LAST ATTEMPT! Covalent compound naming ! !!

 9 is the element Florine

Florine has 9 electrons as well as the 9 protons that determine its atomic number.

The ground state configuration is the lowest energy configuration.

Answer:

a. Endothermic.

b. [tex]Q_{rxn}=5493.6J[/tex]

c. [tex]\Delta H_{rxn}=35.0kJ/mol[/tex]

Explanation:

Hello there!

In this case, according to the given information for this calorimetry problem, it turns out possible for us to proceed as follows:

a. Due to the fact that the temperature of water goes from 23.0 °C to 18.0 °C, we infer this reaction is endothermic as the ammonium nitrate absorbed heat from the water.

b. Here, we consider the following heat equation:

[tex]Q_{rxn}=-Q_{water}[/tex]

Whereas we solve for the heat of reaction by means of the mass of the solution (both water and ammonium nitrate), the specific heat of the solution (we assume it is equal to that of water) and the temperature change:

[tex]Q_{rxn}=-m_{solution}C_{solution}(T_f-T_i)\\\\Q_{rxn}=-(12.6g+250.g)(4.184\frac{J}{g\°C} )(18.0\°C-23.0\°C)\\\\Q_{rxn}=5493.6J[/tex]

c. Here, we divide the previously calculated heat by the moles of ammonium nitrate (molar mass = 80.043 g/mol) to obtain the enthalpy of reaction per mole of this compound:

[tex]n_{NH_4NO_3}=12.6g*\frac{1mol}{80.043 g}=0.157mol\\\\\Delta H_{rxn}=\frac{5493.6J}{0.157mol} =34898.7J/mol\\\\\Delta H_{rxn}=35.0kJ/mol[/tex]

Regards!

Answer:

Option D. 17.5

Explanation:

Equiibrium is: CO + 2H₂  ⇄  CH₃OH

1 mol of CO is in equibrium with 2 moles of hydrogen in order to make, methanol.

Initially we have 0.42 moles of CO and 0.42 moles of H₂

If 0.29 moles of CO remained, (0.42 - 0.29) = 0.13 moles have reacted.

So in the equilibrium we may have:

0.29 moles of CO, and (0.42 - 0.13 . 2) = 0.16 moles of H₂

Ratio is 1:2, if 0.13 moles of CO haved reacted, (0.13 . 2) moles have reacted of hydrogen

Finally 0.13 moles of methanol, are found after the equilibrium reach the end.

Let's make expression for KC: [Methanol] / [CO] . [Hydrogen]²

0.13 / (0.29 . 0.16²)

Kc = 17.5

Answer:

The answer is "170.9 mm Hg".

Explanation:

[tex]\text{Mass of acetone = volume} \times density[/tex]

                          [tex]= 70.0 \times 0.791\\\\ = 55.37\ g\\[/tex]

[tex]\text{Moles of acetone} = \frac{mass}{molar\ mass}\\\\[/tex]

                            [tex]=\frac{55.37}{58.08}\\\\ = 0.9533\ mol[/tex]

[tex]\text{Mass of ethyl acetate = volume} \times density[/tex]

                                   [tex]= 73.0 \times 0.900\\\\ = 65.7\ g[/tex]  

[tex]\text{Moles of ethyl acetate = mass} \times\ molar\ mass[/tex]

                                    [tex]= \frac{65.7}{88.105} \\\\= 0.7457\ mol[/tex]

[tex]\text{Mole fraction of acetone x(acetone)} = \frac{0.9533}{(0.9533 + 0.7457)}\ = 0.5611\\\\[/tex] [tex]\text{Mole fraction of ethyl acetate x(ethyl acetate)} =\frac{0.7457}{(0.9533 + 0.7457) }= 0.4389[/tex]

Applying Raoult's law: [tex]\text{Vapor pressure = x(acetone)P(acetone) + x(ethyl acetate)P(ethyl acetate)}\\\\= 0.5611 \times 230.0 + 0.4389 \times 95.38\\\\ = 170.9\ mm \ Hg\\[/tex]

The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 75.0 mL of ethyl acetate (C4H8O2).

The vapor pressure of the stored mixture is: 170.03 mmHg

In the given information, there is some information that is still missing.

