A sample of gas in a cylinder as in the example in Part A has an initial volume of 48.0 L , and you have determined that it contains 1.80 moles of gas. The next day you notice that some of the gas has leaked out. The pressure and temperature remain the same, but the volume has changed to 12.0 L . How many moles of gas (n2) remain in the cylinder

Answer:

(2) Mg²⁺ loses 2 electrons.

Explanation:

Let's consider the following half-reaction.

Mg ⇒ Mg²⁺ + 2 e⁻

Magnesium is a metal that loses 2 electrons to fulfill the octet rule. When magnesium loses electrons, the oxidation number increases from 0 to +2. This kind of half-reaction is known as oxidation, and has to be accompanied by a reduction of another species.

Answer:

a

Explanation:

Matter can be volume or density. So, this concludes that it is when it takes up space.

Answer: A.

Explanation:

it takes up space

Answer:

1.076 mol (corrected to 2 d.p.)

Explanation:

Take the atomic mass of Ga be 69.7.

since no. of moles = mass/ molar mass,

no. of moles of Ga used = 100.0 / 69.7

= 1.43472023 mol

From the balanced equation, the mole ratio of Ga:S2 = 4:3, which means every 4 moles of  Ga can react completely with 3 moles of S2.

So, let the no. of moles of Sulphur required be y.

[tex]\frac{4}{3} =\frac{ 1.43472023}{y}[/tex]

4 y =  1.43472023 x 3

y = 1.076 mol (corrected to 2 d.p.)

Answer:

[tex]V=27992L=28.00m^3[/tex]

Explanation:

Hello,

In this case, the combustion of methane is shown below:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

[tex]n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4[/tex]

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

[tex]PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3[/tex]

Best regards.

Answer:

THE ENTHALPY OF SOLUTION IS 3153.43 J/MOL OR 3.15 KJ/MOL.

Explanation:

1. write out the variables given:

Mass of Calcium chloride = 13.6 g

Change in temperature = 31.75°C - 25.00°C = 6.75 °C

Density of the solution = 1.000 g/mL

Volume = 100.0 mL = 100.0 mL

Specific heat of water = 4.184 J/g °C

Mass of the water = unknown

2. calculate the mass of waterinvolved:

We must first calculate the mass of water in the bomb calorimeter

Mass = density  * volume

Mass = 1.000 * 100

Mass = 0.01 g

3. calculate the quantity of heat evolved:

Next is to calculate the quantity of heat evolved from the reaction

Heat = mass * specific heat of water * change in temperature

Heat = mass of water * specific heat *change in temperature

Heat = 13.6 g * 4.184 * 6.75

Heat = 13.6 g * 4.184 J/g °C * 6.75 °C

Heat = 384.09 J

Hence, 384.09J is the quantity of heat involved in the reaction of 13.6 g of calcium chloride in the calorimeter.

4. calculate the molar mass of CaCl2:

Next is to calculate the molar mas of CaCl2

Molar mass = ( 40 + 35.5 *2) = 111 g/mol

The number of moles of 13.6 g of CaCl2 is then:

Number of moles of CaCl2 = mass / molar mass

Number of moles = 13.6 g / 111 g/mol

Number of moles = 0.1225 mol

So 384.09 J of heat was involved in the reaction of 1.6 g of CaCl2 in a calorimter which translates to 0.1225 mol of CaCl2..

5. Calculate the enthalpy of solution in kJ/mol:

If 1 mole of CaCl2 is involved, the heat evolved is therefore:

Heat per mole = 384.09 J / 0.1225 mol

Heat = 3 135.43 J/mol

The enthalpy of solution is therefore 3153.43 J/mol or 3.15 kJ/mol.

Answer:

See explanation

Explanation:

In this case, we have to check the neighbors of each carbon in the molecule. In propane, we have two types of carbons (see figure 1) (blue and red ones). The red carbons are equivalent. (Both have the same neighbors). Now, we can analyze each carbon:

Blue carbon

In the blue carbon, we have 6 hydrogens neighbors (three on each methyl). If we follow the n+1 rule, we will have:

6+1= 7

For the blue carbon, we will have a Septet.

Red carbons

In the red carbon, we have only 2 neighbors (the carbon in the middle only have 2 hydrogens). If we follow the n+1 rule, we will have:

2+1=3

For the red carbon, we will have a triplet.

See figure 2

I hope it helps!

