How would you monitor the progress of a neutralization reaction? Question 2 options: We will use a funnel to separate the solid as it forms We will use a balance to see the changes in mass We will use a thermometer to check the changes in temperature We will use an acid-base indicator to see changes in color depending on the pH

Answer:

a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly.

Answer: Mass of methyl 4-fluoro-3-nitrobenzoate = 199 mg;

Mass of 4-chlorobenzylamine = 282 mg

Explanation: The mass and mol of a molecule is related by its molar mass, which is given in g/mol.

The molar mass of methyl 4-fluoro-3-nitrobenzoate, which has molecular formula: [tex]C_{8}H_{6}FNO_{4}[/tex] is:

[tex]C_{8}H_{6}FNO_{4}[/tex]  = 12.8 + 6.1 + 19 + 14 + 16.4 = 199 g/mol

Since it is asking in mg: MM = 199.10³mg/mol

For 4-chlorobenzylamine, with molecular formula [tex]C_{7}H_{8}ClN[/tex]:

[tex]C_{7}H_{8}ClN[/tex] = 12.7 + 8.1 + 35 + 14 = 141 g/mol

In mg: MM = 141.10³mg/mol

The reaction requires 1 mmol of [tex]C_{8}H_{6}FNO_{4}[/tex] , then its mass is:

m = 1.10⁻³ mol * 199.10³mg/mol = 199 mg

For [tex]C_{7}H_{8}ClN[/tex], it requires 2mmol:

m = 2.10⁻³ mol * 141.10³ mg/mol = 282 mg

For the SNAr reaction, it is necessary 199 mg of methyl 4-fluoro-3-nitrobenzoate and 282 mg of 4-chlorobenzylamine

The mass of 1 mmol methyl 4-fluoro-3-nitrobenzoate is 199.14 mg and the mass of 2 mmol of 4-chlorobenzyl amine is 283.2 mg

To determine the masses of both reagents,

First we will determine their molar masses

Molar mass = 199.14 g/mol

Now, for the mass of 1 mmol of methyl 4-fluoro-3-nitrobenzoate

Using the formula

Mass = Number of moles × Molar mass

Mass = 1 mmol × 199.14 g/mol

Mass = 199.14 mg

Molar mass = 141.6 g/mol

Now, for the mass of 2 mmol of 4-chlorobenzyl amine

Mass = 2 mmol × 141.6 g/mol

Mass = 283.2 mg

Hence, the mass of 1 mmol methyl 4-fluoro-3-nitrobenzoate is 199.14 mg and the mass of 2 mmol of 4-chlorobenzyl amine is 283.2 mg

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Answer:

sodium

Explanation:

(Na) atom is paramagnetic and sodium is a na atom.

Answer:

= 0.2 mL.

Explanation:

Given a 0.5 M solution of NaOH as stock solution, 10.0mL of 0.010M can be prepared via dilution with distilled water, by using the formula:

[tex]C_{1} V_{1} = C_{2} V_{2}[/tex]

where C1 and V1 are initial concentration and volume respectively; same as C2 & V2 for fina.

Let C1 = 0.5M, V2 = ?

C2 = 0.010M; V2 = 10mL

⇒Volume of stock solution to be diluted, V2

= [tex]\frac{10}{0.5}[/tex] × 0.010

= 0.2 mL.

Glasswares used would be pipette (for smaller volume experiment) and measuring cylinder. 0.2mL would be measured and then made upto the 10mL mark of the measuring cylinder.

I hope this was a detailed explanation given the missing details of "Trial 1" in the question.

Answer:

Kc = 166.7

[Fe³⁺] =  0.18 M

[SCN⁻] = 2×10⁻⁴ M

Explanation:

In the reaction of Fe³⁺ and SCN⁻, it is formed a complex:

Fe³⁺  +  SCN⁻  ⇄  FeSCN²⁺             Kc

Let's make the expression for Kc →  [FeSCN²⁺] / [Fe³⁺] . [SCN⁻]

5.7×10⁻⁵ / 9.5×10⁻⁴. 3.6×10⁻⁴  = 166.7

We determine the mmoles, we add from each reactant:

