From the graph of Density vs. Concentration, created in Graph 1, what was the relationship between the concentration of the sugar solution and the density of the sugar solution?

Answer:

466 torr

Explanation:

Step 1: Given data

Step 2: Calculate the final pressure

Since we have a gas changing at a constant temperature, we can calculate the final pressure using Boyle's law.

P₁ × V₁ = P₂ × V₂

P₂ = P₁ × V₁ / V₂

P₂ = 626 torr × 5.00 L / 6.72 L

P₂ = 466 torr

Answer:

D

Explanation:

Addition of 0.05 M HCl, will react with all of the C2H3O2- from LiAc which will give 0.05 M more HAc. So there will be no Acetate ion left to make the solution buffer. Hence, the correct option for the this question is d, which is adding 0.050 moles of HCl.

The action that destroys the buffer is option c. adding 0.050 moles of HCl.

It is a solution of a weak acid and salt.

Here, The buffer will destroy at the time when either HC2H3O2 or NaC2H3O2 should not be present in the solution.

The addition of equal moles of HCl finishly reacts with equal moles of NaC2H3O2. Due to this,  there will be only acid in the solution.

Since

moles of HC2H3O2 = 1*0.250 = 0.250

moles of NaC2H3O2 = 1*0.050 = 0.050.

moles of HCl is added = 0.050

Now

The reaction between HCl and NaC2H3O2

[tex]HCl + NaC_2H_3O_2 \rightarrow HC_2H_3O_2 + NaCl[/tex]

Now

BCA table is

            NaC2H3O2  HCl       HC2H3O2

Before 0.050 0.050 0.250

Change -0.050 -0.050 +0.050

After 0 0 0.300

Now, the solution contains the acid (HC2H3O2 ) only.

Therefore addition of 0.050 moles of HCl will destroy the buffer.

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Answer:

A. The atomic mass is 243 amu, and the atomic number is 95.

Answer:

1. Zinc sulfide : about the same solubility, no common ion is found.

2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.

3. Lead iodide  : about the same solubility, no common ion is found.

4. Silver hydroxide : about the same solubility, no common ion is found.

Explanation:

Hello,

In this case, we first must remember that adding a common ion (which is related with the dissolving solid) decreases the solubility of the insoluble solid due to the fact Le Chatelier's principle states the reaction will shift leftwards (reactants) to reestablish equilibrium, therefore, we have:

1. Zinc sulfide : about the same solubility, no common ion is found.

2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.

3. Lead iodide  : about the same solubility, no common ion is found.

4. Silver hydroxide : about the same solubility, no common ion is found.

Best regards.

Answer: [tex]\Delta G^{0}[/tex] = 168.12 kJ

Explanation: Gibbs Free Energy, at any time, is defined as the enthalpy of the system minus product of temperature and entropy of the reaction, i.e.:

[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]

Enthalpy is defined as internal heat existent in the system. It is calculated as:

[tex]\Delta H^{0} = \Sigma H^{0}_{product} - \Sigma H^{0}_{reagent}[/tex]

Using Enthalpy Formation Table:

[tex]\Delta H^{0} = [3*(-299.9)+(-524.7)] - (-1277)[/tex]

[tex]\Delta H^{0} = 62,6 kJ[/tex]

Entropy is the degree of disorder in the system. It is found by:

[tex]\Delta S^{0} = \Sigma S^{0}_{products} - \Sigma S^{0}_{reagents}[/tex]

Calculating:

[tex]\Delta S^{0} = (-321.7) + 3(-10.8) - 0[/tex]

[tex]\Delta S^{0} = -354.1J[/tex]

And so, Gibbs Free energy will be:

[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]

[tex]\Delta G^{0} = 62600 - [298.(-354.1)][/tex]

[tex]\Delta G^{0} = 168121.8 J[/tex]

Rounding to the nearest kJ:

[tex]\Delta G^{0}[/tex] = 168.12 kJ

Answer:

The correct answer is CaO > LiBr > KI.

Explanation:

Lattice energy is directly proportional to the charge and is inversely proportional to the size. The compound LiBr comprises Li+ and Br- ions, KI comprises K+ and I- ions, and CaO comprise Ca²⁺ and O²⁻ ions.  

