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Answer:

b. Binary compounds of carbon and hydrogen

Explanation:

Before proceeding, Hydrocarbons refers to organic chemical compounds composed exclusively of hydrogen and carbon atoms. This means the only elements present in an hydrocarbon are;

- Carbon

- Hydrogen

Looking through the options;

- Option A: This is wrong because the hydroxyl group contains oxygen and hydrocarbons contain only hydrogen and carbon.

- option B: This is correct. Binary compounds refers to compounds with just two elements.

- option C: This is wrong because water contains oxygen and hydrocarbons contain only hydrogen and carbon.

- option D: Carbon atoms can contain other elements so this option is wrong.

- option E: This also wrong because we had already gotten the correct option.

Answer:

A. 10.29 g of C2H6.

B. C2H4.

C. 3.14 g of H2.

Explanation:

Step 1:

We'll begin by writing the balanced equation for the reaction. This is given below:

C2H4 + H2 —> C2H6

Step 2:

Determination of the masses of C2H4 and H2 that reacted and the mass of C2H6 produced from the balanced equation.

This is illustrated below:

Molar mass of C2H4 = (12x2) + (4x1) =28 g/mol

Mass of C2H4 from the balanced equation = 1 x 28 = 28 g.

Molar mass of H2 = 2x1 = 2 g/mol

Mass of H2 from the balanced equation = 1 x 2 = 2 g.

Molar mass of C2H6 = (12x2) + (6x1) = 30 g/mol.

Mass of C2H6 from the balanced equation = 1 x 30 = 30 g

From the balanced equation above,

28 g of C2H4 reacted with 2 g of H2 to produce 30 g of C2H6.

Step 3:

Determination of the limiting reactant. This can be obtained as follow:

From the balanced equation above,

28 g of C2H4 reacted with 2 g of H2.

Therefore, 9.6 g of C2H4 will react with = (9.6 x 2)/28 = 0.69 g of H2.

From the calculations made above, we can see that only 0.69 g out of 3.83 g of H2 given is needed to react completely with 9.6 g of C2H4.

Therefore, C2H4 is the limiting reactant and H2 is the excess reactant.

A. Determination of the maximum mass of ethane, C2H6 produced from the react.

In this case, the limiting reactant will be used because it will produce the maximum yield of the reaction since all of it is consumed in the reaction.

The limiting reactant is C2H4 and the maximum mass of C2H6 can be obtained as follow:

From the balanced equation above,

28 g of C2H4 reacted to produce 30 g of C2H6.

Therefore, 9.6 gof C2H4 will react to produce = (9.6 x 30)/28 = 10.29 g of C2H6.

Therefore, 10.29 g of ethane, C2H6 were produced from the reaction.

B. The limiting reactant is ethylene with formula C2H4. Please refer to step 3 above for details.

C . Determination of the mass of the excess reactant that remained after the reaction.

The excess reactant is H2, please refer to step 3 above for details and the mass that remained after the reaction can be obtained as follow:

Mass of H2 given = 3.83 g

Mass of H2 that reacted = 0.69 g

Mass of H2 remaining =.?

Mass of H2 remaining = mass of H2 given – mass of H2 that reacted.

Mass of H2 remaining = 3.83 – 0.69

Mass of H2 remaining = 3.14 g

Therefore, 3.14 g of H2 remained after the reaction.

Answer:

rate of recrystallization = 4.99 × 10⁻³ min⁻¹

Explanation:

For Avrami equation:

[tex]y = 1-e ^{(-kt^n)} \\ \\ e^{(-kt^n)} = 1-y\\ \\ -kt^n = In(1-y) \\ \\ k = \dfrac{-In(1-y)}{t^n}[/tex]

To calculate the value of k which is a dependent variable for the above equation ; we have:

[tex]k = \dfrac{-In(1-0.40)}{200^{2.5}}[/tex]

[tex]k = 9.030 \times 10 ^{-7}[/tex]

The time needed for 50% transformation can be determined as follows:

[tex]y = 1-e ^{(-kt^n)} \\ \\ e^{(-kt^n)} = 1-y\\ \\ -kt^n = In(1-y) \\ \\ t =[ \dfrac{-In(1-y)}{k}]^{^{1/n}}[/tex]

[tex]t_{0.5} =[ \dfrac{-In(1-0.4)}{9.030 \times 10^{-7}}]^{^{1/2.5}}[/tex]

= 200.00183 min

The rate of reaction for Avrami equation is:

[tex]rate = \dfrac{1}{t_{0.5}}[/tex]

[tex]rate = \dfrac{1}{200.00183}[/tex]

rate = 0.00499 / min

rate of recrystallization = 4.99 × 10⁻³ min⁻¹

Answer:

Helium

Explanation:

Though all noble gases are stable, helium only has two electrons and one shell so it basically doesn't have room for others.