The parameters that we are being given include:

The  first step we need to take is to determine the mass and number of moles  of each compound (i.e. for acetone and ethyl acetate)

For us to do that:

We need the density of acetone and ethyl acetate, which is not given:

Assuming that at a standard condition of vapour pressure:

Then;

Using the relation:

[tex]\mathbf{Density = \dfrac{Mass}{volume}}[/tex]

Mass of acetone = Density of acetone × volume of acetone

Mass of acetone = 0.791  g/mL  × 70.0 mL

Mass of acetone = 55.37 g

Mass of ethyl acetate = Density of ethyl acetate  × volume of ethyl acetate

Mass of ethyl acetate = 0.900 g/mL  ×  75.0 mL

Mass of ethyl acetate = 67.5 g

At standard conditions;

Now, using the formula for calculating the numbers of moles which can be expressed as:

[tex]\mathbf{Number \ of \ moles = \dfrac{mass}{molar \ mass}}[/tex]

For acetone:

[tex]\mathbf{Number \ of \ moles = \dfrac{55.37 \ g}{58.08 \ g/mol}}[/tex]

[tex]\mathbf{Number \ of \ moles =0.95334 \ mol}[/tex]

For ethyl acetate:

[tex]\mathbf{Number \ of \ moles = \dfrac{67.5 \ g}{88.11 \ g/mol}}[/tex]

[tex]\mathbf{Number \ of \ moles =0.76609 \ mol}[/tex]

Now, we will determine the mole fraction of each compound.

The mole fraction describes the ratio a certain constituent of a mixture to the total amount of all the constitutent in the mixture.

Using the formula:

[tex]\mathbf{mole \ fraction = \dfrac{n_A}{n_A+n_B+...n_N}}[/tex]

For Acetone:

[tex]\mathbf{mole \ fraction = \dfrac{0.95334}{0.95334+0.76609}}[/tex]

[tex]\mathbf{mole \ fraction =0.5545 }[/tex]

For ethyl acetate:

[tex]\mathbf{mole \ fraction = \dfrac{0.76609}{0.76609+0.95334}}[/tex]

[tex]\mathbf{mole \ fraction =0.4455}[/tex]

Finally, we can compute determine the vapour pressure of the stored mixture using Raoult's Law.

Raoult's Law posits that the constituent of a partial pressure in a mixture of a liquid is proportional to the mole fraction of that constituent in the mixture provided the temperature is constant.

∴ For the stored mixture = Vapor pressure of acetone + vapour pressure of ethyl acetate.

where:

For acetone;

Vapor pressure = 0.5545 × 230 mmHg

Vapour pressure = 127.54 mmHg

For ethyl acetate:

Vapour pressure = 0.4455 × 95.38 mmHg

Vapour pressure ==42.49 mmHg

Thus, the vapor pressure of the stored mixture is

= (127.54 + 42.49 ) mmHg

= 170.03 mmHg

Therefore, we can conclude that the vapour pressure of the stored mixture is 170.03 mmHg

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Answer:

1. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.

2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.

3. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.

4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.

5. FADH2 is reoxidized to FAD reducing NAD+ to NADH.

Explanation:

The oxidation of pyruvate to AcetylCoA is catalyzed by the pyruvate dehydrogenase complex. The reaction is an irreversible oxidative decarboxylation process in which the carboxyl group of pyruvate is removed as a molecule of carbon dioxide, CO₂, while the remaining two carbons are attached to a CoASH molecule to form acetylCoA.

The pyruvate dehydrogenase complex contains three enzymes - Pyruvate dehydrogenase known as E₁, dihydrolipoyl transacetylase known as E₂, and dihydrolipoyl dehydrogenase known as E₃. It also requires five coenzymes namely: thiamine pyrophosphate (TPP), flavine adenine dinucleotide (NAD), coenzyme A (CoA-SH), nicotinamide adenine dinucleotide (NAD) and lipoate.

Oxidative decorbyxylation of pyruvate takes place in the pyruvate dehydrogenase complex in five steps:

1. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.

2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.

3. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.

4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.

5. FADH2 is reoxidized to FAD reducing NAD+ to NADH.

Answer:

1/2 the distance

Explanation:

If the viscosity of a solution is quadrupled then the distance of collection of diffusing molecules would be half over the same amount of time. The viscosity of the molecules is dependent on density of the liquid. It is independent to the volume of the liquid.