18 ml . 0.2M = 3.6 mmoles of Fe³⁺

2 ml . 0.002M = 4×10⁻³ mmoles of SCN⁻

General form of the dilution equation is:

Concentrated [C] . Concentrated Volume = Diluted [C] . Diluted Volume

Total volume = 20mL

[Fe³⁺]: 3.6 mmoles /20mL = 0.18 M

[SCN⁻]: 4×10⁻³ mmoles /20mL = 2×10⁻⁴ M

The value should be 1.67 x 10^2

The initial concentration should be 0.18 M and 2.0 x 10^(-4) M

The value is

= 5.7 x 10^(-5)/(9.5 x 10^(-4) x 3.6 x 10^(-4))

= 167

= 1.67 x 10^2

we know that

Initial moles  = volume x concentration

So,

= 18/1000 x 0.200

= 0.0036 mol

Now

Initial moles  = volume x concentration

= 2/1000 x 0.0020

= 4.0 x 10^(-6) mol

So,

Total volume should be

= 18 + 2

= 20 mL

= 0.02 L

Now

Initial concentration   

= moles /total volume

= 0.0036/0.02

= 0.18 M

Now

Initial concentration

= moles  /total volume

= 4.0 x 10^(-6)/0.02

= 2.0 x 10^(-4) M

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Answer:

Fluorine, Chlorine, Bromine, or Iodine

Explanation:

These all have an ALMOST full valence shell. And they need one more electron so they'd react with hydrogen

Answer:

its chlorine

Explanation:

just trust me do i look like i would lie too you ;-)

btw i just took the test :-)

Answer:

91.29 g of Na₃PO₄ are needed to produce 1.67 moles of sodium ions.

Explanation:

The formula for sodium phosphate is Na₃PO₄

Molar mass of sodium phosphate = 164 g/mol

The dissociation of one mole of sodium phosphate produces 3 moles of sodium ions;

Na₃PO₄ ------> 3Na⁺ + PO₄³

Number of moles of Na₃PO₄ that will produce 1.67 moles of Na⁺ = 1/3 * 1.67 = 0.556 moles  of Na₃PO₄

Mass of 0.556 moles of Na₃PO₄ = 0.556 moles * 164 g/mol = 91.29 g

Therefore, 91.29 g of Na₃PO₄ are needed to produce 1.67 moles of sodium ions.

Answer:

See explanation

Explanation:

In this case, we have to keep in mind the valence electrons for each atom:

N => 5 electrons

O => 6 electrons

If the formula is [tex]N_2O_4[/tex], we will have in total:

[tex](5*2)+(6*4)=34~electrons[/tex]

Additionally, we have to remember that each atom must have 8 electrons. So, for oxygens 5 and 3 we will have 3 lone pairs and 1 bond (in total 8 electrons. For oxygens, 6 and 4 we will have 2 lone pairs and 2 bonds (in total 8 electrons) and for nitrogens 1 and 2 we will have 4 bonds (in total 8 electrons).

To find the hybridization, we have to count the atoms and the lone pairs around the nitrogen. We have 3 atoms and zero lone pairs. If we take into account the following rules:

[tex]Sp^3~=~4[/tex]

[tex]Sp^2~=~3[/tex]

[tex]Sp~=~2[/tex]

With this in mind, the hybridization of nitrogen is [tex]Sp^2[/tex].

See figure 1

I hope it helps!

The central nitrogen atoms in N2O4 are both sp2 hybridized.

The Lewis structure shows the number of electron pairs that surround the atoms in a molecule as dots. It is quite easy to determine the number of valence electrons in a molecule simply by observing its Lewis dot structure.

The molecule N2O4 has 34 valence electrons as shown in its dot electron structure. The central nitrogen atoms in N2O4 are both sp2 hybridized as shown. The formal charges on each atom in N2O4 are also shown.

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Answer:

[tex]\triangle G = -911.296 \ kJ[/tex]

Explanation:

ΔG = ΔH-TΔS

Where ΔH = -176 kJ = -176000 J , T = 25°C + 273 = 298 K , ΔS = -284.8 JK⁻¹

=> [tex]\triangle G =-176000 - (298)(-284.8)[/tex]

=> [tex]\triangle G = -176000+84870.4[/tex]

=> [tex]\triangle G = -91129.6 \ J[/tex]

=> [tex]\triangle G = -911.296\ kJ[/tex]

Since the value is negative, the reaction is spontaneous under standard conditions at 298 K and the reactants have more free energy than the products.