Answer:

The number of oxide ions as the nearest neighbors of  [tex]{\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}[/tex] ions are known to be as six

Explanation:

The regularity of a crystal structure leads to the idea of space lattice.In order to explain this concept, let us consider a crystal of NaCl, It consists of a perfectly regular arrangement of sodium ions and chlorine ions.

If we represent the position of each Na+ in the crystal by a point marked x the result will be a regular three dimensional network of points. This will be the space lattice of Na+ in the crystal NaCl. The symmetry of the combined lattice determined the symmetry of the crystal as a whole.

The space lattice of a crystal may be considered as built up of a three dimensional basic pattern  called unit cell. The unit cell is a repeat unit which generates the whole pattern in three dimensions of the unit cell.

In Solid MgO , the crystal structure which is used to predict the properties of the material, have the same structure as that of NaCl.

The obtain the structure of a face centered cubic FCC unit cell where the ions occupy the corner of the cube and the center of each face of the cube.

The number of oxide ions as the nearest neighbors of  [tex]{\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}[/tex] ions are known to be as six. As a result of that , the coordination number of [tex]{\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}[/tex] ions is six.

Answer:

Atomic mass of Ag-107 = 106.94 amu

Explanation:

Let the mass of Ag-107 be y

Since the mass ratio of Ag-109/Ag-107 is 1.0187, the mass of Ag-109  is 1.0187 times heavier than the mass of Ag-107

Mass of Ag-109  = 1.0187y

Relative atomic mass of Ag = sum of (mass of each isotope * abundance)

Relative atomic mass of Ag = 107.87

Abundance of Ag-107 = 51.839% = 0.51839

Abundance of Ag-109 = 41.161% = 0.48161

107.87 = (y * 0.51839) + (1.018y * 0.48161)

107.87 = 0.51838y + 0.49027898y

107.87 = 1.00865898y

y = 107.87/1.00865898

y = 106.94 amu

Therefore, atomic mass of Ag-107 = 106.94 amu

simple distillation can be used when the temperature difference between the boiling points of two miscible liquid is at least 25°c. the temperature difference between the boiling points of kerosene and petrol is 25c. hence, this mixture can separated using simple distillation.

simple distillation

Answer:

[tex]r_B=2M/s[/tex]

Explanation:

Hello,

In this case, since the average rate of reaction is related with the consumption of A which has an stoichiometric coefficient of 1, the rate of formation of B will be:

[tex]r_B=2*1M/s\\\\r_B=2M/s[/tex]

By cause of the stoichiometric coefficient of B which doubles the average rate.

Best regards.

Answer:

The Henry's law constant for argon is [tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]

Explanation:

Henry's Law indicates that the solubility of a gas in a liquid at a certain temperature is proportional to the partial pressure of the gas on the liquid.

C = k*P

where C is the solubility, P the partial pressure and k is the Henry constant.

So, being the concentration [tex]C=\frac{ngas}{V}[/tex]  

where ngas is the number of moles of gas and V is the volume of the solution, you must calculate the number of moles ngas. This is determined by the Ideal Gas Law: P*V=n*R*T where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. So [tex]n=\frac{P*V}{R*T}[/tex]

In this case:

Then:

[tex]n=\frac{1 atm*5.16*10^{-2} L}{0.082 \frac{atm*L}{mol*K} *298K}[/tex]

Solving:

n= 2.11 *10⁻³ moles

So: [tex]C=\frac{ngas}{V}=\frac{2.11*10^{-3} moles}{1 L} =2.11*10^{-3} \frac{moles}{L}= 2.11*10^{-3} M[/tex]

Using Henry's Law and being C=CAr and P =PAr:

2.11*10⁻³ M= k* 1 atm

Solving:

[tex]k=\frac{2.11*10^{-3} M}{1 atm}[/tex]

You get:

[tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]

The Henry's law constant for argon is [tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]

The Henry's law constant for argon gas in 1 litre of water is 2.1 × 10⁻³M/atm.