Answer:

The degree of compactness of a substance

Answer:

Aerate solution

Explanation:

aerate solution is the best way to remove CO2 from water (Carbon dioxide in the water that does not form bicarbonates is “uncombined” and can be removed by aeration).

Answer:

See explanation

Explanation:

For a substance to dissolve in another, there must be some sort of interaction between the substances.

Recall that like dissolves like. That is, polar substances dissolve polar substances and non polar substances dissolve nonpolar substances.

Hydrocarbons are nonpolar hence they do not dissolve in polar sulphuric acid. Methyl benzoate is polar hence it dissolve in polar sulphuric acid.

The equation showing the ions is depicted in the image attached to this answer.

Answer:

5.37% w/w is the mass percent of vinegar assuming a molarity of 0.8935mol/L

Explanation:

Assuming the molarity of vinegar is 0.8935mol/L:

Mass percent is defined as 100 times the ratio between mass of solute (In this case, acetic acid), and the mass of the solution

To solve this question we need to find the mass of acetic acid from the moles using the molar mass and the mass of the solution from the volume in liters using the density:

Mass Acetic acid -Molar mass: 60.052g/mol-

0.8935mol * (60.052g / mol) = 53.656g Acetic Acid

Mass Solution:

1L = 1000mL * (1.00g/mL) = 1000g Solution

Mass Percent:

53.656g Acetic Acid / 1000g Solution * 100 =

The mass percent of acetic acid in the original sample of vinegar of molarity 0.8935mol/L is 5.37% w/w.

Mass percent of any solute present in any solution will be calculated as the:

Mass % of solute = (mass of solute / mass of solution) × 100

Let the molarity of vinegar = 0.8935mol/L

Means 0.8935 moles of vinegar present in the 1 liter of the solution.

Now we calculate mass from moles as:

n = W/M, where

W = required mass

M = molar mass = 60.052g /mol

W = (0.8935mol)(60.052g/mol) = 53.656g

Mass of solution = 1L = 1000mL×(1.00g/mL) = 1000g Solution

Then the mass % of acetic acid:

Mass % = (53.656g / 1000g) × 100 = 5.37% w/w

Hence the required % mass is 5.37% w/w.

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Answer:

we don't understand why humans have only 46 chromosomes

Answer:

46 chromosomes is what we don't understand

Answer:

ITS ANSWER IS

OPTION B. ATOMIC NUMBER

HI HAVE A NICE DAY

Answer:

Moles of water are 0.868

Explanation:

Answer:

Hence the correct option is an option (b) Sr4, Cl,Br−,Na+.

Explanation:    

Bromine and chlorine belong to an equivalent group. As we go down the group the dimensions increases which too there's a charge on the bromine atom. therefore the size of the Br- is going to be larger in comparison to the chlorine atom.

Sr atom is within the second group, and also it's below the above-mentioned atoms.so Sr is going to be the larger one among all the atoms.

Sodium and chlorine belong to an equivalent period .size decrease from left to right. but due to the charge on sodium its size decreases and there's an opportunity that Na+ size could be adequate for Cl.      

Here we finally assume that two atoms are of an equivalent size (Na+ and Cl) which are less in size compared to the opposite two(Sr and Br-) during which one is greater (Sr)and the opposite is smaller(Br-).

Answer:

Explanation:

The best way to explain this is to use an example

[tex]I\frac{125}{53} + e \frac{0}{-1} ====> Te\frac{125}{52}[/tex]

You have to understand what happened. A electron was shot into the nucleus of the Iodine. That electron change the entire composition of the nucleus resulting in 52 protons. The mass remained the same (125) but the nucleus was altered. The chemical became 125 52 Tellurium.  But what is important is that it takes a tremendous amount of energy to disrupt a nucleus, and a new chemical is born from that disruption.

Answer:

B.Lone pair in pyrrolidine ring is localized and, therefore, is expected to be more reactive.

Explanation:

There are two nitrogen atoms bearing lone pairs of electrons in the structure of nicotine as shown in the image attached.

One nitrogen atom is found in the pyrrolidine ring. The lone pair on this nitrogen atom is localized hence it is more reactive than the lone pair of electrons found on the nitrogen atom in the pyridine ring which is delocalized a shown in the image attached to this answer.

