Which of the following conditions favor the SN1 mechanism as opposed to the SN2 mechanism?

I. tertiary alkyl halide
II. primary alkyl halide
III. polar solvent

a. only I
b. only Il
c. I and Ill
d. Il and Ill

Answer:

Most of the atom is empty space. The rest consists of a positively charged nucleus of protons and neutrons surrounded by a cloud of negatively charged electrons. The nucleus is small and dense compared with the electrons, which are the lightest charged particles in nature.

Answer:

The segment of this graph that shows a decreasing velocity is the EF segment. AB and DE segments show that whatever was moving, in these segments was stopped. The segment BC shows an increase in velocity - the metres travelled were always more as the time went by.

Explanation:

Answer:

Can you attach a picture?

Explanation:

Answer:

A carbon atom bound to three atoms (two single bonds, one double bond) is sp2 hybridized and forms a flat trigonal or triangular arrangement with 120° angles between bonds. Notice that acetic acid contains one sp2 carbon atom and one sp3 carbon atom.

Answer:

100 g of water has the highest number of moles

Explanation:

Recall that the number of moles is obtained as given mass/formula weight

For HCl;

number of moles = 100g/36.5g/mol = 2.7 moles

For H2O;

number of moles = 100g/18g/mol = 5.5 moles

For MgCO3

number of moles = 100g/84.3 g/mol = 1.2 moles

For AlCl3

number of moles = 100g/133.3g/mol = 0.75 moles

For NaCl

number of moles = 100g/58.4 g/mol = 1.7 moles

The enthalpy change of combustion of methane : -892 kJ/mol

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation  

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)  

Determination of the enthalpy of formation of a reaction can be through a calorimetric experiment, based on the principle of Hess's Law, enthalpy of formation table, or from bond energy data  

The value of ° H ° can be calculated from the change in enthalpy of standard formation:  

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)  

Reaction for combustion of Methane (CH₄) :

CH₄ + 2O₂ → CO₂ + 2H₂O.

The elements in standard conditions are not included in the enthalpy calculations because the enthalpy of those elements under the standard conditions is zero(∆Hf °O₂ =0)

∆Hf ° CO₂ = -394 kj/mol

∆Hf ° H₂O=-286 kj/mol

∆Hf ° CH₄=-74 kJ/mol

The enthalpy change of combustion of methane :

[tex]\tt \Delta H_{rxn}=\Delta H_f~CO_2+2.\Delta H_fH_2O-(\Delta H_fCH_4)\\\\\Delta H_{rxn}=(-394-2.-286)-(-74)\\\\\Delta H_{rxn}=-892~kJ/mol[/tex]

Answer:

-655

Explanation:

ye

Answer: The oxidation state of a free element (uncombined element) is zero. For a simple (monoatomic) ion, the oxidation state is equal to the net charge on the ion. For example, Cl– has an oxidation state of -1. When present in most compounds, hydrogen has an oxidation state of +1 and oxygen an oxidation state of −2.

Hope this helps...... Stay safe and have a Merry Christmas!!!!!!!!!!! :D

Explanation:

Problem #1: If you dilute 175 mL of a 1.6 M solution of LiCl to 1.0 L, determine the new concentration of the solution.

Solution:

M1V1 = M2V2

(1.6 mol/L) (175 mL) = (x) (1000 mL)

x = 0.28 M

Note that 1000 mL was used rather than 1.0 L. Remember to keep the volume units consistent.

Problem #2: You need to make 10.0 L of 1.2 M KNO3. What molarity would the potassium nitrate solution need to be if you were to use only 2.5 L of it?

Solution:

M1V1 = M2V2

(x) (2.5 L) = (1.2 mol/L) (10.0 L)

x = 4.8 M

Please note how I use the molarity unit, mol/L, in the calculation rather than the molarity symbol, M.

Problem #3: How many milliliters of 5.0 M copper(II) sulfate solution must be added to 160 mL of water to achieve a 0.30 M copper(II) sulfate solution?

Solution:

M1V1 = M2V2

(5.00 mol/L) (x) = (0.3 mol/L) (160 + x)

5x = 48 + 0.3x

4.7x = 48

x = 10. mL (to two sig figs)

The solution to this problem assumes that the volumes are additive. That's the '160 + x' that is V2.