Answer:

H2<Ne<O2<NH3

Explanation:

Intermolecular forces refer to the force of attraction between molecules of a substance in any given state of matter whether solid, liquid or gas. Molecules in a substance must be held together by intermolecular forces of attraction. The magnitude of these intermolecular forces of attraction depends on many factors.

For H2, He and O2, the intermolecular force present in these gases are London forces. As the relative molecular mass of individual gas molecules becomes greater, London forces increases significantly with molecular mass. This explains the sequence shown in the answer.

NH3 has the strongest intermolecular interaction because it contains hydrogen bonds since nitrogen is an electronegative element. This a greater intermolecular interaction than dispersion forces.

Intermolecular forces are those forces that bind the molecules of the substance and the polarity of molecules. These forces range from the strongest to the weakest in ion-dipole, hydrogen bonding, and dipole to dipole.

Hence the form the strongest to the weakest is NH3, 02, Ne and H2.

Learn more about the substances in order of increasing the strength of intermolecular forces.

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Answer:

(a) The diode voltage,  [tex]V_D =[/tex]  0.776 V

(b) The diode current, [tex]I_D =[/tex] 3.81 x 10⁻²⁰ A

Explanation:

Given;

saturation current in diode, [tex]I_s[/tex] = 10⁻¹⁷ A

nonideality factor, n = 1.05

(a) the diode voltage

Given diode current, [tex]I_D[/tex] = 70 μA = 7 x 10⁻⁶ A

Diode voltage is calculated as;

[tex]V_D = nV_Tln(1+ \frac{I_D}{I_S} )[/tex]

Where;

[tex]V_T[/tex] is thermal voltage at 25°C = 0.025

[tex]V_D = 1.05 * 0.025 ln(1+ \frac{70*10^{-6}}{1*10^{-17}})\\\\V_D = 0.02625ln(1+ 7*10^{12})\\\\V_D = 0.776 \ V[/tex]

b) the diode current for VD = 0.1 mV

[tex]V_D = nV_Tln(1 +\frac{I_D}{I_S} )\\\\ln(1 +\frac{I_D}{I_S} ) = \frac{V_D}{nV_T} \\\\ln(1 +\frac{I_D}{I_S} ) = \frac{0.1*10^{-3}}{1.05*0.025} \\\\ln(1 +\frac{I_D}{I_S} ) = 0.00381\\\\1 +\frac{I_D}{I_S} = e^{0.00381}\\\\1+ \frac{I_D}{I_S}= 1.00381\\\\ \frac{I_D}{I_S}=1.00381 - 1\\\\ \frac{I_D}{I_S}= 0.00381\\\\I_D = 0.00381(I_S)\\\\I_D = 0.00381(10^{-17})\\\\I_D = 3.81*10^{-20} \ A[/tex]

Answer:

It is an example of elimination reaction through the E2 mechanism.

Explanation:

The reaction between 1-iodobutane and pyridine is an example of an E2 (bimolecular elimination) elimination reaction.

Pyridine acts predominantly as a base and the given alkyl halide is a primary alkyl halide. Both of these two factors facilitate the E2 mechanism.

Here, both H and Cl are eliminated in a single step to produce 1-butene as the product of the reaction.

The reaction mechanism and the structure of the product are shown below.

The mechanism by which 1-iodobutane reacts with pyridine is by the E2 mechanism.

The E2 mechanism process (Bimolecular Elimination) is a one-step reaction mechanism whereby carbon-hydrogen (C-H) and carbon-halogen (C-X) bonds split to generate a double bond. (C = C πbond).

The following characteristics of the E2 reaction are:

From the information given:

Pyridine functions primarily as a base, and the alkyl iodide in question is a primary alkyl halide that helps in the E2 mechanism.

In this case, both H and Cl are removed in a single step, yielding 1-butene as the byproduct of the reaction.