Answer:

number of ions = 12.04 x 10^²³

Explanation:

n = number of ions/Avogadro's constant

2 = number of ions/6.02 x 10^²³

number of ions= 2 x 6.02 x 10^²³

number of ions = 12.04 x 10^²³

Answer and Explanation:

The buffer solution is composed by sodium acetate (CH₃COONa) and acetic acid (CH₃COOH). Thus, CH₃COOH is the weak acid and CH₃COO⁻ is the conjugate base, derived from the salt CH₃COONa.

If we add a strong base, such as barium hydroxide, Ba(OH)₂, the base will dissociate completely to give OH⁻ ions, as follows:

Ba(OH)₂ ⇒ Ba²⁺ + 2 OH⁻

The OH⁻ ions will react with the acid (CH₃COOH) to form the conjugate base CH₃COO⁻.

Initial number of moles of CH₃COOH = 0.40 mol/L x 1 L = 0.40 mol

Initial number of moles of CHCOO⁻= 0.31 mol/L x 1 L = 0.31 mol

moles of OH- added: 2 OH-/mol x 0.100 mol/L x 1 L = 0.200 OH-

According to this, the following are the answers to the sentences:

a. The number of moles of CH₃COOH will remain the same ⇒ FALSE

The number of moles of CH₃COOH will decrease, because they will react with OH⁻ ions

b. The number of moles of CH₃COO⁻ will increase ⇒ TRUE

Moles of CH₃COO⁻ will be formed from the reaction of the acid (CH₃COOH) with the base (OH⁻ ions)

c. The equilibrium concentration of H₃O⁺ will decrease ⇒ FALSE

The equilibrium concentration of OH⁻ is increased

d. The pH will decrease⇒ FALSE

pKa for acetic acid is 4.75. We add the moles of base to the acid concentration and we remove the same number of moles from the conjugate base in the Henderson-Hasselbach equation to calculate pH:

[tex]pH= pKa + log \frac{[conjugate base + base]}{[acid - base]}[/tex]

pH = 4.75 + log (0.31 mol + 0.20 mol)/(0.40 mol - 0.20 mol) = 5.15

Thus, the pH will increase.

Answer:

[PCl₅] = 0.5646M

[PCl₃] = 0.1174M

[Cl₂] = 0.1174M

Explanation:

In the reaction:

PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)

K equilibrium is defined as:

K = 2.44x10⁻² = [PCl₃] [Cl₂] / [PCl₅]

The initial moles of each compound when volume is 15.3L are:

PCl₅ = 0.300mol/L×15.3L = 4.59mol

Cl₂ = 8.55x10⁻²mol/L×15.3L = 1.308mol

PCl₃ = 8.55x10⁻²mol/L×15.3L = 1.308mol

At 8.64L, the new concentrations are:

[PCl₅] = 4.59mol / 8.64L = 0.531M

[PCl₃] = 1.308mol / 8.64L = 0.151M

[Cl₂] = 1.308mol / 8.64L = 0.151M

At these conditions, reaction quotient, Q, is:

Q = [0.151M] [0.151M] / [0.531M]

Q = 4.29x10⁻²

As Q > K, the reaction will shift to the left producing more reactant, that means equilibrium concentrations are:

[PCl₅] = 0.531M + X

[PCl₃] = 0.151M - X

[Cl₂] = 0.151M - X

Where X is reaction coordinate.

Replacing in K expression:

2.44x10⁻² = [0.151M - X] [0.151M - X] / [0.531M + X]

1.296x10⁻² + 2.44x10⁻²X = 0.0228 - 0.302X + X²

0 = 9.84x10⁻³ - 0.3264X + X²

Solving for X:

X = 0.293 → False solution. Produce negative concentrations

X = 0.0336M → Right solution.

Replacing:

[PCl₅] = 0.531M + 0.0336

[PCl₃] = 0.151M - 0.0336

[Cl₂] = 0.151M - 0.0336

Answer:

8 mol

Explanation:

Step 1: Given data

Moles of copper (II) sulfate: 2 mol

Chemical formula of copper (II) sulfate: CuSO₄

Step 2: Establish the molar ratio of copper (II) sulfate to oxygen

According to the chemical formula, the molar ratio of copper (II) sulfate to oxygen is 1:4.

