which of the following structures can differentiate plant and animal cells​

Answer:

2. At pH near the pI, nearly all the molecules carry no net charge

Explanation:

The general equilibrium for a molecule with multiple ionizable groups could be:

H₃A⁺ ⇄ H₂A + H⁺ ⇄HA⁻ + H⁺⇄A²⁻ + H⁺

Where each equilibrium has its own pKa

The equilibriums in which H₂A are involved (The neutral molecule), are used to determine the isoelectrical point.

The isoelectrical point is defined as the value of pH in which the molecule has no net charge.

Thus, true option is:

Answer:

-973 KJ

Explanation:

The balanced reaction equation is;

N2H4(aq) + 2Cl2(g) + 4OH^-(aq)---------> 4Cl-(aq) + 4H ^+(aq) + 4OH^-(aq) + N2(g)

Reduction potential of hydrazine = -1.16 V

Reduction potential of chlorine = 1.36 V

From;

E°cell= E°cathode - E°anode

E°cell= 1.36 - (-1.16)

E°cell= 2.52 V

∆G°=- nFE°cell

n= number of moles of electrons = 4

F= Faraday's constant = 96500 C

E°cell = 2.52 V

∆G°=- (4 × 96500 × 2.52)

∆G°= -972720 J

∆G°= -972.72 KJ

Answer:

C+O2=CO2

Explanation:

no.of atoms on reactant side= no.of atoms on product side

The correct answer for this question is that a statement that can be proven by observation or measurement is known as a fact. The other possible answers here simply refer to idea that different individuals can place value on, but that are not necessarily known to be true.

Answer:

48 molecules of CO₂

Explanation:

I think you made a mistake in your question.  The formula for propane is C₃H₈, not C₃H₃.  But, I will give you the answer for both cases.

For C₃H₃:

First you have to balance the equation.

4 C₃H₃  +  15 O₂  ⇒  12 CO₂  +  6 H₂O

Next, you need to use the mole ratios between C₃H₃ and CO₂ to find the amount of molecules of CO₂ you will produce with the given amount of C₃H₃.

(16 mol's C₃H₃) × (12 mol's CO₂/4 mol's C₃H₃) = 48 mol's CO₂

You will get 48 molecules of CO₂.

For C₃H₈:

Balance the equation.

C₃H₈  +  5 O₂  ⇒  3 CO₂  +  4 H₂O

Use the mole ratios between C₃H₈ and CO₂.

(16 mol's C₃H₈) × (3 mol's CO₂/1 mol's C₃H₈) = 48 mol's CO₂

You will get 48 molecules of CO₂ for this equation as well.

Answer:

Cl2/FeCl3

Explanation:

The electrophilic chlorination of aromatic compounds proceeds in the presence of Cl2/FeCl3.

The function of the Lewis acid FeCl3 is to 'activate' the chlorine molecule towards electrophilic substitution. This occurs by the formation of a 'chlorine cation' which attacks the aromatic ring followed by deprotonation of the aromatic ring to retain the aromaticity of the compound.

Answer:

0.1025 g/L

Explanation:

From the question above,

The solubility of oxygen gas in water at 25°c and 1.0 atm pressure is 0.041g/L

Henry's law states at a constant temperature the solubility of a gas is proportional to the pressure

Since the temperatures are both similar (25°c) then, the solubility of oxygen in water at 2.5 atm can be calculated as follows

= 0.041/1.0 × 2.5

= 0.041×2.5

= 0.1025 g/L

Hence the solubility of oxygen in water at 2.5 atm and 25°c is 0.1025 g/L

Answer:

( CH_3) _2CHCH_2CH_2CO_2CH_2CH

Answer:

19.5 mL

Explanation:

Answer:

The 9-v transistor battery is a direct current source, while the power supply from the wall socket is an alternating current source.

Explanation:

A direct current (DC) is one that flows continuously without any minimum and maximum values. This is majorly obtained from generators, batteries or cells etc.

An alternating current (AC) is a type that flows in a sinusoidal manner, having minimum and maximum value with respect to its flowing frequency. An example is the one obtained from the wall electrical source , generators etc.

An alternating current is not mainly used for electrolysis because of its nature. It would cause a switch of the electrodes in every cycle. Therefore, a direct current source is used.

The 9-v transistor battery is a direct current source, while the power supply from the wall is an alternating current source.

Answer:

0.394g

Explanation:

At 78°C, you can dissolve, in 10mL of ethanol, 0.424g of the compound will dissolve. Then, you must filter the solution.

You take the filtrate of the solution in ethanol and cool it to 0°C. Om mL, only 0.03g of the compound will remain in solution. The other part is the purified compound.