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Answer:

A 50:50 mixture of ethylene glycol and water is effective both summer and winter extremes in temperature because of high boiling point of 106°C and low freezing point of about -37°C. In summer when the average daily temperature rises to about 22°c, the mixture will be effective in keeping the automobile engine cool. Also in winter, when the average temperature falls below 0°C, the mixture will be effective as an antifreeze as it remains a liquid well below 0°C.

Explanation:

Ethylene glycol or antifreeze is an organic compound which is used in automobile engines as a coolant and also as an anti-freezing agent, however it does not conduct heat effectively as water due to its lower heat capacity. It has a freezing point of -12.9°C and boiling point of 197.3°C.

Water is also used as a coolant in automobile engine but it has a limited range due to its boiling point of 100°C. It is also not a good anti-freezing agent due to it high freezing point of 0.°C

However, when ethylene glycol is mixed with water in a ratio of  50:50, the property of the mixture is enhanced to both serve as a coolant and as an antifreeze. The boiling point is elevated to about 106°C while its freezing point is lowered to about -37°C.

This temperature range is effective for both summer and winter temperatures. In summer when the average daily temperature rises to about 22°c, the mixture will be effective in keeping the the automobile engine cool. Also in winter, when the average temperature falls below 0°C, the mixture will be effective as an antifreeze as it remains a liquid well below 0°C.

Answer:

CH3CH=NH2+>CH3CH2NH3+

Explanation:

A close examination of both structures will reveal that they are both amines hence they must have the polar N-H bond.

Electrons usually move towards the nitrogen atom and this makes both compounds acidic. We must also remember that some features of a compound may make it more acidic than another of close resemblance. Being more acidic may imply that the proton of the N-H is more easily lost.

CH3CH=NH2+ has an sp2 hybridized carbon atom in its structure which is known to be very electronegative due to increasing s character of the bond. It will withdraw electron density from the N-H bond thereby leading to a greater acidity of CH3CH=NH2+ in comparison to CH3CH2NH3+

The molar mass of stearic acid is 18*AC+36*AH+2*AO=18*12+36*1+16*2=284g/mol.

n=m/M=2.831/284=0.01 moles

C18H36O2+27O2-->18CO2+18H2O

we have 18*0.01=0.18 moles of CO2

18*0.01=0.18 moles of H2O

0.01*948=9.48kJ from stearic acid

0.18*393.5=70.83kJ from CO2

0.18*241.826=43.52kJ from H2O

9.48+70.83+43.52=123.83kJ

123.83*4.184=518.10kcal

Answer:

the answer is in the diagram

Explanation:

when 2,4-dimethylpent-2-ene undergo electrophilic addition reaction in the presence of HBr to form 2-bromo-2,4-dimethylpentane, it firstly lead to an intermediate carbocation

A carbocation can be describe as an organic molecule, which serves as an intermediate, that has a carbon atom bearing a positive charge and three bonds instead of four

Answer:

C3H7OH → C3H6 + H20

Explanation:

If we look at the reactant and the product we will realize that the reactant is an alcohol while the product is an alkene. The reaction involves acid catalysed elimination of water from an alcohol.

Water is a good leaving group, hence an important synthetic route to alkenes is the acid catalysed elimination of water from alcohols. Hence the conversion represented by C3H7OH → C3H6 + H20 is an elimination reaction in which water is the leaving group.

Answer:

B and C. Just finished my lesson on Edge.

Answer: stay away from doors and windows.

Explanation:

to aviod geting hit by glass

Answer:

Stay away from doors and windows.

Explanation:

Always stay in the center of the room during a tornado storm. Avoid windows, doors, and corners. If you’re near a window, the glass can shatter and hurt you.

Answer:

2 moles of HNO3

Explanation:

The equation seems to be balanced correctly. The problem is we done know what you started with. We will assume it is 3 moles of NO2.

If that is the case then 2 moles of HNO3 will be produced.

Answer:

c iodine

Explanation:

fluorine is a halogen group element like Bromine, Iodine,Astatine,Chloride

Answer:

1.00x10^-9

Explanation:

Answer:

[tex]\rm Cl_2 + 2\; KBr \to Br_2 + 2\; KCl[/tex].

One chlorine molecule reacts with two formula units of (aqueous) potassium bromide to produce one bromine molecule and two formula units of (aqueous) potassium chloride.

