At 0.70 atmosphere pressure, a certain mass of a gas has a volume of 3.8 L. What will be the volume if the pressure is decreased to 0.409 atmosphere pressure?

Answer:  The nature of an amphoteric substance is amphoprotic; it has ability to donate proton or gain proton. So the amphotric substances are those which can react as both acid or base in a reaction. Many oxides of metals when subjected to a reaction mixture can acts as both an acid or either a base, they are called amphoteric oxides. some of the examples of these amphoteric oxide is zinc oxide and lead oxide.

Hence, the correct option here is (c )

Answer:

To know when a central atom in a lewis structure will have more than 8 electrons it is important to know where is the element at the periodic table and the orbital configuration of the atom.

Explanation:

The octet rule establishes that an atom could win, lose, or share electrons with other atoms till every atom have eight valence electrons. However exist exceptions to the octet rule. Some elements like Be, B and Al are stable with only six valence electrons because these three elements are small and with low electronegativity. On the other hand, some elements from the 3d orbital could expand their octet and still be stable, like S in SF6. This happens because elements from 3d orbital have enough space to suit more electrons.

Answer:

[tex]V=\dfrac{2}{9\ \mu}R^2g(\rho_d-\rho_f)[/tex]

Explanation:

Given that

Radius =R

[tex]Density\ of\ droplet=\rho_d[/tex]

[tex]Density\ of\ fluid=\rho_f[/tex]

When drop let will move downward then so

[tex]F_{net}=F_{weight}-F_{b}-F_d[/tex]

Fb = Bouncy force

Fd = Drag force

We know that

[tex]F_b=\dfrac{4\pi }{3}R^3\ \times \rho_f\times g[/tex]

[tex]F_{weight}=\dfrac{4\pi }{3}R^3\ \times \rho_d\times g[/tex]

[tex]F_{d}=6\pi \mu\ R\ V[/tex]

μ=Dynamic viscosity of fluid

V= Terminal velocity

So at the equilibrium condition

[tex]F_{net}=F_{weight}-F_{b}-F_d[/tex]

[tex]0=F_{weight}-F_{b}-F_d[/tex]

[tex]F_{weight}=F_{b}+F_d[/tex]

[tex]\dfrac{4\pi }{3}R^3\ \times \rho_d\times g=\dfrac{4\pi }{3}R^3\ \times \rho_f\times g+6\pi \mu\ R\ V[/tex]

So

[tex]V=\dfrac{2}{9\ \mu}R^2g(\rho_d-\rho_f)[/tex]

This is the terminal velocity of droplet.

Explanation:

It is known that pinch point temperature difference is defined as the difference between the temperature of exhaust existing the evaporator and the temperature of water evaporation.

     [tex]\Delta T_{p.p}[/tex] = Exhaust existing the evaporator temperature - Temperature of water evaporation

Since, it is given that exhaust existing the evaporator temperature = [tex]1000^{o}F[/tex]

           Temperature of water evaporation = [tex]800^{o}F[/tex]

Hence, calculate [tex]\Delta T_{p.p}[/tex] using the above formula as follows.

                   [tex]\Delta T_{p.p}[/tex] = Exhaust existing the evaporator temperature - Temperature of water evaporation

                                  = [tex]1000^{o}F[/tex] - [tex]800^{o}F[/tex]

                                  = [tex]200^{o}F[/tex]

Thus, we can conclude that the temperature difference at the pinch point in the generator is [tex]200^{o}F[/tex].

Answer: Freezing point of a solution will be [tex]-1.16^0C[/tex]

Explanation:

Depression in freezing point is given by:

[tex]\Delta T_f=i\times K_f\times m[/tex]

[tex]\Delta T_f=T_f^0-T_f=(5.50-T_f)^0C[/tex] = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

[tex]K_f[/tex] = freezing point constant = [tex]5.12^0C/m[/tex]

m= molality

[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]

Weight of solvent (benzene)= 1480 g =1.48 kg

Molar mass of solute (octane) = 114.0 g/mol

Mass of solute (octane) = 220 g

[tex](5.50-T_f)^0C=1\times 5.12\times \frac{220g}{114.0 g/mol\times 1.48kg}[/tex]

[tex](5.50-T_f)^0C=6.68[/tex]

[tex]T_f=-1.16^0C[/tex]

Thus the freezing point of a solution will be [tex]-1.16^0C[/tex]

Answer: The moles of silver nitrate reacted is 5.72 moles, moles of iron (II) nitrate produced is 2.86 moles and moles of silver chloride produced is 5.72 moles

Explanation:

We are given:

Moles of iron (II) chloride = 2.86 moles

For the given chemical equation:

[tex]FeCl_2(aq.)+2AgNO_3(aq.)\rightarrow Fe(NO_3)_2(aq.)+2AgCl(s)[/tex]

By Stoichiometry of the reaction:

1 mole of iron (II) chloride reacts with 2 moles of silver nitrate.