Step 3: Calculate the moles of O in 2 mol of CuSO₄

[tex]2molCuSO_4 \times \frac{4molO}{1molCuSO_4} = 8molO[/tex]

Answer:

+2

Explanation:

Because the charge of the chloride ion is negative, that means that the charge of the magnesium ion must be positive since cations and anions go together, not cation and cation nor anion and anion. Using the "reverse criss-cross method", since the subscript of Mg is 1, that means that this is the lowest whole number ratio so we don't need to worry about simplifying. Therefore, since the charge of Cl is 2, the answer is +2.

Answer:

It's harder to melt sodium chloride because of its density and stability.

Explanation:

The bond between Na+ and Cl- is an ionic bond, meaning its an ionic compound. Ionic compounds have more inter molecular forces(forces involving multiple molecules). Chlorine has much less of these forces than NaCl and isn't as dense.

Answer:

It's harder to melt sodium chloride because of its density

Explanation:

Answer:

HCl is limiting reactant

Theoretical yield: 23.8g Cl₂

Actual yield: 17.6g C₂

Explanation:

Based on the reaction:

4HCl(aq)+MnO2(s)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

4 moles of HCl reacts per mole of MnO₂ to produce 1 mole of MnCl₂ and Cl₂ and 2 moles of water.

To find the limiting reactant you must know the moles of each reactant and knowing that 4 moles of HCl reacts per mole of MnO₂ you can sikve this problem, thus:

Moles HCl (Molar mass: 36.46g/mol): 48.9g ₓ (1mol / 36.46g/mol) =

1.341 moles HCl

Moles MnO₂ (Molar mass: 86.937g/mol): 36.9g ₓ (1mol / 86.937g) =

0.424 moles MnO₂

For a complete reaction of 0.424 moles of MnO₂ you require:

0.424moles MnO₂ ₓ (4 moles HCl / 1 mole MnO₂) = 1.696 moles of HCl.

As you have just 1.341 moles of HCl. HCl is limiting reactant.

Theoretical yield means, in the reaction, that 4 moles of HCl will produce 1 mole of Cl₂. As moles of HCl are 1.341:

1.341 moles HCl ₓ (1 mole Cl₂ / 4 moles HCl) = 0.33525 moles Cl₂

In grams (Molar mass Cl₂: 70.9g/mol):

Theoretical yield: 0.33525 moles Cl₂ ₓ (70.9g / mol) = 23.8g Cl₂

As yield of reaction is 74.7%, the real mass of Cl₂ you obtain (Actual yield) is:

23.8g Cl₂ ₓ 74% = 17.6g C₂

Answer:

The molar mass of the unknown gas is [tex]\mathbf{ 51.865 \ g/mol}[/tex]

Explanation:

Let assume that  the gas is  O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires  time t₂  =  6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

[tex]R \ \alpha \ \dfrac{1}{\sqrt{d}}[/tex]

[tex]R = \dfrac{k}{d}[/tex]  where K = constant

If we compare the rate o diffusion of two gases;

[tex]\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}[/tex]

Since the density of a gas d is proportional to its relative molecular mass M. Then;

[tex]\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}[/tex]

Rate is the reciprocal of time ; i.e

[tex]R = \dfrac{1}{t}[/tex]

Thus; replacing the value of R into the above previous equation;we have:

[tex]\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}[/tex]

We can equally say:

[tex]{\dfrac{t_2}{t_1}}= {\sqrt{\dfrac{M_2}{M_1}}[/tex]

[tex]{\dfrac{6.34}{4.98}}= {\sqrt{\dfrac{M_2}{32}}[/tex]

[tex]M_2 = 32 \times ( \dfrac{6.34}{4.98})^2[/tex]

[tex]M_2 = 32 \times ( 1.273092369)^2[/tex]

[tex]M_2 = 32 \times 1.62076418[/tex]

[tex]\mathbf{M_2 = 51.865 \ g/mol}[/tex]

Answer:

1) dispersion forces

2) ion-dipole forces

Explanation:

For the formation of a solution of the substances, CH3CH2CH2CH2CH3 and CH3CH2CH2CH2CH2CH3, dispersion forces are involved. Recall that the both are organic substances hence they are both molecular in nature. They are not only molecular, they are nonpolar molecules. This implies that only dispersion forces are the possible intermolecular forces.

For the formation of a solution of LiNO3 and H2O, the dipoles of water interacts strongly with the ions Li^+ and NO3^- thereby separating the ions in solution. The positive dipole of water interacts with NO3^- while the negative dipole of water interacts with Li^+.

Therefore, the interaction between nonpolar molecules is dispersion forces while between polar molecules is ion-dipole forces.