Its mass (Theoretical yield) is:

0.424g - 0.030g =

Answer:

24.84 g

Explanation:

The balanced reaction equation is;

Cr2O3(s) + 3H2S(g) ⟶Cr2S3(s) + 3H2O(l)

We must first determine the limiting reactant

For chromium III oxide;

Amount of chromium III oxide = mass/molar mass = 70.5g/ 151.99 g/mol = 0.46 moles

If 1 mole of chromium III oxide yields 3 moles of water

0.46 moles of chromium III oxide yields 0.46 × 3 = 1.38 moles of water

For hydrogen sulphide

Amount of hydrogen sulphide = mass/molar mass = 90g/ 34 gmol-1 = 2.64 moles

If 3 moles of H2S yields 3 moles of water

2.64 moles of H2S yields 2.64 × 3/3 = 2.64 moles of water

Hence chromium III oxide is the limiting reactant.

Mass of water produced= 1.38 moles × 18gmol-1= 24.84 g

Hello. This question is incomplete. The full question is:

Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 ∘C . Cu(s) || Cu2+(0.14 M) ‖‖ Fe2+(0.0040 M) || Fe(s) E∘Cu2+/Cu=0.339 VE∘Fe2+/Fe=−0.440 V Ecell= V Is the electrochemical cell spontaneous or not spontaneous as written at 25 ∘C ? not spontaneous spontaneous

Answer:

Ecell = -0.825v

Negative results indicate that the electrochemical cell is not spontaneous.

Explanation:

You will find the result by doing the following calculation:

Cu2 + = 0.14M

Fe2 + = 0.0040M

Eº Cu2 + / Cu = 0.339V

Eº Fe2 + / Fe = -0.440V

Electron change ............ n = 2

Cu // Cu2 + (0.14M) // Fe2 + (0.0040M) / Fe (s)

Cu + Fe2 + ----> Fe (s) + Cu2 + (aq)

You should now use the formula =

Eºcell = Eº anode - Eº cathode ------> Eº = Eº (Fe2 + / Fe) - Eº (Cu2 + / Cu)

Eº Cell = -0.440 - 0.339 = -0.779V

E cell = EºCell - (0.0591 / n) log (Cu2 + / Fe2 +)

E cell = -0.779 - (0.0591 / 2) log (0.14 / 0.0040)

Ecell = -0.825

Answer:  Chemical

Explanation:

When paper is burned, it creates a chemical change.

Answer:

Explanation:

Hey, there!!

The answer is chemical change.

When the paper is burned it changes into an ash. Which is a chemical change. { as it can't be reversed }.

Hope it helps...

Answer:

The products have a higher heat content than the reactants

Explanation:

Recall that the change in enthalpy for a reaction is obtained from

∆Hreaction= enthalpy of products - enthalpy of reactants

For an exothermic reaction, enthalpy of reactants > enthalpy of products, hence ∆Hreaction is negative.

This implies that the heat content of reactants is greater than that of products for an exothermic reaction. Hence the option selected in the answer is false as written.

Answer:

I. 2.79x10⁻³

II. 0.155m

III. 6.49mL of the solution you must add from the stock solution

Explanation:

A solution of 0.90% by mass NaCl means that, in 100g of solution, you have 0.90g of NaCl. In other words, there are 0.90g of NaCl + 99.1g of water.

I. Mole fraction of NaCl is the ratio between moles of NaCl and total moles of solution.

Moles NaCl:

0.90g * (1mol / 58.44g) = 0.0154 moles NaCl

Moles water:

99.1g water * (1mol / 18.01g) = 5.50 moles water.

Total moles = 5.50 +0.0154 moles = 5.518 moles

Mole fraction NaCl: 0.0154 / 5.518 moles =

II. Molality is the ratio between moles of solute (NaCl) and kg of solvent.

There are 99.1g of water = 0.0991kg.

Molality is: 0.0154 moles / 0.0991kg

II. 100mL of a 0.01M are:

100mL = 0.100L * (0.01 mol / L) = 1x10⁻³ moles of NaCl

In 100g = 100mL (Density of 1g/mL) there are 0.0154 moles. To have 1x10⁻³ moles you need:

1x10⁻³ moles of NaCl * (100mL / 0.0154 moles) =

I. Mole fraction of NaCl is 2.79x10⁻³

II. Molarity of NaCl solution is 0.155m

III. 6.49mL of the stock solution required to prepare 100 mL of 0.01 M NaCl  solution.​

What does % by mass mean?

A solution of 0.90% by mass NaCl means that, in 100g of solution, you have 0.90g of NaCl.

Or

There is 0.90g of NaCl + 99.1g of water.