So, 2.86 moles of iron (II) chloride will react with = [tex]\frac{2}{1}\times 2.86=5.72mol[/tex] of silver nitrate

Moles of silver nitrate reacted = 5.72 moles

By Stoichiometry of the reaction:

1 mole of iron (II) chloride produces 1 mole of iron (II) nitrate

So, 2.86 moles of iron (II) chloride will produce = [tex]\frac{1}{1}\times 2.86=2.86mol[/tex] of iron (II) nitrate

Moles of iron (II) nitrate produced = 2.86 moles

By Stoichiometry of the reaction:

1 mole of iron (II) chloride produces 2 moles of silver chloride

So, 2.86 moles of iron (II) chloride will produce = [tex]\frac{2}{1}\times 2.86=5.72mol[/tex] of silver chloride

Moles of silver chloride produced = 5.72 moles

Hence, the moles of silver nitrate reacted is 5.72 moles, moles of iron (II) nitrate produced is 2.86 moles and moles of silver chloride produced is 5.72 moles

When 2.86 moles of FeCl2 react, they consume 5.72 moles of AgNO3, produce 2.86 moles of Fe(NO3)2, and produce 5.72 moles of AgCl.

In the balanced equation FeCl2 (aq) + 2AgNO3(aq) --> Fe(NO3)2(aq) + 2AgCl(s), each FeCl2 molecule reacts with 2 AgNO3 molecules. This means that for every mole of FeCl2, 2 moles of AgNO3 are consumed. Therefore, if we have 2.86 moles of FeCl2, it will consume 2 x 2.86 = 5.72 moles of AgNO3.

In this reaction, for every mole of FeCl2 consumed, 1 mole of Fe(NO3)2 is produced. Therefore, if we have 2.86 moles of FeCl2, 2.86 moles of Fe(NO3)2 will be produced.

In addition, for every mole of FeCl2 consumed, 2 moles of AgCl are produced. Therefore, if we have 2.86 moles of FeCl2, 2 x 2.86 = 5.72 moles of AgCl will be produced.

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Answer:

The answer is c. Express Scripts Company

Explanation:

DrugDigest is a website tool that helps you to find information about medicaments  you consume or you see for example you introduce the first three letters of the drug in the website and you will have a detailed description of brands, supplied information eg. if the medicament comes by pills or by syrup and also the concentration of the medicament and recommendations ( under which case you can take the medicine, effects, side effects)  

Answer:

[Ca²⁺] = 0.068 M

Explanation:

The concentration of the calcium ion will be equal to the amount of calcium in CaCl₂, divided by the total volume:

C = n/V

When CaCl₂ dissociates in water, one mole of calcium ion is produced for every mole of CaCl₂, so the molar ratio of CaCl₂ to Ca²⁺ is 1:1. The moles of Ca²⁺ are calculated as follows:

(0.18 mol/L)(30.00 mL) = 5.4 mmol CaCl₂ = 5.4 mmol Ca²⁺

The total volume is (50.00 mL + 30.00 mL) = 80.00 mL

Thus, the concentration of Ca²⁺ is:

C = n/V = (5.4 mmol)/(80.00 mL) = 0.068 M

Answer:

Math expression: [tex]=\frac{0.77\ g/ml*5200\ ml}{1\ g} *45.0\ kJ[/tex]

Explanation:

Given:

Energy produced per gram of gasoline = 45.0 kJ

Density of gasoline = 0.77 g/ml

Volume of gasoline = 5.2  L=5200 ml

To determine:

The amount of energy produced by burning 5.2 L gasoline

Calculation set-up:

1. Calculate the mass (m) of gasoline given the density (d) and volume (v)

[tex]m = d*v\\\\m = 0.77 g/ml*5200 ml[/tex]

2. Calculate the amount of energy produced

[tex]=\frac{0.77\ g/ml*5200\ ml}{1\ g} *45.0\ kJ=180180 kJ[/tex]

Answer:

30 m/v %

Explanation:

First, we need to understand the definition of mass/volume percentage (m/v %). This is a way of expressing concentration of a substance and it means grams of a determined compound per 100 mL of solution:

m/v% = g of X substance/100 mL of solution

Having said that, if we have 0.600 g of NaOH per 2.00 mL of solution, then in 100 mL of solution we will have:

2.00 mL solution ---- 0.600 g of NaOH

100 mL solution---- x=(100 mL × 0.600 g NaOH)/ 2.00 mL = 30 g  = 30 %m/v

So, the concentration of a solution containing 0.600 g of NaOH per 2.00 mL is a 30 m/v% solution.-

Explanation:

It is known that number of moles of a substance equal to mass divided by its molar mass.

