pressure and subtract from total pressure

Class A foam is a specialized firefighting foam that is primarily used for extinguishing fires involving ordinary combustible materials, such as wood, paper, fabric, and plastics.

It is designed to enhance the effectiveness of water by reducing surface tension and increasing its ability to penetrate and wet these materials.

Class A foam can be used in various firefighting scenarios, including structural fires, wildland fires, and vehicle fires. It is particularly effective in situations where water alone may not be sufficient to control or extinguish the fire.

The use of Class A foam can improve firefighting operations by increasing the efficiency of water application, reducing water usage, and enhancing fire suppression capabilities. It helps to cool down the fire, minimize heat transfer, and reduce the generation of smoke and hazardous gases.

Overall, Class A foam is a valuable tool in the firefighting arsenal and can greatly aid in the extinguishment of fires involving ordinary combustible materials. Its application should be done in accordance with proper firefighting protocols and guidelines.

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The percent by mass of nitrogen in ammonia (NH3) is approximately 82.15%

Calculating the mass of nitrogen to the total mass of the compound and then expressing the result as a percentage will allow us to determine the percent by mass of nitrogen in NH3 (ammonia).

Ammonia's molecular structure, NH3, indicates that it is made up of one nitrogen atom (N) and three hydrogen atoms (H). We must take both the molar masses of nitrogen and ammonia into account when calculating the percent by mass of nitrogen.

Nitrogen's (N) molar mass is roughly 14.01 g/mol. The molar masses of nitrogen and hydrogen are added to determine the molar mass of ammonia (NH3). Since hydrogen's molar mass is around 1.01 g/mol, ammonia's molar mass is:

(3 mol H 1.01 g/mol) + (1 mol N 14.01 g/mol) = 17.03 g/mol = NH3.

Now, we can use the following formula to get the nitrogen content of ammonia in percent by mass:

(Mass of nitrogen / Mass of ammonia) / 100% is the percentage of nitrogen by mass.

Ammonia weighs 17.03 g/mol and contains 14.01 g/mol of nitrogen by mass. By entering these values, we obtain:

(14.01 g/mol / 17.03 g/mol) 100% 82.15 % of nitrogen by mass

Ammonia (NH3) has a nitrogen content that is roughly 82.15 percent by mass.

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