Problem #4: What volume of 4.50 M HCl can be made by mixing 5.65 M HCl with 250.0 mL of 3.55 M HCl?

Solution:

Here is the first way to solve this problem:

M1V1 + M2V2 = M3V3

(3.55) (0.250) + (5.65) (x) = (4.50) (0.250 + x)

Where x is volume of 5.65 M HCl that is added

(0.250 + x) is total resultant volume

0.8875 + 5.65x = 1.125 + 4.50 x

1.15x = 0.2375

x= 0.2065 L

Total amount of 4.50 M HCl is then (0.250 + 0.2065) = 0.4565 L

Total amount = 456.5 mL

Here is the second way to solve this problem:

Since the amount of 5.65 M added is not asked for, there is no need to solve for it.

M1V1 + M2V2 = M3V3

(3.55) (250) + (5.65) (x − 250) = (4.50) (x)

That way, x is the answer you want, the final volume of the solution, rather than x being the amount of 5.65 M solution that is added.

Problem #5: A 40.0 mL volume of 1.80 M Fe(NO3)3 is mixed with 21.5 mL of 0.808M Fe(NO3)3 solution. Calculate the molar concentration of the final solution.

Answer:

b no

Explanation:

because it is decomposing into two elements

Answer:

yes

Explanation:

Answer: a: extended or controlled release

Explanation:

Answer:

wheres the image

Explanation:

Answer:

1

Explanation:

The full question is shown in the image attached.

If we look at the structure closely, we will discover that there is only one C=C double bond in the molecule.

It is worthy of note that hydrogen (H2) only reduces the C=C double bond. it does not affect any other bond in the molecule.

Hence, only one molecule of H2 is required to reduce one C=C bond present.

Answer:

[tex]T_S=-1.09\°C[/tex]

Explanation:

Hello!

In this case, since the freezing point depression for a solution is computed via:

[tex](T_S-T_W)=-imK_f[/tex]

Whereas TW is the freezing temperature of water, TS that of the solution, i the van't Hoff's factor (3 for K2S as it ionizes properly), m the molality of the solution and Kf the freezing point constant of water. Thus, we plug in to obtain:

[tex](T_S-0\°C)=-3*0.195m*1.86\frac{\°C}{m}\\\\T_S=-1.09\°C[/tex]

Best regards!

The correct answer is -1.088. (Don't forget the negative sign).

Answer:

Uracil

Explanation:

The base that will NOT combine with 2-deoxyribose to form a nucleic acid is Uracil.

2-deoxyribose is a pentose sugar found in the DNA (Deoxyribonucleic acid). It is devoid of oxygen in its 2' position. The bases found in DNA are Adenine, Guanine, Cytosine and Thymine. Adenine, Guanine, and Cytosine are also found in RNA (Ribonucleic acid). Thymine is not present in RNA, it is only found in DNA. The base found in RNA is Uracil which in turn is not present in DNA. The five carbon sugar present in RNA is ribose which combines with Uracil.

Answer:

b. Beta emission, beta emission

Explanation:

A factor to consider when deciding whether a particular nuclide will undergo this or that type of radioactive decay is to consider its neutron:proton ratio (N/P).

Now let us look at the N/P ratio of each atom;

For B-13, there are 8 neutrons and five protons N/P ratio = 8/5 = 1.6

For Au-188 there are 109 neutrons and 79 protons N/P ratio = 109/79=1.4

For B-13, the N/P ratio lies beyond the belt of stability hence it undergoes beta emission to decrease its N/P ratio.

For Au-188, its N/P ratio also lies above the belt of stability which is 1:1 hence it also undergoes beta emission in order to attain a lower N/P ratio.

Radioactive decay refers to the loss of energy by an unstable atomic nucleus as radiation. The substance having unstable nuclei is radioactive.

The correct answer is:

Option B. Beta emission, beta emission

The radioactive decay of 13B and 188 Au can be explained as:

1. Neutron proton ratio refers to the particular nuclei that will undergo what type of radioactive decay.

2. In the B-13 element, the N/P ratio is beyond the stability of the atom, thus, it will emit beta emission to reduce the N/P ratio.