Henry's law of gas states that solubility of a gas in any liquid at particular temperature is directly proportional to the partial pressure of the gas.

C∝P

C = kP, where

k = Henry's constant

P = partial pressure of gas

C is the solubility and it is present in the form of concentration and will be calculated as:
C = n/V

n = no. of moles

V = volume

And moles of the gas will be calculated by using the ideal gas equation as:

PV = nRT

n = (1)(5.16×10⁻²) / (0.082)(298) = 2.1 × 10⁻³ moles

And Concentration in liquid will be:

C = 2.1 × 10⁻³mol / 1L = 2.1 × 10⁻³ M

Now we put all these values in the first equation to calculate the value of k as:

k = (2.1 × 10⁻³M) / (1atm) = 2.1 × 10⁻³M/atm

Hence required value of k is 2.1 × 10⁻³M/atm.

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Explanation:

I hope it helps you

good luck

Answer for the question

Answer:Conduction materials include metals, electrolytes, superconductors, semiconductors, plasmas and some nonmetallic conductors such as graphite and Conductive polymers. Copper has a high conductivity.

Explanation:

Answer: The oxidation number of oxygen in this compound is -2.

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.  

In [tex]Na_2O[/tex], sodium is having an oxidation state of +1 called as [tex]Na^+[/tex] cation and oxygen is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]Na_2O[/tex]

The cations and anions being oppositely charged attract each other through strong coloumbic forces and form an ionic bond.

Answer:

Sodium Oxide: Na2O, Barium Phosphide: Ba3P2

Explanation:

Just did the question

Answer:

d, 40 dm3.

Explanation:

According to Avogadro's law, the mole ratio of chemicals in a reaction is equal to the ratio of volumes of chemicals reacted (for gas).

From the equation, the mole ratio of N2 : H2 : NH3 = 1 : 3 : 2, meaning 1 mole of N2 reacts completely with 3 moles of H2 to give 2 moles of NH3, the ratio of volume required is also equal to 1 : 3 : 2.

Considering both N2 and H2 have 30dm3 of volume, but 1 mole of N2 reacts completely with 3 moles of H2, so we can see H2 is limiting while N2 is in excess. Using the ratio, we can deduce that 10dm3 equals to 1 in ratio (because 3 moles ratio = 30dm3).

With that being said, all H2 has reacted, meaning there's no volume of H2 left.  2 moles of NH3 is produced, meaning the volume of NH3 produced = 10 x 2 = 20 dm3. (using the ratio again)

1 mole of N2 has reacted, meaning from the  30dm3, only 10 dm3 has reacted. This also indicate that 20 dm3 of N2 has not been reacted.

So at the end, the mixture contains 20dm3 of NH3, and 20 dm3 of unreacted N2. Hence, the answer is d, 40 dm3.

The name of the molecule which is given below is 2-pentene.

Alkenes are the organic compounds which are composed of carbon and hydrogen atoms, in which double bond is present.

In the given diagram:

So the compound is 2 pentene.

Hence, 2 pentene is the name of the compound.

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Answer:

too low

Explanation:

If our aim is to recover the naphthalene and measure its mass after separation, then we must not allow any vapour to escape.

Naphthalene is a sublime substance, it can be separated by sublimation. It changes directly from solid to gas. This vapour must be kept securely so that none of it escapes. If part of the naphthalene vapour happens to escape accidentally, then the measured mass of naphthalene will be too low compared to the mass of naphthalene originally present in the mixture.

Answer:

They blow away from poles to the equator.

Explanation:

Hello,

In this case, we must take into account that global wind systems are formed by the constant increase in the temperature of the Earth’s surface. Thus, they drive the oceans’ surface currents. In such a way, we can say wind is the basic movement of air from an area of higher pressure to an area of lower pressure, for that reason they blow away from the poles to the equator.

Best regards.

The statement that describes the global winds is they travel over short distances.

Wind is a pattern or type of the movement of the natural air or any other composition of gases over to the relative position of the planet's surface.

Global winds are those winds which can travel in a straight path and originated due to global convention currents. Global winds always move from west to east direction and travels short distances only.

Hence, option (2) is correct.