Dispersion Forces:

They are the weakest form of intermolecular force. The attraction between molecules is temporary.

For example- Interaction between [tex]\bold { CH_3CH_2CH_2CH_2CH_3 }[/tex] and [tex]\bold { CH_3CH_2CH_2CH_2CH_2CH_3 }[/tex]

Ion-dipole forces:

The attraction between molecules is due to the interaction between opposite ions of molecules.

For example- Interaction between [tex]\bold { LiNO_3 }[/tex] and [tex]\bold { H_2O }[/tex]

Therefore, the interaction between nonpolar molecules is dispersion forces while between polar molecules is ion- dipole forces.

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Answer:

D.They have properties similar to those of their component elements.

Explanation:

When elements are joined, the atoms lose their individual properties and have different properties from the elements they are made of .

Answer:

The colour of the orange solution becomes yellow.  

Explanation:

1. Before adding NaOH

Assume the picture showed a beaker of potassium chromate and one of potassium dichromate.

Both solutions are involved in the same equilibrium:

[tex]\rm\underbrace{\hbox{2CrO$_{4}^{2-}$(aq)}}_{\text{yellow}} +2H^{+}(aq) \rightleftharpoons \, \underbrace{\hbox{Cr$_{2}$O$_{7}^{2-}$}}_{\text{orange}} + H_{2}O[/tex]

The first beaker contains mostly chromate ions with a few dichromate ions.

The position of equilibrium lies to the left and the solution is yellow.

The second beaker contains mostly dichromate ions with a few chromate ions.

The position of equilibrium lies to the right and the solution is orange.

2. After adding NaOH

According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

Beaker 1

If you add OH⁻ to the equilibrium solution, it removes the H⁺ (by forming water).

The system responds by having the dichromate react with water to replace the H⁺.  

At the same time, the system forms more of the yellow chromate ion.

The position of equilibrium shifts to the left.

However, the solution is already yellow, so you see no change in colour.

Beaker 2

The reaction is the same as in Beaker 1.

This time, however, as the dichromate ion disappears, do does its orange colour.

Also, the yellow chromate is being formed and its yellow colour appears .

The colour changes from orange to yellow.

Complete Question

The complete question is shown on the first uploaded image

Answer:

The  initial concentration is  [tex]C_f = 0.0022 \ M[/tex]

Explanation:

From the question we are told that

    The volume of solution A is  [tex]V_i = 10.0 mL[/tex]

    The concentration of A is [tex]C_i = 0.0200 \ M[/tex]

    The volume of solution B  is  [tex]V_B = 10.0mL[/tex]

    The volume of water is  [tex]V_{w } = 70.0 mL[/tex]

Generally the law of dilution is mathematically represented as

             [tex]C_i * V_i = C_f * V_f[/tex]

Where  [tex]C_f[/tex] is the concentration of  the mixture

            [tex]V_f[/tex] is the volume of the mixture which is mathematically evaluated as

            [tex]V_f = 10 + 10 + 70[/tex]

           [tex]V_f = 90mL[/tex]

So  

      [tex]C_f = \frac{C_i * V_i}{ V_f}[/tex]

substituting values

       [tex]C_f = \frac{0.0200 * 10 }{90}[/tex]

       [tex]C_f = 0.0022 \ M[/tex]

Note the mixture obtained is  [tex]KIO_3[/tex]

Answer: molar ratio: 0.58

Explanation:

This is a buffer problem. We would use the equation [tex]pH=pKa+log(\frac{base}{acid} )[/tex]. Since we were given Ka and pH, we can find the ratio.

pKa=-log(Ka)

pKa=-log(1.8×10⁻⁵)

pKa=4.74

Now that we have pKa, we can plug in our values and find the molar ratio. Instead of writing base/acid, let's use x.

4.5=4.74+log(x)

-0.24=log(x)

10⁻⁰⁻²⁴=x

x=0.58

Answer: The Answer should be B. 0.887 (:

Explanation:

The mole fraction of the solution is 0.887.

Mass percent of LiCl present =  23.0%

Mass of the LiCl can be obtained from;

23 = x/100 × 100/1

x = 23 g

Number of moles of LiCl = 23g/42 g/mol = 0.55 moles

Number of moles of water = (100 - 23) g/18 g/mol = 4.3 moles

Total number of moles = 0.55 + 4.3 = 4.85 moles

Mole fraction of water = 4.3 moles/4.85 moles = 0.887

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Answer:

double replacement

Explanation:

Double replacement reactions are where both cations and anions of reactants switch.

