which primitive organic molecule was essential to form lipid bilayer?
a)protenoid
b)phospholipid
c)autocatalytic RNA
d)aminoacids​

Answer:

a. 1s2 2s2 2p6 3s2 3p6 4s2 3d 9 - the element is zinc and this is an excited state configuration

b. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f 14 5d7 -the element is iridium and this is a ground state configuration

Explanation:

The ground state is the lowest energy state of an atom while excite states of Zn atom are higher energy states.

The configuration, 1s2 2s2 2p6 3s2 3p6 4s2 3d 9 applies to zinc atom in excited state while the configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f 14 5d7 applies to iridium atom in ground state.

Answer:

yes

Explanation:

the components of solutions fixed

Answer:

lower the activation energy needed to start the reaction.

Explanation:

The activation energy is defined as the energy barrier that stands between reactants and products.

An enzyme is a biological catalyst. Catalysts are known to lower the activation energy of a reaction.

Hence, a catalyst lowers the activation energy of the reaction. The lower the activation energy of a reaction, the faster the reaction is expected to be.

Answer:

See explanation and image attached

Explanation:

The reaction of HBr with 1-butene is an addition reaction. The HBr adds across the double bond to yield a saturated halogenoalkane.

In the absence of peroxides, the reaction proceeds in accordance with Markovnikov rule which states that;''the negative part of the addendum, is attached to the carbon atom with the least number of hydrogen atoms attached.'' This occurs in the first reaction shown in the image.

In the presence of peroxides, the reaction proceeds in an anti-Markovnikov manner to yield the product shown in the second reaction.

i don't know if this is for school or what. but I believe that your destiny is set out but you do have the power to change it. you decide what path you want to go down with your actions

Question Completion with Options:

O coarse...few...rapid

O fine...few...slow

O fine...multiple...rapid

O coarse...few...slow

O fine...multiple...slow

Answer:

The choice that best completes the sentence is:

O coarse...few...slow

Explanation:

Transformations near the melting temperature develop coarse grains because few nucleation sites are formed and the rate of the grain growth is usually slow.  This is because of the process that starts with  recrystallization, recovery, and nucleation before growth can occur.  While recrystallization enables the grain to increase in size at high temperature, nucleation gives the grain the energy to irreversibly grow into larger-sized nucleus.

Answer: The layers are ordered by density, with the least dense layer on top, and the densest layer on the bottom.

Explanation:

Plato

Answer:

[tex]{ \sf{A. \: 2OH {}^{ - } _{(aq)} +Ca {}^{2 + } _{(aq)} −>Ca(OH) _{2(s)} }}[/tex]

The chemical equations depict a balanced ionic equation is  2OH−+Ca2+−>Ca(OH)2. option A is correct.

An ionic equation is a chemical equation in which the electrolytes in an aqueous solution are expressed as dissociated ions. The ions in aqueous solutions are stabilized by ion-dipole interactions with water molecules

Strong acids, strong bases, and soluble ionic compounds (usual salts) exist as dissociated ions in an aqueous solution, and Weak acids and bases and insoluble salts are usually written using their molecular formulas because only a small amount of them dissociates into ions.

Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) → AgCl(s) + Na+(aq) + NO3-(aq) is an ionic equation example.

Therefore, the balanced ionic reaction will be  2OH−+Ca2+−>Ca(OH)2. option A is correct.

Learn more about ionic reactions, here:

https://brainly.com/question/13887096

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Answer:

[tex]\%diff=24.0\%[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to set up the van der Waals' equation as shown below:

[tex]p=\frac{RT}{v-b}-\frac{a}{v^2}[/tex]

Thus, we secondly calculate the molar volume as:

[tex]v=\frac{0.8311L}{9.998mol} =0.083L/mol[/tex]

Then, we plug in the entire variables in the vdW equation to get such pressure:

[tex]p=\frac{0.08206\frac{atm*L}{mol*K}*502.7K}{0.08313L/mol-0.03219L/mol}-\frac{1.345L*atm/mol}{(0.08313L/mol)^2}\\\\p=615.2atm[/tex]

And the ideal gas pressure:

[tex]p=\frac{0.08206\frac{atm*L}{mol*K}*502.7K}{0.08313L/mol}\\\\p=496.2atm[/tex]

Finally, the percent difference:

[tex]\%diff=\frac{|496.2atm-615.2atm|}{496.2atm} *100\%\\\\\%diff=24.0\%[/tex]

Regards!

Explanation:

here's the answer to your question

Answer:

0.8017

Explanation:

Find the molar Mass of KF

K = 39

F = 19

Total = 58

Note: these numbers are approximate. Use your periodic table to get the exact numbers.

mols = given mass / molar mass

given mass = 46.5

molar mass = 58

mols = 46.5 / 58

mols = 0.8017

Explanation:

here's the answer to your question

Answer:

4 boiling point elevation

Answer:

See explanation and image attached

Explanation:

The reaction of (R)-2-chloro-3-methylbutane with potassium tert-butoxide yields a monosubstituted alkene .