With the increase in the charge, there will be an increase in lattice energy. In the given case, the lattice energy of CaO will be the highest due to the presence of +2 and -2 ions. K⁺ ions are larger than Li⁺ ion, and I⁻ ions are larger than Br⁻ ion.  

The distance between Li⁺ and Br⁻ ions in LiBr is less in comparison to the distance between K⁺ and I⁻ ions in KI. As a consequence, the lattice energy of LiBr is greater than KI. Therefore, CaO exhibits the largest lattice energy, while KI the smallest.  

Arranging the chemical compounds in order of decreasing magnitude of lattice energy, we have:

c. CaO.

a. LiBr

b. KI

Lattice energy can be defined as a measure of the energy required to dissociate one (1) mole of an ionic compound into its constituent anions and cations, in the gaseous state.

Hence, it is typically used to measure the bond strength of ionic compounds.

Generally, lattice energy is inversely proportional to the size of the ions and directly proportional to their electric charges.

Lithium bromide (LiBr) comprises the following ions:

Potassium iodide (KI) comprises the following ions:

Calcium oxide (CaO) comprises the following ions:

From the above, we can deduce that there is an increase in the charge possessed by the ionic chemical compounds and as such this would result in an increase in the lattice energy.

In order of decreasing magnitude of lattice energy, the chemical compounds are arranged as:

I. CaO.

II. KI.

III. LiBr.

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Explanation:

An atom undergoes alpha decay by losing a helium atom.

So when bismuth undergoes alpha decay, we have;

²¹⁰₈₃Bi --> ⁴₂He + X

Mass number;

210 = 4 + x

x = 206

Atomic number;

83 = 2 + x

x = 81

The element is Thallium. The symbol is Ti.

For the second part;

X --> ⁴₂He + ²³⁴₉₀Th

Mass number;

x = 4 + 234 = 238

Atomic Number;

x = 2 + 90 = 92

The balanced nuclear equation is;

²³⁸₉₂U --> ⁴₂He + ²³⁴₉₀Th

Answer:

The oxidation state of silver in [tex]\rm Ag_2O[/tex] is [tex]+1[/tex].

The oxidation state of sulfur in [tex]\rm SO_2[/tex] is [tex]+4[/tex].

Explanation:

The oxidation states of atoms in a compound should add up to zero.

There are two silver [tex]\rm Ag[/tex] atoms and one oxygen [tex]\rm O[/tex] atom in one formula unit of [tex]\rm Ag_2O[/tex]. Therefore:

[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].

The oxidation state of oxygen in most compounds (with the exception of peroxides and fluorides) is [tex]-2[/tex]. Silver oxide [tex]\rm Ag_2O[/tex] isn't an exception. Therefore:

[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times (-2) = 0\end{aligned}[/tex].

Solve this equation for the (average) oxidation state of [tex]\rm Ag[/tex]:

[tex]\text{Oxidation state of $\rm Ag$} = 1[/tex].

Similarly, because there are one sulfur [tex]\rm S[/tex] atom and two oxygen [tex]\rm O[/tex] atoms in each [tex]\rm SO_2[/tex] molecules:

[tex]\begin{aligned}&\rm 1\times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].

The oxidation state of [tex]\rm O[/tex] in [tex]\rm SO_2[/tex] is also [tex]-2[/tex], not an exception, either.

Therefore:

[tex]\begin{aligned}&\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times (-2) = 0\end{aligned}[/tex].

Solve this equation for the oxidation state of [tex]\rm S[/tex] here:

[tex]\text{Oxidation state of $\rm S$} = 4[/tex].

Answer:

We can have: Calcium, strontium, or barium

Explanation:

In this case, we have to remember the solubility rules for sulfate [tex]SO_4~^-^2[/tex] and the chloride [tex]Cl^-[/tex]:

Sulfate

All sulfate salts are SOLUBLE-EXCEPT those also containing: Calcium, silver, mercury (I), strontium, barium or lead.([tex]Ca^+^2~,Ag^+~,Hg_2^+^2~,Sr^+2~,Ba^+^2~,Pb^+^2[/tex]), which are NOT soluble.

Chloride

All chloride salts as SOLUBLE-EXCEPT those also containing: lead, silver, or mercury (I). ([tex]Pb^+^2~,Ag^+~,Hg_2~^+^2[/tex]), which are NOT soluble.