Explanation:

Start by finding the formula for each of the compound.

Therefore, the ratio between [tex]\rm K[/tex] atoms and [tex]\rm Br[/tex] atoms in potassium bromide is supposed to be one-to-one. That corresponds to the empirical formula [tex]\rm KBr[/tex]. Similarly, the ratio between

The formula for chlorine gas is [tex]\rm Cl_2[/tex], while the formula for bromine gas is [tex]\rm Br_2[/tex].

Write down the equation using these chemical formulas.

[tex]\rm ?\; Cl_2 + ?\; KBr \to ?\;Br_2 + ?\; KCl[/tex].

Start by assuming that the coefficient of compound with the largest number of elements is one. In this particular equation, both [tex]\rm KBr[/tex] and [tex]\rm KCl[/tex] features two elements each.

Assume that the coefficient of [tex]\rm KCl[/tex] is one. Hence:

[tex]\rm ?\; Cl_2 + 1 \; KBr \to ?\;Br_2 + ?\; KCl[/tex].

Note that [tex]\rm KBr[/tex] is the only source of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms among the reactants of this reaction.

There would thus be one [tex]\rm K[/tex] atom and one [tex]\rm Br[/tex] atom on the reactant side of the equation.

Because atoms are conserved in a chemical equation, there should be the same number of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms on the product side of the equation.

In this reaction, [tex]\rm Br_2[/tex] is the only product with [tex]\rm Br[/tex] atoms.

One [tex]\rm Br[/tex] atom would correspond to [tex]0.5[/tex] units of [tex]\rm Br_2[/tex].

Similarly, in this reaction, [tex]\rm KCl[/tex] is the only product with [tex]\rm K[/tex] atoms.

One [tex]\rm K[/tex] atom would correspond to one formula unit of [tex]\rm KCl[/tex].

Hence:

[tex]\displaystyle \rm ?\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].

Similarly, there should be exactly one [tex]\rm Cl[/tex] atom on either side of this equation. The coefficient of [tex]\rm Cl_2[/tex] should thus be [tex]0.5[/tex]. Hence:

[tex]\displaystyle \rm \frac{1}{2}\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].

That does not meet the requirements, because two of these coefficients are not integers. Multiply all these coefficients by two (the least common multiple- LCM- of these two denominators) to obtain:

[tex]\displaystyle \rm 1\; Cl_2 + 2 \; KBr \to 1\;Br_2 + 2\; KCl[/tex].

Answer:

a. New alpha- 1,6 linkages can only form if the branch has a free reducing end

b. The number of sites for enzyme action on a glycogen molecule is increased through alpha- 1,6 linkages

c. At least four glucose residues separate alpha-1,6 linkages

e. The reaction that forms alpha-1,6 linkages is catalyzed by a branching enzyme.

Explanation:

Glycogen i is the main storage polysaccharide in animals. It a homoplymer of (alpha-1-->4)-linked subunits of glucose molecules, with alpha-1--->6)-linked branches.

The alpha-1,6 branches are formed by the glycogen-branching enzyme which catalyzes the transfer of about 7 glucose residues from the non-reducing end of a glycogen branch having at least 11 residues to the C-6 hydroxyl group of a glucose residue which lies inside the same glycogen chain or another glycogen chain, thereby forming a new branch. This ensures that there are at least four glucose residues separating alpha-1,6 linkages.

The effect of branching is that it makes the glycogen molecule more soluble and also increases the number of non-reducing ends, thereby increasing the number of sites for the action of the enzymes glycogen phosphorylase and glycogen synthase.

Answer:

See explanation

Explanation:

In molecules with the same number of electron groups but different molecular geometries, the bond angles differ markedly owing to the presence of lone pairs on the central atom. Recall that lone pairs of electrons take up more space around the central atom and causes more repulsion thus squeezing the bond angle and making it less than the value expected on the basis of the Valence Shell Electron Pair Repulsion Theory.

As the number of bonding groups increases, the bond angle increases since the repulsion due to lone pairs of electrons is being progressively removed by increase in the number of bonding groups.

For 5 electron group molecules, the axial groups are oriented at a bond angle of 90° while the equatorial groups are oriented at a bond angle of 120°. In the presence of lone pairs, the equatorial bonds are removed because the equatorial bonds often have a greater bond length than the axial bonds.