3. Similarly, the N/P ratio of the Au-188 is above the stability of the atom, thus, it undergoes the beta emission to reduce the N/P ratio.

Thus, the correct answer is Option B.

To know more about radioactive decay, refer to the following link:

https://brainly.com/question/14077220

Answer:

e)  400 mg of a radioactive substance with a half-life of 8 hours after decaying for 8 hours

a) 800 mg of a radioactive substance with a half-life of 2 years after decaying for 4 years.

Explanation:

a) 800 mg of a radioactive substance with a half-life of 2 years after decaying for 4 years.

Number of half lives passed = Time elapsed/ half life

Number of half lives passed = 4 year/2 year

Number of half lives passed = 2

at time zero = 800 mg

At first half life = 800 mg/2 = 400 mg

At 2nd half life = 400 mg/2 = 200 mg

b) 1,000 mg of a radioactive substance with a half-life of 5 years after decaying for 20 years

Number of half lives passed = Time elapsed/ half life

Number of half lives passed = 20 year/ 5 year

Number of half lives passed = 4

at time zero = 1000 mg

At first half life = 1000 mg/2 = 500 mg

At 2nd half life = 500 mg/2 = 250 mg

At 3rd half life = 250 mg/2 =  125 mg

At 4th half life = 125 mg/2 = 62.5 mg

c) 300 mg of a radioactive substance with a half-life of 1 year after decaying for 1 year

Number of half lives passed = Time elapsed/ half life

Number of half lives passed = 1 year/ 1 year

Number of half lives passed = 1

at time zero = 300 mg

At first half life = 300 mg/2 = 150 mg

d) 600 mg of a radioactive substance with a half-life of 6 hours after decaying for 18 hours

Number of half lives passed = Time elapsed/ half life

Number of half lives passed = 18 hours / 6 hours

Number of half lives passed = 3

at time zero = 600 mg

At first half life = 600 mg/2 = 300mg

At 2nd half life = 300 mg/2 = 150 mg

At 3rd half life = 150 mg/2 =  75 mg

e)  400 mg of a radioactive substance with a half-life of 8 hours after decaying for 8 hours

Number of half lives passed = Time elapsed/ half life

Number of half lives passed = 8 hours / 8 hours

Number of half lives passed = 1

at time zero = 400 mg

At first half life = 400 mg/2 = 200 mg

Answer:

the answer is applied force

Answer:

21.46 kg.

Explanation:

Hello!

In this case, when a variety of measurements are said to contribute to a total, we understand we need to add them all up in order to get that total; in such a way, since the given seafood is composed by 15 kg of salmon, 4.7 kg of crab and 1.76 kg of oysters, the total mass turns out:

[tex]m_T=15kg+4.7kg+1.76kg\\\\m_T=21.46kg[/tex]

Best regards!

Answer:

BBBBBBBB ITS B

Explanation:

Answer:

2.5471951319999997

Explanation:

Answer:

A. Tertiary alcohol

Explanation:

Secondary alcohols are oxidized to ketones - and that's it. For example, if you heat the secondary alcohol propan-2-ol with sodium or potassium dichromate(VI) solution acidified with dilute Sulphuric acid, you get propanone formed. Playing around with the reaction conditions makes no difference whatsoever to the product.

Answer:

0.47 g

Explanation:

From the question given above, we obtained the following information:

4.20 g of nitrogen occupies 100 L.

We can obtain the mass of nitrogen that occupied 11.2 L by doing the following:

4.20 g of nitrogen occupies 100 L.

Therefore, Xg of nitrogen will occupy 11.2 L i.e

Xg of nitrogen = (11.2 × 4.2) / 100

Xg of nitrogen = 0.47 g

From the calculation made above, 0.47 g of nitrogen will occupy 11.2 L

Answer:

the celery would change colour!

Explanation:

Answer:

D

Explanation:

Answer:

C

Explanation:

Because molecules of solids don't move

while molecules of liquids and gases move.

I Hope this helps!!.

Phan

convex

concave

duplex