If we the salt formed a precipitated with the sulfate anion, we will have as possibilities "Calcium, silver, mercury (I), strontium, barium or lead". If We dont have any precipitated with the Chloride anion we can discard "Silver, mercury (I),  lead" and our possibilities are:

"Calcium, strontium, or barium".

I hope it helps!

Answer:

CO2 will diffuse more rapidly.

Explanation:

From Graham's law of diffusion, we understood that the rate of diffusion of a gas is inversely proportional to the square root of its density as shown below:

Rate (R) & 1/√Density (d)

R & 1/√d

But, the density of a gas is directly proportional to the relative molecular mass (M) of the gas.

Thus, we can say that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. This can be represented mathematically as:

Rate (R) & 1/√Molar mass (M)

R & 1/√M

From the above illustration, we can say that the lighter the gas, the faster the rate of diffusion and the heavier the gas, the slower the rate of diffusion.

Now, to answer the question given above,let us determine the molar mass of Cl2 and CO2.

This is illustrated below:

Molar mass of Cl2 = 2 x 35.5 = 71 g/mol

Molar mass of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol

Summary

Gas >>>>>> Molar mass

Cl2 >>>>>> 71 g/mol

CO2 >>>>> 44 g/mol

From the illustration above, we can see that CO2 is lighter than Cl2.

Therefore, CO2 will diffuse more rapidly.

Answer: CO2

Explanation:

Answer:

It has 2

Explanation:

The significant figures are 1 and 5!

Hope this helps:)

Answer:

3BaO + Al₂(SO₄)₃  →  Al₂O₃+ 3BaSO₄

Explanation:

Yes! A reactiin occurs between barium hydroxide and auminium sulphate.

barium sulfate (BaSO4) and aluminum hydroxide (Al(OH)3) are the products obtained in this reaction.

The reaction is given by the equation below;

3BaO + Al₂(SO₄)₃  →  Al₂O₃+ 3BaSO₄

Answer:

Following are the solution to this equation:

Explanation:

In the given-question, an attachment file of the choices was missing, which can be attached in the question and its solution can be defined as follows:

In the given question "Option (iii)" is correct, which is defined in the attachment file.

When 2,2,3-trimethylcyclohexanone reacts with hydroxylamine it will produce the 2,2,3-trimethylcyclohexanoxime.

The given question is incomplete, the complete question is:

How does the spontaneity of the process below depend on temperature? PCI5(9)+H2O(g)POCI3(g) +2HCI(g) -126 kJ mol1, AS = 146 J K-'mol1 ΔΗ Select the correct answer below: nonspontaneous at all temperatures spontaneous at all temperatures spontaneous at high temperatures and nonspontaneous at low temperatures spontaneous at low temperatures and nonspontaneous at high temperatures

Answer:

The correct answer is spontaneous at all the temperatures.

Explanation:

Gibbs Free energy is an essential relation that determines the spontaneity of any reaction, that is, ΔG = ΔH - TΔS

When ΔG is less than zero, that is, negative, the reaction is considered to be in spontaneous state. Based on the given information, ΔH = -126 kJ/mol

= -126000 J/mol, it is negative

ΔS = 146 J/K/mol, it is positive

Now, ΔG = ΔH-TΔS

= (-ve) - T (+ve), Thus, when ΔH, is -ve, ΔS is +ve, -TΔS is -ve, the ΔG will be -ve. Therefore, reaction will be spontaneous at all the temperatures.  

Answer:

Cu2 + 2Mg-> 2Cu+ Mg2

Explanation:

Balance the equation and make sure both the reactant and the products are the same

Hope it will be helpful

[tex]Cu^{+2} + 2Mg[/tex]  -> [tex]2Cu + Mg^+2[/tex]  is the correct half-reactions.

A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total oxidation numbers is the same for both the reactants and the products.

[tex]Cu^{+2} + 2Mg[/tex]  -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.

Magnesium is oxidized because its oxidation state increased from 0 to +2 while Cu is reduced because its oxidation state decreased from +2 to 0.

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Answer:

Reaction of Chlorine with Hydrogen Chlorine and Hydrogen mixed together explodes when exposed to sunlight, which produces Hydrogen Chloride. In the dark away from sunlight, no reaction occurs, so light energy is required for a reaction. Cl2 + H2 = 2 HCl Reaction of Chlorine with Non-Metals Chlorine directly combines with most non-metals.