Mathematically,         No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

It is given that no. of oles present are 27.2 mol and molar mass of N is 14 g/mol.

Therefore, calculate mass of N as follows.

                 No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

                       27.2 mol = [tex]\frac{mass}{14 g/mol}[/tex]

                      mass = 380.8 g

Thus, we can conclude that mass of N is 380.8 g.

Answer:

pH of solution is 3.80

Explanation:

[tex]pH=pK_{a}(formic acid)+log(\frac{C_{formate}}{C_{formic acid}})[/tex]

where, C stands for concentration

So, [tex]pH=3.75+log(\frac{0.055}{0.049})=3.80[/tex]

Answer:

a) Reaction is not at equilibrium

b) Reaction will move towards backward direction

Explanation:

[tex]PCl_5(g) \rightarrow PCl_3(g) + Cl_2(g)[/tex]

Equilibrium constant = 26

[tex]Reaction\ quotient (Q) = \frac{[p_{PCl_3}]\times [p_{Cl_2}]}{[p_{PCl_5}]}[/tex]

[tex][p_{PCl_5}] = 0.012 atm[/tex]

[tex][p_{PCl_3}]= 0.90 atm[/tex]

[tex][p_{Cl_2}]= 0.45 atm[/tex]

[tex]Reaction\ quotient (Q) = \frac{[p_{PCl_3}]\times [p_{Cl_2}]}{[p_{PCl_5}]}[/tex]

[tex]Reaction quotient (Q) =\frac{0.90\times 0.45} {0.012} = 33.75[/tex]

As reaction quotient (Q) is more than equilibrium constant, so reaction is not at equilibrium and reaction will move towards backward direction.

Answer:

Explanation:

The equilibrium relevant for this problem is:

H₂PO₄⁻ ↔ HPO₄⁻² + H⁺

The Henderson–Hasselbalch (H-H) equation is needed to solve this problem:

pH= pka + [tex]log\frac{[A^{-} ]}{[HA]}[/tex]

In this case, [A⁻] = [HPO₄⁻²], [HA] = [H₂PO₄⁻], pH = 7.4; from literature we know that pka=7.21.

We use the H-H equation to describe [HPO₄⁻²] in terms of  [H₂PO₄⁻]:

[tex]7.4=7.21+log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]} \\0.19=log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]} \\10^{0.19}= \frac{[HPO4^{-2} ]}{[H2PO4^{-} ]} \\1.549*[H2PO4^{-} ]=[HPO4^{-2} ][/tex]

The problem tells us that the concentration of phosphate is 1 mM, which means:

[HPO₄⁻²] + [H₂PO₄⁻] = 1 mM = 0.001 M

In this equation we can replace [HPO₄⁻²] with the term expressed in the H-H eq:

1.549 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.001 M

2.549 * [H₂PO₄⁻] = 0.001 M

[H₂PO₄⁻] = 3.923 * 10⁻⁴ M

With the value of [H₂PO₄⁻] we can calculate [HPO₄⁻²]:

[HPO₄⁻²] + 3.923 * 10⁻⁴ M = 0.001 M

[HPO₄⁻²] = 6.077 * 10⁻⁴ M

With the concentrations, the molecular weight, and the volume, we calculate the mass of each reagent:

Explanation:

The electronegativity values of the 5d series are very high because the size of the 5d orbitals are much larger as compared to the size of the 3d and 4d orbitals. As a consequence of this, the shielding capacity of 5d elements are low or these elements are not effective at shielding the nuclear charge . Thus, this causes the increase in the effective nuclear charge which makes the electronegativity values to increase steeply from the Lutetium (1.27) to the Gold (2.54).

Answer: NaCl (s) → NaCl (aq)

Explanation:

Entropy is often associated with the disorder or randomness of a system. Therefore, in each reaction, it is necessary to evaluate if the disorder increases or decreases to understand what happens to the entropy:

1) KCl (aq) + AgNO₃ (aq) → KNO₃ (aq) + AgCl (s) - In this reaction, we have only aqueous reactants in the beginning and in the product we have a precipitate. The solid state is more organised than the liquid, consequently, the entropy decreases.