We need to find:

I. Mole fraction of NaCl:

Mole fraction can is the ratio of moles of Solute  to the total number of moles of the solutes and the solvent.

Moles of  NaCl(solute) = 0.90g * (1mol / 58.44g) = 0.0154 moles NaCl

Moles of  water(solvent) = 99.1g water * (1mol / 18.01g) = 5.50 moles water.

Total moles = 5.50 +0.0154 moles = 5.518 moles

Mole fraction of NaCl =  [tex]\frac{0.0154}{ 5.518} =2.79 * 10^{-3}[/tex]

II. Molality is a measure of the number of moles of solute(NaCl) in a solution corresponding to 1 kg or 1000 g of solvent

Since, 99.1g of water = 0.0991kg.

Molality of NaCl = [tex]\frac{0.0154 moles}{0.0991 kg}=0.155m[/tex]

III. 100mL of a 0.01M is:

100mL = 0.100L * (0.01 mol / L) = 1x10⁻³ moles of NaCl

In 100g = 100mL (Density of 1g/mL) there are 0.0154 moles. To have 1x10⁻³ moles you need:

1x10⁻³ moles of NaCl * [tex]\frac{100 mL}{0.0154 moles}[/tex] = 6.49mL of the solution you must add from the stock solution

Learn more:

https://brainly.com/question/15063496

Answer:

0.0074 cL

7.4 x 10⁻³ cL

Explanation:

Step 1: Find conversion

1 liter (L) = 100 centiliters (cL)

Step 2: Set up dimensional analysis

[tex]0.74cL(\frac{1L}{100cL} )[/tex]

Step 3: Multiply and cancel out units

cL and cL cancel out.

We are left with L.

0.74/100 = 0.0074 cL or 7.4 x 10⁻³ cL

Answer:

millileter????

Answer:

By definition:

Example:

If I study hard in school, I will get good grades.

Answer: the time taken for the radioactivity of a specified isotope to fall to half its original value.

Explanation:

Answer:

Half-life is the time required for a quantity to reduce to half of its initial value. The term is commonly used in nuclear physics to describe how quickly unstable atoms undergo, or how long stable atoms survive, radioactive decay.

Answer:

C. 3

Explanation:

12.0 - 3 significant figures 1,2 and 0, because 0 is after decimal point.

Chemistry is vital in forensic science (used to solve crimes). With chemistry, forensic scientists demonstrate what happened on a molecular level. Chemical analysis can reveal how long ago a person died, what weapon was used, the kind of tape or rope used to tie someone up, etc. They also analyze drugs/controlled substances taken from scenes and people to identify and sometimes quantify these materials.

Answer:

16.5

Explanation:

Answer:

is your answer is 16.5 km

Given :

Initial volume , [tex]v_1=50\ ml[/tex] .

Final volume , [tex]v_2=300 \ ml[/tex] .

Initial molarity , [tex]M_1=0.45 \ M[/tex] .

To Find :

Molarity of  300 ml solution .

Solution :

We know , for constant moles .

[tex]M_1v_1=M_2v_2\\\\M_2=\dfrac{M_1v_1}{v_2}\\\\M_2=\dfrac{50\times 0.45}{300}\ M\\\\M_2=0.075\ M[/tex]

Therefore , the molarity of the resulting solution is 0.075 M.

Hence , this is the required solution .

Answer:

0.875 Pounds

Explanation:

Formula  is

Just divide the mass value by 16

This question is incomplete, the complete question is;

An electrochemistry lab was conducted. What voltage would have been observed if you had switched the position of the electrodes but not the solutions for any of the electrochemical cells? (e.g. placed Cu electrode with Zn²+ and Zn with Cu²+).

Clearly explain your answer, include what would have happened in each cell

Answer:

E°cell = +1.10V

Explanation:

At lower reduction potential, electrode acts as an anode ( -ve )

At higher reduction potential, electrode acts as a cathode ( +ve )

NOW

standard reduction of Cu⁺² / Cu electrode,

E° cu⁺²/cu = 0.34V

standard reduction potential of zn⁺² / zn electrode,

E° zn⁺/zn = -0.76V

Zn has lower reduction potential, therefore it acts as an anode and standard reduction potential of Cu has higher reduction potential so it acts as a cathode.

now we know that

E°cell = E°cathode - E°anode

we substitute

E°cell = 0.34V - ( -0.76V)

E°cell = +1.10V

Therefore there is no flow of e- in the wire when the copper electrode placed in Zn²⁺ ( electrode reverse). When zinc electrode placed in copper solution, copper deposition in zinc electrode takes place, so zinc is converted into Zn ion and no flow of e- will occur in outer circuit.

∴ The voltage becomes zero