Explanation:

I hope this helps bro

Answer:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases.

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases.

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases.

Explanation:

Hello.

In this case, we can understand a higher entropy when more disorder is present and a lower entropy when less disorder is present, thus:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases since iodine has the greatest molar mass (254 g/mol) and fluorine the least molar mass (38 g/mol).

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases since hydrogen peroxide weights 34 g/mol as well as hydrogen sulfide but the peroxide has more bonds (more complex, higher entropy).

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases since diamond has a well-ordered structure and amorphous carbon has a very disordered one.

Best regards.

Answer:

A double bond is formed when two pairs of electrons are shared between the two participating atoms. It is represented by two dashes (=). It is represented by two dashes (=). Double covalent bonds are much stronger than a single bond, but they are less stable

Explanation:

Answer:

A. Sharing valence electrons between atoms.

Explanation:

This is the definition of a covalent bond. Option B describes ionic bonds, Option C describes intermolecular forces, and Option D is wrong because then there wouldn't be any mention of them in our high school chemistry textbooks :).  

Answer:

a. Gly-Lys + Leu-Ala-Cys-Arg + Ala-Phe

b. Glu-Ala-Phe + Gly-Ala-Tyr

Explanation:

In this case, we have to remember which peptidic bonds can break each protease:

-) Trypsin

It breaks selectively the peptidic bond in the carbonyl group of lysine or arginine.

-) Chymotrypsin

It breaks selectively the peptidic bond in the carbonyl group of phenylalanine, tryptophan, or tyrosine.

With this in mind in "peptide a", the peptidic bonds that would be broken are the ones in the "Lis" and "Arg" (See figure 1).

In "peptide b", the peptidic bond that would be broken is the one in the "Phe" (See figure 2). The second amino acid that can be broken is tyrosine, but this amino acid is placed in the C terminal spot, therefore will not be involved in the hydrolysis.

Answer:

Zeros located at the end of significant figures are significant.

Explanation:

Hope it will help :)

Answer:

The final temperature of the entire water sample after all the ice is melted, is 12,9°C. We should realize that if there is no loss of heat in our system, the sum of lost or gained heat is 0.  It is logical to say that the temperature has decreased because the ice gave the water "heat" and cooled it

Thats all i know

Answer:

(a) when a reaction system reaches a state of equilibrium, the concentration of the products is equal to the concentration of the reactants

Determining whether the statements about equilibrium is True or False

A) The concentration of the products is equal to the concentration of the reactants at equilibrium : TRUE

B) When a system is at equilibrium, Keq = 1 : TRUE

C) The rates of the forward reaction and the reverse reaction are equal at equilibrium :  TRUE

D) Adding a catalyst to a reaction system will shift the position of equilibrium to the right : FALSE

In a chemical reaction at equilibrium the value of Keq will be equal to 1 because the concentration of the products is equal to the concentration of the reactants in the chemica reaction. Also at equilibrium the rate of forward reaction is same as the rate of reverse reaction.

A catalyst can only affect the rate of reaction and not the amount of product ( yield of reaction).

Hence we can conclude that the answers to your questions are as listed above.

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Answer:

[tex]\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}[/tex]

Explanation:

1. Density from mass and volume

[tex]\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}[/tex]

2. Volume from density and mass

[tex]V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}[/tex]

3. Mass from density and volume

[tex]\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}[/tex]

4. Density by displacement

Volume of water + object = 24.6 mL

Volume of water                = 12.8 mL

Volume of object               = 11.8 mL

[tex]\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}[/tex]

Your drawing showing water displacement using a graduated cylinder should resemble the figure below.

Answer:

The answer is "Tertiary carbon".

Explanation:

Accent to the results, the carbon-bromine bond is weak, whenever, the bromine is connected to tertiary carbon so, bonding energy is separation for methyl-carbon, which is connected to the bromine = 70 kcal/mol and for the primary energy to the secondary energy is=  68 kcal/mol, and for tertiary CO2 = 65 kcal/mol.

The stronger the energy dissociating connection and the weaker, its power dissociation connection and its weaker bond becomes connecting with a tertiary carbon, that's why "Tertiary carbon" is the correct answer.