For eg: AB + CD -----> AD + BC (Here we see both things are replaced)

Synthesis reactions is where two simple things make one complex thing.

So A + B --- > AB

Combustion is usually reaction where Oxygen is reactant but here we don't have that.

Single replacement only replaces either cation or anion.

So AB + C .------> AC + B

Answer:

Replacement

Explanation:

Answer:

1. stability factor

2. steric hindrance factor

Explanation:

stability of ethane is lesser to that of two tert-butyl, so ethane will be more reactive and faster.

ethane is less hindered and more reactive, while two tert-butyl is more hindered and less reactive

Answer:

pH = 7.29

Explanation:

Ka of butanoic acid is 1.54x10⁻⁵

To obtain the pH of the solution you must use H-H equation for butanoic acid:

pH = pKa + log₁₀ [C₃H₇COO⁻] / [C₃H₇COOH]

Where pKa is defined as -log Ka = 4.81

Now, you need to find [C₃H₇COO⁻] and [C₃H₇COOH] concentrations (Also, you can find moles of each substance and replace them in the equation.

Butanoic acid reacts with LiOH, producing C₃H₇COO⁻, thus:

C₃H₇COOH + LiOH → C₃H₇COO⁻ + H₂O + Li⁺

Moles of both reactants, C₃H₇COOH and LiOH are:

C₃H₇COOH = 0.0300L ₓ (0.1000mol / L) = 0.003000moles of C₃H₇COOH

LiOH = 0.0299L ₓ (0.1000mol / L) = 0.00299 moles of LiOH.

That means moles of C₃H₇COO⁻ produced are 0.00299 moles.

And moles of C₃H₇COOH that remains in solution are:

0.00300 - 0.00299 = 0.00001 moles of C₃H₇COOH

Replacing in H-H equation:

pH = pKa + log₁₀ [C₃H₇COO⁻] / [C₃H₇COOH]

pH = 4.81 + log₁₀ [0.00299moles] / [0.00001moles]

Answer:

CH3CH2CH2COOH<CH2(F)CH2CH2COOH<CH3CH(F)CH2COOH<CH2(F)CH(F)CH2COOH

Explanation:

We know that the presence of highly electronegative elements in carboxylic acid molecules lead to -I inductive effect. This implies that electrons are withdrawn along the chain towards the electronegative element. As electrons are withdrawn towards the electronegative element, the electron cloud of the carbonyl- hydrogen bond in the acid weakens and the hydrogen can now be easily lost as a proton, that is , the molecule becomes more acidic.

The -I inductive effect increases with increase in the number of electronegative elements present in the molecule and the proximity of the electronegative element to the carbonyl group. The closer the electronegative element is to the carbonyl group, the greater the acidity of the molecule since the -I inductive effect dies out with increasing distance from the carbonyl group. Also, the more the number of electronegative elements in the molecule, the greater the - I inductive effect and the greater the acidity of the molecule, hence the answer.

Answer:

Formula for effective nuclear charge is as follows. So, for He atom value of S = 0.30 because the electrons are present in 1s orbital. Therefore, calculate the effective nuclear charge for helium as follows. Thus, we can conclude that the effective nuclear charge for helium is 1.7

Explanation:

The effective nuclear charge experienced by a 1s electron in helium is +1.70.

Answer:

the answer to this question is C

The electron configuration of the iodide ion is 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶. The correct option is option C.

The arrangement of an atom's or molecule's electrons in their respective atomic or molecular orbitals is known as its electron configuration; for instance, the electron configuration of the neon atom is 1s2 2s2 2p6.

According to electronic configurations, electrons move individually within each orbital while interacting with the average field produced by all other orbitals. The corrosion potential and reactivity of an atom may be calculated from its electron configuration. The electron configuration of the iodide ion is  1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶.

Therefore, the correct option is option C.

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Answer:

Option C. Will always.

Explanation:

A spontaneous reaction is a reaction that occurs without an external supply of heat.

This implies that spontaneous reaction will always occur as no external supply of heat is needed.

Answer:

The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10.

Explanation:

The element nitrogen is in the Group 15 in the Periodic Table, so it tends to gain 3 electrons (3 negative charges) to fill its valance shell with 8 electrons.

The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10 (the original 7 plus the 3 gained). It is isoelectronic with the gas Neon, which accounts for its stability.