Follow the rules of Bronsted Lowry theory

So

#a

NH_2-

Add a H

#b

HnO_3

Take a H

#1

Conjugate acid means one H+ is added

sw

#2

Conjugate base means donate a H+

Answer:

The product is aqueous [tex]CO(HCl)_2[/tex] and [tex]O_2(g)[/tex].

Explanation:

Given:

⇒ [tex]2HClO_4(aq) +CO(s)[/tex]

then,

The reaction will be:

⇒ [tex]2HClO_4(aq)+CO(s) \rightarrow CO(HCl)_2 +O_2 (g)[/tex]

In the above reaction, we can see that

The products is:

aqueous [tex]CO(HCl)_2[/tex] and [tex]O_2(g)[/tex]

Thus the above is the correct answer.

Answer:

1.0188 J/g*C

Explanation:

Using the formula; Q = m × c × ∆T

Q(water) = -Q(metal)

m × c × ∆T (water) = -{m × c × ∆T (metal)}

According to this question,

mass of metal = 25g

initial temp of metal = 99°C

mass of water = 50g

initial temp of water = 10.55°C

final temperature of the system = 20.15°C

c of water = 4.184 J/g*C

50 × 4.184 × (20.15 - 10.55) = 25 × c × (20.15 - 99)

209.2 × 9.6 = 25c × -78.85

2008.32 = -1971.25c

c = 2008.32 ÷ 1971.25

c of metal = 1.0188 J/g*C

Answer:

0.21 L.

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 17.5°C = 17.5°C + 273 = 290.5 K

Initial volume (V₁) = 173.8 mL

Final temperature (T₂) = 78 °C = 78 °C + 273 = 351 K

Final volume (V₂) =?

V₁/T₁ = V₂/T₂

173.8 / 290.5 = V₂ / 351

Cross multiply

290.5 × V₂ = 173.8 × 351

290.5 × V₂ = 61003.8

Divide both side by 290.5

V₂ = 61003.8 / 290.5

V₂ = 210 mL

Finally, we shall convert 210 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

210 mL = 210 mL × 1 L / 1000 mL

210 mL = 0.21 L

Thus, the final volume of the balloon is 0.21 L.

Answer:

[tex]\boxed {\boxed {\sf 0.775 \ L}}[/tex]

Explanation:

We are asked to find the final volume of a balloon given a change in temperature. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula for this law is:

[tex]\frac{V_1}{T_1}= \frac{V_2}{T_2}[/tex]

The initial volume is 173.8 milliliters and the initial temperature is 17.5 degrees Celsius.

[tex]\frac {173.8 \ mL}{17.5 \textdegree C}= \frac{V_2}{T_2}[/tex]

The balloon is heated to a final temperature of 78.0 degrees Celsius, but the volume is unknown.

[tex]\frac {173.8 \ mL}{17.5 \textdegree C}= \frac{V_2}{78.0 \textdegree C}[/tex]

We are solving for the final volume, so we must isolate the variable V₂. It is being divided by 78.0 degrees Celsius. The inverse of division is multiplication, so we multiply both sides by 78.0 °C.

[tex]78.0 \textdegree C *\frac {173.8 \ mL}{17.5 \textdegree C}= \frac{V_2}{78.0 \textdegree C} * 78.0 \textdegree C[/tex]

[tex]78.0 \textdegree C *\frac {173.8 \ mL}{17.5 \textdegree C}=V_2[/tex]

The units of degrees Celsius cancel.

[tex]78.0 *\frac {173.8 \ mL}{17.5}=V_2[/tex]

[tex]78.0 *9.931428571 \ mL= V_2[/tex]

[tex]774.6514286 \ mL =V_2[/tex]

We are asked to give the volume in liters, so we must convert out units. Remember that 1 liter contains 1000 milliliters.

[tex]\frac { 1 \ L}{1000 \ mL}[/tex]

[tex]774.6514286 \ mL * \frac{ 1 \ L}{1000 \ mL}[/tex]

[tex]774.6514286 * \frac{ 1 \ L}{1000}[/tex]

[tex]0.7746514286 \ L[/tex]

The original values of volume and temperature have 3 and 4 significant figures. We always round our answer to the least number of sig figs, which is 3. This is the thousandths place for the number we calculated. The 6 in the ten-thousandths place tells us to round the 4 up to a 5.

[tex]0.775 \ L[/tex]

The final volume is approximately 0.775 liters.