Answer:

pH of solution B is 5

Explanation:

A weak acid, HA, is in equilibrium with water as follows:

HA(aq) + H₂O(l) ⇄ A⁻(aq) + H₃O⁺(aq)

Where Ka (10^-pKa = 1x10⁻⁹) is:

Ka = 1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]

Where concentrations of this species are equilibrium concentrations

As initial concentration of HA is 0.1M, the equilibrium concentrations of the species are:

[HA] = 0.1M - X

[A⁻] = X

[H₃O⁺] = X

Where X is the amount of HA that reacts until reach the equilibrium, X is reaction coordinate.

Replacing in Ka expression:

1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]

1x10⁻⁹ = [X] [X] / [0.1 - X]

1x10⁻¹⁰ - 1x10⁻⁹X = X²

1x10⁻¹⁰ - 1x10⁻⁹X - X² = 0

Solving for X:

X = -0.00001 → False solution, there is no negative concentrations.

X = 1x10⁻⁵ → Right solution.

As [H₃O⁺] = X

[H₃O⁺] = 1x10⁻⁵M

And pH = -log[H₃O⁺]

pH = 5

Answer:

Ratio of Vrms of argon to Vrms of hydrogen = 0.316 : 1

Explanation:

The root-mean-square speed measures the average speed of particles in a gas, and is given by the following formula:  

Vrms = [tex]\sqrt{3RT/M}[/tex]

where R is molar gas constant = 8.3145 J/K.mol, T is temperature in kelvin, M is molar mass of gas in Kg/mol

For argon, M = 40/1000 Kg/mol = 0.04 Kg/mol, T = 4T , R = R

Vrms = √(3 * R *4T)/0.04 = √300RT

For hydrogen; M = 1/1000 Kg/mol = 0.001 Kg/mol, T = T, R = R

Vrms = √(3 * R *T)/0.001 = √3000RT

Ratio of Vrms of argon to that of hydrogen = √300RT / √3000RT = 0.316

Ratio of Vrms of argon to that of hydrogen = 0.316 : 1

Answer:

4.22

Explanation:

We know from the question, that the pOH of the solution is 9.78. Now the pOH is defined as -log [OH^-].

If the pOH of a solution is given, one may obtain the pH of such solution from the formula;

pH + pOH =14

Hence we can write;

pH = 14-pOH

pH = 14 - 9.78 = 4.22

Hence the pH of the solution is 4.22.

Answer:

A. Ka = [CO2] / [C] [O2]^1/2

B. Kb = [CO2] / [CO] [O2]^1/2

Explanation:

Equilibrium constant is simply defined as the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.

Now, we shall obtain the expression for the equilibrium constant for the reaction as follow:

A. Determination of the expression for equilibrium constant Ka.

This is illustrated below:

C(s) + 1/2 O2(g) <==> CO(g)

Ka = [CO2] / [C] [O2]^1/2

B. Determination of the expression for equilibrium constant Kb.

This is illustrated below:

CO(g) + 1/2 O2(g) <==> CO2(g)

Kb = [CO2] / [CO] [O2]^1/2

Answer:

0.423 m.

Explanation:

The following data were obtained from the question:

Mass of glycine (NH2CH2COOH) = 141.5 g

Mass of water = 4.456 kg

Molality =.?

Next, we shall determine the number of mole in 141.5 g of glycine (NH2CH2COOH.

This is illustrated below:

Mass of glycine (NH2CH2COOH) = 141.5 g

Molar mass of glycine (NH2CH2COOH) = 14 + (2x1) + 12 + (2x1) + 12 + 16 + 16 + 1 = 75 g/mol

Mole of glycine (NH2CH2COOH) =.?

Mole = mass /Molar mass

Mole of glycine (NH2CH2COOH) = 141.5/75

Mole of glycine (NH2CH2COOH) = 1.887 moles

Finally, we shall determine the molality of the solution as follow:

Molality is simply defined as the mole of solute per kilogram of water. Mathematically it is expressed as:

Molality = mole / mass (kg) of water

With the above formula, we can obtain the molality of the solution as follow:

Mole of glycine (NH2CH2COOH) = 1.887 moles

Mass of water = 4.456 kg

Molality =.?

Molality = mole /mass (kg) of water

Molality =1.887/4.456

Molality = 0.423 m

Therefore, the molality of the solution is 0.423 m