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Answer:

Approximately [tex]10.88[/tex].

Explanation:

[tex]\rm OBr^{-}[/tex] can act as a weak Bronsted-Lowry base:

[tex]\rm OBr^{-}\; (aq) + H_2O\; (l) \rightleftharpoons HOBr\; (l) + OH^{-}\; (aq)[/tex].

(Side note: the state symbol of [tex]\rm HOBr[/tex] in this equation is [tex]\rm (l)[/tex] (meaning liquid) because [tex]\rm HOBr[/tex] is a weak acid.)

However, the equilibrium constant of this reaction, [tex]K_\text{eq}[/tex], isn't directly given. The idea is to find [tex]K_\text{eq}[/tex] using the [tex]\rm pH[/tex] value at the half-equivalence point. Keep in mind that this system is at equilibrium all the time during the titration. If temperature stays the same, then the same [tex]K_\text{eq}[/tex] value could also be used to find the [tex]\rm pH[/tex] of the solution before the acid was added.

At equilibrium:

[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].

At the half-equivalence point of this titration, exactly half of the base, [tex]\rm OBr^{-}[/tex], has been converted to its conjugate acid, [tex]\rm HOBr[/tex]. Therefore, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should both be equal to one-half the initial concentration of [tex]\rm OBr^{-}[/tex].

As a result, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should be the same. The expression for [tex]K_\text{eq}[/tex] can thus be simplified:

[tex]\begin{aligned}& K_\text{eq} \\&= \frac{\left(\text{half-equivalence $[\rm HOBr\; (l)]$}\right)\cdot \left(\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\right)}{\text{half-equivalence $[\rm OBr^{-}\; (l)]$}}\\ &=\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\end{aligned}[/tex].

In other words, the [tex]K_\text{eq}[/tex] of this system is equal to the [tex]\rm OH^{-}[/tex] concentration at the half-equivalence point. Assume that [tex]\rm p\mathnormal{K}_\text{w}[/tex] the self-ionization constant of water, is [tex]14[/tex]. The concentration of [tex]\rm OH^{-}[/tex] can be found from the [tex]\rm pH[/tex] value:

[tex]\begin{aligned}& \text{half-equivalence $[\rm OH^{-}\; (aq)]$} \\ &= 10^{\rm pH - p\mathnormal{K}_\text{w}}\;\rm mol \cdot L^{-1} \\ &= 10^{7.75 - 14}\; \rm mol \cdot L^{-1}\\ &= 10^{-6.25}\; \rm mol \cdot L^{-1}\end{aligned}[/tex].

Therefore, [tex]\begin{aligned} K_\text{eq} &= 10^{-6.25}\end{aligned}[/tex].

Again, since [tex]\rm KOBr[/tex] is a soluble salt, all that [tex]0.200\; \rm M[/tex] of [tex]\rm KOBr[/tex] in this solution will be in the form of [tex]\rm K^{+}[/tex] and [tex]\rm OBr^{-}[/tex] ions. Before any hydrolysis takes place, the concentration of [tex]\rm OBr^{-}[/tex] should be equal to that of [tex]\rm KOBr[/tex]. Therefore:

[tex]\text{$[\rm OBr^{-}\; (aq)]$ before hydrolysis} = 0.200\; \rm M[/tex].

Let the equilibrium concentration of [tex][\rm OH^{-}\; (aq)][/tex] be [tex]x\; \rm M[/tex]. Create a RICE table for this reversible reaction:

[tex]\begin{array}{c|ccccccc} & \rm OBr^{-}\; (aq) &+&\rm H_2O\; (l)& \rightleftharpoons & \rm HOBr\; (l)& + & \rm OH^{-}\; (aq) \\ \textbf{I}& 0.200\; \rm M & & & & 0 \; \rm M & & 0\; \rm M \\ \textbf{C} & -x\; \rm M & & & & +x \; \rm M & & +x\; \rm M \\ \textbf{E}& (0.200 + x)\; \rm M & & & & x \; \rm M & & x\; \rm M \end{array}[/tex].

Assume that external factors (such as temperature) stays the same. The [tex]K_\text{eq}[/tex] found at the half-equivalence point should apply here, as well.

[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].

At equilibrium:

[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x}[/tex].