Since the base is bulky, the Hoffman product predominates because attack occurs at the less hindered carbon atom to yield the major product as shown.

The alkene reacts with HBr at the secondary carbon atom to yield a carbocation intermediate which is flat and planar. Attack on either face of the carbocation yields a racemic mixture of the (2R) and (2S) products.

Rearrangement of the carbocation to yield a tertiary carbocation gives the 2-bromo-2-methyl butane product as shown in the image attached.

Answer:

The hydrogen sample has a pressure of 0.957 atmospheres.

Explanation:

Let consider that the hydrogen sample behaves ideally, the equation of state for ideal gases is:

[tex]P\cdot V = n\cdot R_{u}\cdot T[/tex] (1)

Where:

[tex]P[/tex] - Pressure, in atmospheres.

[tex]V[/tex] - Volume, in liters.

[tex]n[/tex] - Molar quantity, in moles.

[tex]T[/tex] - Temperature, in Kelvin.

[tex]R_{u}[/tex] - Ideal gas constant, in atmosphere-liters per mole-Kelvin.

If we know that [tex]V = 19.1\,L[/tex], [tex]n = 0.684\,mol[/tex], [tex]T = 326\,K[/tex] and [tex]R_{u} = 0.082\,\frac{atm\cdot L}{mol\cdot K}[/tex], then the pressure of the hydrogen sample is:

[tex]P = \frac{n\cdot R_{u}\cdot T}{V}[/tex]

[tex]P = \frac{(0.684\,mol)\cdot \left(0.082\,\frac{atm\cdot L}{mol\cdot K} \right)\cdot (326\,K)}{19.1\,L}[/tex]

[tex]P = 0.957\,atm[/tex]

The hydrogen sample has a pressure of 0.957 atmospheres.

Answer:

(a) m = 0.327 m.

(b) m = 4.57 m.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by firstly considering the fact that the molality is computed by dividing the moles of solute by the kilograms of solvent, in this case water; in such a way, we proceed as follows:

(a) We firstly calculate the moles of 36.2 grams of sucrose as its molar mass is 342.3 g/mol:

[tex]\frac{36.2g}{342.3g/mol} =0.106mol[/tex]

Next, the kilograms of water in this case are 0.323 kg so that the molality will be:

[tex]m=\frac{0.106mol}{0.323kg}\\\\m=0.327m[/tex]

(b) In this case, we directly realize that the kilograms of water are now 1.889 kg so that the molality will be:

[tex]m=\frac{8.63mol}{1.889kg}=4.57m[/tex]

Clearly, the both of them in molal, m, units.

Regards!

Answer:B) 1s 2s 2p 3s 3p 4s 3d

Explanation:

The ground state electron configuration shows how the electrons in the atomic orbitals of an atom are in their lowest , most stable energy arrangements and since Electrons must be filled following the Aufbau's principle(electrons fill lowest energy shells first)

Now, Titanium lies in period IV and  group 4 of the periodic table with 22 as its atomic number

Thus, the ground-state electron configuration of a neutral atom of titanium is 1s²2s²2p⁶3s²3p⁶4s²3d².  

The question is incomplete, the complete question is shown in the image attached to this answer.

Answer:

a) Br^-

b) FeCl3

c) slower

d) see the first attached image

Explanation:

Aromatic compounds undergo electrophilic substitution sections in the presence of the appropriate electrophile.

In the reaction above, the Br^- nucleophile attacks the Lewis acid FeCl3. Recall that the nitro group is meta directing hence the incoming Br^+ electrophile is directed towards the meta position as shown in the image attached.

Note that the nitro group deactivates the ring towards electrophilic substitution hence the reaction is slower with nitrobenzene than with unsubstituted benzene.

Answer:

1.1 × 10²¹ NH₃ molecules

Explanation:

From the given information:

We were being told that the number of the hydrogen (H) atoms present in the sample of NH3 = 3.3 × 10²¹ hydrogen.

However, it signifies that each molecule of ammonia harbors 3hydrogen (H) atoms.

Hence,  the number of  molecules of NH₃ present;

[tex]\mathsf{=\dfrac{3.3\times 10^{21}}{3} \ molecules \ of \ {NH_3}}[/tex]

= 1.1 × 10²¹ NH₃ molecules

Answer:

Production of some products is highly concentrated in a few countries, China, the leading producer of wheat and ramie in 2013, produces 6% of the world's ramie fiber but only 17% of the world's wheat.

Answer:

Modern windshields are generally made of laminated safety glass, a type of treated glass, which consists of, typically, two curved sheets of glass with a plastic layer laminated between them for safety, and bonded into the window frame.