In the tetrahedral geometry, four groups are bonded to the central atom while in a bent molecular geometry, only two groups are bonded to the central atom with two lone pairs present in the molecule.

Answer:

760 mmHg

Explanation:

Step 1: Given data

Step 2: Calculate the atmospheric pressure

Since air is a gaseous mixture, the atmospheric pressure is equal to the sum of the gases that compose it.

P = pN₂ + pO₂ + pAr + pt = 592 mmHg + 160 mmHg + 7 mmHg + 1 mmHg = 760 mmHg

Answer:

Catalyst

Explanation:

For the reaction:

[tex]Cl_2_(_g_)~+~3F_2_(_g_)->2ClF_3_(_g_)[/tex]

We have a main observation: When platinum is added the reaction goes faster. With this in mind, we have to remember the kinetic equilibrium theory. In figure 1, we have an energy diagram. In which we have an specific energy for the reagents and the products. When the reaction takes place, the reaction has to must go through an energy peak. This energy peak is called "activation energy". When platinum is added the activation energy decreases and the reaction can go faster. Therefore, platinum is a "catalyst", a substance with the ability to reduce the activation energy.

I hope it helps!

Answer:

The answer to your question is given below.

Explanation:

Aluminium has atomic number of 13. Thus, the electronic configuration of aluminium can be written as:

Al (13) —› 1s² 2s²2p⁶ 3s²3p¹

The orbital diagram is shown on the attached photo.

Answer: screen shot

Explanation:

Answer:

Explanation:

Calcium bicarbonate  dissolved in hard water can easily be removed by heating the hard water . On heating , it decomposes to give calcium carbonate which is insoluble and therefore can be filtered out .

Ca( HCO₃)₂  =  CaCO₃ + CO₂ + H₂O.

In this way hardness of water is removed .

Answer:

The answer is 6 because the p sublevel holds 3 orbitals and since each orbital can hold 2 electrons, the answer is 3 * 2 = 6.

Answer:

6 electrons

Explanation:

Each principal energy level above the first contains one s orbital and three p orbitals. A set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. So the answer is 6 electrons.

Answer:

[tex][H^+]=0.000285[/tex]

[tex]pH=3.55[/tex]

Explanation:

In this, we can with the ionization equation for the hydrazoic acid ([tex]HN_3[/tex]). So:

[tex]HN_3~<->~H^+~+~N_3^-[/tex]

Now, due to the Ka constant value, we have to use the whole equilibrium because this is not a strong acid. So, we have to write the Ka expression:

[tex]Ka=\frac{[H^+][N_3^-]}{[HN_3]}[/tex]

For each mol of [tex]H^+[/tex] produced we will have 1 mol of [tex]N_3^-[/tex]. So, we can use "X" for the unknown values and replace in the Ka equation:

[tex]Ka=\frac{X*X}{[HN_3]}[/tex]

Additionally, we have to keep in mind that [tex]HN_3[/tex] is a reagent, this means that we will be consumed. We dont know how much acid would be consumed but we can express a subtraction from the initial value, so:

[tex]Ka=\frac{X*X}{0.004-X}[/tex]

Finally, we can put the ka value and solve for "X":

[tex]2.2X10^-^5=\frac{X*X}{0.004-X}[/tex]

[tex]2.2X10^-^5=\frac{X^2}{0.004-X}[/tex]

[tex]X= 0.000285[/tex]

So, we have a concentration of 0.000285 for [tex]H^+[/tex]. With this in mind, we can calculate the pH value:

[tex]pH=-Log[H^+]=-Log[0.000285]=3.55[/tex]

I hope it helps!

The [H+] and pH of a 0.0040 M hydrazoic acid solution is 0.000296648 and  3.527759

What information do we have?

Hydrazoic acid solution = 0.0040 M

Ka of hydrazoic acid = 2.20 × 10⁻⁵

We know that weak acids

[H+] = √( Ka × C)

[H+] = √( 2.2 × 10⁻⁵ × 0.0040)

[H+] = 0.000296648

So,

pH = -log [H+]

pH = -log [0.000296648]

Using log calculator

pH = 3.527759

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