Assume that [tex]x[/tex] is much smaller than [tex]0.200[/tex], such that the denominator is approximately the same as [tex]0.200[/tex]:

[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x} \approx \frac{x^2}{0.200}[/tex].

That should be equal to the equilibrium constant, [tex]K_\text{eq}[/tex]. In other words:

[tex]\displaystyle \frac{x^2}{0.200} \approx K_\text{eq} = 10^{-6.25}[/tex].

Solve for [tex]x[/tex]:

[tex]x \approx 3.35\times 10^{-4}[/tex].

In other words, the [tex]\rm OH^{-}[/tex] before acid was added was approximately [tex]3.35\times 10^{-4}\; \rm M[/tex], which is the same as [tex]3.35\times 10^{-4}\; \rm mol \cdot L^{-1}[/tex]. Again, assume that [tex]\rm p\mathnormal{K}_\text{w} = 14[/tex]. Calculate the [tex]\rm pH[/tex] of that solution:

[tex]\begin{aligned}\rm pH &= \rm p\mathnormal{K}_\text{w} + \log [\mathrm{OH^{-}}] \approx 10.88\end{aligned}[/tex].

(Rounded to two decimal places.)

Answer:

B

Explanation:

The Ca+2 ion means that 2 electrons have been given away. So when you try and find the answer, you have to count backwards from Calcium. When you do, you get K+ first and then Argon which is either column 8 orc column 18 depending on your periodic table.

The element you hit is Argon.

The answer is B

Answer:

B ar

Explanation:

pen foster answer

Answer: The molar solubility of copper phosphate is [tex]\frac{X}{2}[/tex]

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as [tex]K_{sp}[/tex]

The equation is given as:

[tex]Cu_3(PO_4)_2(s)\rightarrow 3Cu^{2+}(aq)+2PO_4^{3-}[/tex]

By stoichiometry of the reaction:

1 mole of  [tex]Cu_3(PO_4)_2[/tex] gives 3 moles of [tex]Cu^{2+}[/tex] and 2 moles of [tex]PO_4^{3-}[/tex]

When the solubility of  [tex]Cu_3(PO_4)_2[/tex] is S moles/liter, then the solubility of [tex]Cu^{2+}[/tex]  will be 3S moles\liter and solubility of [tex]PO_4^{3-}[/tex] will be 2S moles/liter.

Molar concentration of [tex]PO_4^{3-}[/tex] = X

Given : 2S = X

Thus S =[tex]\frac{X}{2}[/tex]

Thus the molar solubility of copper phosphate is [tex]\frac{X}{2}[/tex]

Answer:

If the temperature increases the molecular movement as well, and if it increases the same it will happen with the molecular movement.

Pressure, volume and temperature are three factors that are closely related since they increase the temperature, the pressure usually decreases due to the dispersion of the molecules that can be generated, so the volume also increases.

If the temperature drops, the material becomes denser, its molecules do not collide with each other, their volume and pressure increases.

Explanation:

The pressure is related to the molecular density and the movement that these molecules have.

The movement is regulated by temperature, since if it increases, the friction and collision of the molecules also.

On the other hand, the higher the volume, the less pressure there will be on the molecules, since they are more dispersed among themselves.

(in the opposite case that the volume decreases, the pressure increases)

Answer:

Sn1 mechanism reaction

Explanation:

In this case, we have to start with the protonation of the "OH" group by the attack of the lone pairs in the alcohol group to the "H" in the HBr producing a positive charge in the oxygen. Then, water is produced and a carbocation is generated. This carbocation can be stabilized by a hydride shift. We can move a hydrogen atom to the positive charge and we will obtain a tertiary carbocation. Finally, the [tex]Br^- [/tex] will attack to produce the final halide.

See figure 1.

I hope it helps!

Answer:

Explanation:

A nonpolar molecule has no separation of charge, so no positive or negative poles are formed. In other words, the electrical charges of nonpolar molecules are evenly distributed across the molecule. Nonpolar molecules tend to dissolve well in nonpolar solvents, which are frequently organic solvents. The answer is hydrogen cyanide.

Answer:

Silver ion - Lewis acid, Ammonia -  Lewis base

Explanation:

The reaction is given as;

Ag+(aq) + 2NH3(aq) ⇌ [Ag(NH3)]2+(aq)

A lewis acid is an electron pair acceptor. While a lewis base is any substance that that can donate a pair of nonbonding electrons.