Answer:

2Cu2^+ + 2I^- ----> 2Cu^+ + I2

Explanation:

The reaction performed in the experiment is;

2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2

The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.

The oxidation - reduction equation is as follows;

2Cu2^+ + 2e ----> 2Cu^+ reduction half equation

2I^- ----> I2 + 2e. Oxidation half equation

Balanced redox reaction equation;

2Cu2^+ + 2I^- ----> 2Cu^+ + I2

Answer:

attached below

Explanation:

The Four Claisen condensation are grouped into :

Self Claisen condensation is when R = R'

Cross Claisen condensation is when R ≠ R'

attached below are the four Claisen condensation

Answer:

Molarity is 0.075 M.

Explanation:

Moles:

[tex]{ \tt{ = \frac{65.5}{RFM} }}[/tex]

RFM of potassium sulphate :

[tex]{ \tt{ = (39 \times 2) + 32 + (16 \times 4)}} \\ = 174 \: g[/tex]

substitute:

[tex]{ \tt{moles = \frac{65.5}{174} = 0.376 \: moles}}[/tex]

In volume of 5.00 l:

[tex]{ \tt{5.00 \: l = 0.376 \: moles}} \\ { \tt{1 \: l = ( \frac{0.376}{5.00} ) \: moles}} \\ { \tt{molarity = 0.075 \: mol \: l {}^{ - 1} }}[/tex]

Answer:

Have you ever touched a fish? Most fish will feel a bit rough - due to their scales. Some, like sharks, will feel like sandpaper. Even fish with small, smoother scales will feel a bit like that. Amphibians don’t have scales, and most species will be wet to some degree - they have to keep their skin moist or they’ll die. A few groups, like toads and newts, have rougher skin, which is heavier and thicker, which allows them to retain moisture better away from water.

Functionally, the big thing about amphibian skin is that it is semi-permeable. Amphibians can breathe through their skin - all amphibians can get some oxygen through their skin, but some species of salamanders get all their oxygen that way - they have no lungs or gills. The skin can also allow water in - sort of like a paper towel. The bad thing is that other chemicals can pass through the skin, too - pollutants and other chemicals tend to affect amphibians far more than they do other groups.

Amphibians also shed their skin - fish do not. People don’t tend to see frogs shedding their skin often, though, since they eat it to regain nutrients and other resources in the skin.

Finally, since amphibian skin offers no defense against predators in the way that scales do, and limited barrier against disease the way non-amphibian skin does (shedding helps), the skin of many amphibians contain toxins, and some of them have anti-fungal properties (typically due to symbiotic bacteria). Many species have evolved chemical defenses in the skin, while others have glands that produce toxins that can be secreted outside of the skin.

The skin can withstand dessication more than the fish.

They have moist skin used as respiratory surface during deep sleep / hibernation.

They have moist skin due to secretion of mucus by glands under the skin.

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Answer:

a. True.

Explanation:

Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.

weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C

take an example of ethanol:

[tex]{ \bf{CH _{3} CH_{2}OH \: \: \frac{Ag/O_{2} }{500 \degree C} > \: \:CH _{3} CHO}}[/tex]

[tex]{ \sf{CH _{3} CHO \: \: is \: ethanal}} [/tex]

By ozonolysis:

Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.

take an example of propanol:

if it undergoes ozonolysis, it gives ethanal and methanal.

Answer:

A. True

Explanation:

Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.

weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C

take an example of ethanol:

By ozonolysis:

Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.

take an example of propanol:

if it undergoes ozonolysis, it gives ethanal and methanal.

Answer:

The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.

Explanation:

The correct answer is: The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.

The d orbital contains 10 electrons, the s orbital takes 2 electrons and the p orbital takes six electrons.

The orbital in chemistry is defined as a region in space where there is a high probability of finding an electron. There are s, p, d, f orbitals in chemistry which correspond to sharp, principal, diffuse and fundamental.

The electronic configuration of arsenic is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3.

From this electronic configuration, we can see that the 4s and 3d orbitals are half filled while the 4p orbital is half filled.

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Answer:

True

Explanation:

Aromatic compounds undergo substitution rather than addition reactions because the aromatic structure is maintained.

Electrophilic aromatic substitution begins with attack of the electrophile on the aromatic ring to yield a delocalized intermediate called the arenium intermediate. Loss of hydrogen from this intermediate yields the final product.