List three ways in which the octet rule can sometimes fail to be obeyed

Answer:

So confusing but I'll try

Answer:Scientists have identified more than 260,000 kinds of plants. They classify plants according to whether they have body parts such as seeds, tubes, roots, stems, and leaves. The three main groups of plants are seed plants, ferns, and mosses.

Explanation:

hope i help

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:

Nitrogen cycle

Explanation:

Answer:

1.34 mol

Explanation:

Molarity, which is the molar concentration of a solution, can be calculated by dividing the number of moles (n) by the volume (V).

That is;

Molarity (M) = n/V

According to the information provided in this question;

M = 2.40M

V = 56.0 mL = 56/1000 = 0.056 L

Since molarity = n/V

number of moles = M × V

n = 0.056 × 24

n = 1.34 mol

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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Answer:

C

Explanation:

It is run by Elon Musk, who is determined to get to Mars.

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:

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:

0.29 atm

Explanation: add both partial pressure and subtract from total pressure

Answer:

[tex]T_F=17.56\°C[/tex]

Explanation:

Hello there!

In this case, for this calorimetry problem, it is possible for us to realize that the heat lost by the hot silver is gained by the cold far whose specific heat is 3.94 J/g°c, so we can write:

[tex]-Q_{Ag}=Q_{fat}[/tex]

Which can be written in terms of mass, specific heat and temperature as shown below:

[tex]-m_{Ag}C_{Ag}(T_F-T_{Ag})=m_{fat}C_{fat}(T_F-T_{fat})[/tex]

In such a way, solving for the final temperature, we obtain:

[tex]T_F=\frac{m_{Ag}C_{Ag}T_{Ag}+m_{fat}C_{fat}T_{fat}}{m_{Ag}C_{Ag}+m_{fat}C_{fat}}}[/tex]

Then, we plug in the given data to obtain:

[tex]T_F=\frac{14.16g*0.2165J/g\°C*133.5\°C+250g*3.94J/g\°C*17.20\°C}{14.16g*0.2165J/g\°C+250g*3.94J/g\°C} \\\\T_F=17.56\°C[/tex]

Best regards!

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:

12 M

Explanation:

The reaction between HNO₃ and KOH is:

First we calculate how many KOH moles reacted with the diluted HNO₃ sample, using the given volume and concentration:

As 1 KOH mol reacts with 1 HNO₃ mol, in 25 mL of the diluted HNO₃ solution there are 30 HNO₃ mmoles.

With that information in mind we can calculate the HNO₃ concentration in the diluted solution:

Finally we can use the C₁V₁=C₂V₂ formula to calculate the concentration of the original solution:

Answer:

The answer is agriculture.

Explanation:

Answer: Agriculture

Explanation:

I got it right on my exam

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:

C

Explanation:

because then you'd be able to make a group to raid, run with you, and save more people.

Answer:

A)Renewable Energy Sources

Explanation:

Answer:1 2 and 4

Explanation:

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:

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:

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: 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:

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:

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:

They can be replenished fast.