This reaction however is a complexation reaction, where ammonia is reacting with the silver ion.

Silver ion accepts electrons in this reaction, hence it is the lewis acid. The ammonia on the other hand donates the electrons used in bonding so it is the lewis base.

Answer:

0.677 moles

Explanation:

Take the atomic mass of K = 39.1, O =16.0, P = 31.0

no. of moles = mass / molar mass

no. of moles of K3PO4 used = 4.79 / (39.1x3 + 31 + 16x4)

= 0.02256 mol

From the equation, the mole ratio of KOH : K3PO4 = 3 :1,

meaning every 3 moles of KOH used, produces 1 mole of K3PO4.

So, using this ratio, let the no. of moles of KOH required to be y.

[tex]\frac{3}{1} =\frac{y}{0.02256} \\[/tex]

y = 0.02256 x3

y = 0.0677 mol

If you don't find exactly 0.677 moles as one of the options, go for the closest one. A very slight error may occur because of taking different significant figures of atomic masses when calculating.

Answer:

0,

Explanation:

if n was 2, then 1,0,-1

Answer:

A

Explanation:

infer means use data to reach conclusion.

Answer:

Replacing 3 of the H of LiALH4 with OR groups make it a more reactive reducing agents than NaBH4

Explanation:

LiAlH4 and NaBH4 are two well known reducing agents in organic chemistry. These two reducing agents function by transfer of hydrogen to the substrate.

If the hydrogen atoms in LiAlH4 are replaced by the -OR moiety, the new compound will be far less reducing than NaBH4 because the hydrogen atoms necessary to effect the reduction has been removed. Thus, the new compound containing -OR moiety can never be more reducing than NaBH4. This implies that the statement written in the answer is false as written.

Answer:

6.1×10^8

Explanation:

The reaction is;

Sn^2+(aq) + Cd(s) -----> Sn(s) + Cd^2+(aq)

E°cell = E°cathode - E°anode

E°cathode= -0.14 V

E°anode= -0.40 V

E°cell = -0.14-(-0.40)

E°cell= -0.14+0.40

E°cell= 0.26 V

But

E°cell= 0.0592/n log K

E°cell= 0.0592/2 log K

0.26= 0.0296log K

log K = 0.26/0.0296

log K= 8.7838

K= Antilog (8.7838)

K= 6.1×10^8

Answer: C

Explanation:

Right on edge 2020

The Bureau of Land Management was originally created to manage land for grazing, mining, developing oil and gas.

Land management refers to the activities which are done in order to protect and preserve the land as well the resources found on land.

The Bureau of Land Management was created to manage land in the US.

The Bureau of Land Management was originally created to manage land for grazing, mining, developing oil and gas.

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Answer:

Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)

3.52 g of PbCl2

3.76 g of KCl

Explanation:

The equation of the reaction is;

Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)

Number of moles of Pb(NO3)2 =mass/molar mass 5.06g/331.2 g/mol = 0.0153 moles

Number of moles of KCl= mass/ molar mass= 6.03g/74.5513 g/mol= 0.081 moles

Next we obtain the limiting reactant; the limiting reactant yields the least number of moles of products.

For Pb(NO3)2;

1 mole of Pb(NO3)2 yields 1 mole of PbCl2

Therefore 0.0153 moles of Pb(NO3)2 yields 0.0153 moles of PbCl2

For KCl;

2 moles of KCl yields 1 mole of PbCl2

0.081 moles of KCl yields 0.081 moles ×1/2 = 0.041 moles of PbCl2

Therefore Pb(NO3)2 is the limiting reactant.

Theoretical Mass of precipitate obtained = 0.0153 moles of PbCl2 × 278.1 g/mol = 4.25 g of PbCl2

% yield = actual yield/theoretical yield ×100

Actual yield = % yield × theoretical yield /100

Actual yield= 82.9 ×4.25/100

Actual yield = 3.52 g of PbCl2

If 1 mole of Pb(NO3) reacts with 2 moles of KCl

0.0153 moles of Pb(NO3)2 reacts with 0.0153 moles × 2 = 0.0306 moles of KCl

Amount of excess KCl= 0.081 moles - 0.0306 moles = 0.0504 moles of KCl

Mass of excess KCl = 0.0504 moles of KCl × 74.5513 g/mol = 3.76 g of KCl