Explanation:

mark me brain liest if my answer is correct

Answer:

6.4×10¯³ g of O₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CH₄ + 2O₂ —> CO₂ + 2H₂O

Next, we shall determine the masses of CH₄ and O₂ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of CH₄ = 12 + (4×1)

= 12 + 4

= 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ from the balanced equation = 2 × 32 = 64 g

SUMMARY:

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂.

Finally, we shall determine the mass of O₂ needed to react with 1.6×10¯³ g of CH₄. This can be obtained as illustrated below:

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂.

Therefore, 1.6×10¯³ g of CH₄ will react with = (1.6×10¯³ × 64) / 16 = 6.4×10¯³ g of O₂

Thus, 6.4×10¯³ g of O₂ is needed for the reaction.

Answer:

Radical chain initiator

Explanation:

The peroxide here serves as a radical chain initiator. In the field of chemistry the radical initiatives are those substances that are used in industrial processes like polymer synthesis. These initiatives have weak bonds generally and they're mostly used to create free radicals. These radicals are atoms that have odd numbers of electrons. Peroxide is an example of such.

Answer:

Because it is dissolving and has pigment.  

Explanation:

Answer:

The solvated electron is responsible for a great deal of radiation chemistry. Alkali metals dissolve in liquid ammonia giving deep blue solutions which conduct electricity . The blue colour of the solution is due to ammoniated electrons which absorb energy in the visible region of light.

Answer:

Correct ratio of carbon to hydrogen is 2:1:1

Answer:

Its actually 1:2:1

Explanation:

The molecular formula is C6H12O6 because one molecule actually contains 6 C, 12 H, and 6 O atoms. The simplest whole-number ratio of C to H to O atoms in glucose is 1:2:1, so the empirical formula is CH2O.

Answer:

2Li(s) + F2(g)→2LiF(s)

Explanation:

Answer:

C) 5

Step-by-step explaination:

Benzene has 5 resonance structures.

Solution :

A substrate is defined as the chemical species that are being observed in the chemical reaction where the substrate reacts with a reagent and forms a product. It can also be referred to the surface where some other chemical reactions are performed.

Stereochemistry is defined as the study of relative spatial arrangement of the atoms which forms the structure of the molecules and their respective manipulations.

In the context, the products including the stereochemical designations for any chirality centers starting from the  (R)-3-methylhex-1-yne as the substrate are attached below.  

Answer:

V₂ = 0.154 Liters

Explanation:

Pressure => P

Volume   => V

Temperature => T

mass (moles) => n

This problem...

P₁ = 1.00 ATM     P₂ = 3.25 ATM

V₁ = 0.500L       V₂ = ?

T₁ =  constant    T₂ = T₁ = constant

n₁ =  constant     n₂= n₁ = constant

P₁V₁/n₁T₁ = P₂V₂/n₂T₂  => V₂ = V₁(P₁/P₂) = 0.500L (1.00ATM/3.25ATM) = 0.154 Liters

Answer:

Following are the response to the given question:

Explanation:

In the given scenario, When the balance has never been tainted before the KHP is weighted, which can affect the molar concentration of NaOH because its molarity is directly proportional to the weight including its substance. In this question it is the mistake is systemic because it may be corrected by modifying balancing parameters.

Answer:

The activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.

Explanation:

The activity of P-32 can be calculated with the following equation:

[tex] A = \lambda N [/tex]   (1)

Where:

N: is the number of atoms of P-32

λ: is the decay constant

We can find the number of atoms of P-32 as follows:

[tex] N = \frac{N_{A}*m}{M} [/tex]  (2)

Where:

[tex]N_{A}[/tex]: is the Avogadro's number = 6.022x10²³ atoms/mol

m: is the mass of P-32 = 3.5x10⁻³ g

M: is the molar mass of the radionuclide (P-32) = 32 g/mol    

Now, the decay constant is given by:

[tex] \lambda = \frac{ln(2)}{t_{1/2}} [/tex]   (3)

Where:

[tex]{t_{1/2}} [/tex]: is the half-life of P-32 = 14.3 days

Finally, we can find the activity of P-32 by entering equations (2) and (3) into (1):

[tex] A = \lambda N = \frac{ln(2)}{t_{1/2}}*\frac{N_{A}*m}{M} = \frac{ln(2)}{14.3 d*\frac{24 h}{1 d}*\frac{3600 s}{1 h}}*\frac{6.022 \cdot 10^{23} mol^{-1}*3.5 \cdot 10^{-3} g}{32 g/mol} = 3.7 \cdot 10^{13} dis/s [/tex]      

Since a becquerel (Bq) is defined as a disintegration (dis) per second, the activity in Bq is:

[tex] A = 3.7 \cdot 10^{13} Bq [/tex]

And, since a Curie (Ci) is 3.7x10¹⁰ Bq, the activity in Ci is:

[tex] A = 3.7 \cdot 10^{13} Bq*\frac{1 Ci}{3.7 \cdot 10^{10} Bq} = 1.0 \cdot 10^{3} Ci [/tex]

Therefore, the activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.  

I hope it helps you!