2) NaCl (s) → NaCl (aq) - In this case, oposite to the first one, we go from a solid state to an aqueous state. The solvation of the ions Na⁺ and Cl⁻ is random while the solid state is very organised. Therefore, in this reaction the entropy increases.

3) 2NaOH (aq) + CO₂ (g) → Na₂CO₃ (aq) + H₂O (l) - In this reaction, the reactants have higher entropy because of the gas CO₂. Therefore, the entropy decreases.

4) C₂H₅OH (g) → C₂H₅OH (l) -  In this reaction, the reactant is a gas and the product a liquid. Therefore, the entropy decreases.

Answer:

b. halogens

Explanation:

The elements of group 17 are called halogens. These are six elements Fluorine, Chlorine, Bromine, Iodine, Astatine. Halogens are very reactive these elements cannot be found free in nature. Their chemical properties are resemble greatly with each other. As we move down the group in periodic table size of halogens increases that's way fluorine is smaller in size as compared to other halogens elements. Their boiling points also increases down the group which changes their physical states.  

Answer:

k = 0.0198 s⁻¹

t = 74.25 seconds

Explanation:

Given that:

Half life = 35.0 s

[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]

Where, k is rate constant

So,  

[tex]k=\frac {ln\ 2}{t_{1/2}}[/tex]

[tex]k=\frac {ln\ 2}{35.0}\ s^{-1}[/tex]

The rate constant, k = 0.0198 s⁻¹

Using integrated rate law for first order kinetics as:

[tex][A_t]=[A_0]e^{-kt}[/tex]

Where,  

[tex][A_t][/tex] is the concentration at time t

[tex][A_0][/tex] is the initial concentration

Given:

77.0 % is decomposed which means that 0.77 of [tex][A_0][/tex] is decomposed. So,

[tex]\frac {[A_t]}{[A_0]}[/tex] = 1 - 0.77 = 0.23

t = ?

[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]

[tex]0.23=e^{-0.0198\times t}[/tex]

t = 74.25 seconds

When in a reaction heat is a reactant then the reaction is called thermal decomposition reaction. The first-order rate constant is 0.0198 per second and the time required is 74.25 seconds.

The rate constant is the specific rate that is the proportionality constant in a reaction and depicts the relation between the chemical reaction rate and the concentration of the reactants.

Rate constant can be given as,

[tex]\rm t\dfrac{1}{2} = \dfrac{ln 2}{k}[/tex]

The half-life of the reaction is 35 seconds.

Substituting values in the above equation:

[tex]\begin{aligned}\rm k &= \rm \dfrac{ln\;2}{t\frac{1}{2}}\\\\&= \dfrac{\rm ln\2}{35}\\\\&= 0.0198 \;\rm s^{-1}\end{aligned}[/tex]

The time taken can be calculated by the rate law for first order:

[tex]\rm [A_{t}] = [A_{o}]e^{-kt}[/tex]

Here,

Solving for time (t):

[tex]\begin{aligned}\rm \dfrac {[A_{t}]}{[A_{o}]} &= 1 - 0.77\\\\0.23 &= \rm e ^{-0.0198 \times t}\\\\\rm t &= 74.25\;\rm seconds\end{aligned}[/tex]

Therefore, the rate constant is 0.0198 per second and the time taken is 74.25 seconds.

Learn more about rate constant here:

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Answer:

b)boiling water

Explanation:

Explanation:

Acid spills

These types of spills should be neutralized with base like sodium bicarbonate and then must be cleaned up by using paper towel or  sponge. Strong base like sodium hydroxide must not be used to neutralize. Best base to use is sodium bicarbonate which has much less chance of the injury.

Mercury spills

Mercury is found commonly in  the thermometers. If one have mercury spill, it must be cleaned up immediately with  either commercial product like Hg Aborb ™ or by using elemental sulfur. Mercury sponges  can also be purchased that form amalgam with liquid mercury and thus trapping it on surface of sponge.

Answer:

option D - coking or fouling

Explanation:

Coking (not cocking) is the process involving the deposition of small carbon particles (created by simply putting carbon atoms) on a catalyst's accessible surface, leading in a reduction in the area accessible for catalytic activity. This is also sometimes related to fouling catalysts or merely fouling them.

Answer:

a) pH = 2,88

b) pH = 4,58

c) pH = 5,36

d) pH = 8,79

e) pH = 12,10

Explanation:

In a titration of a strong base (KOH) with a weak acid (CH₃COOH) the reaction is:

CH₃COOH + KOH → CH₃COOK + H₂O

a) Here you have just CH₃COOH, thus:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ where ka =1,74x10⁻⁵ and pka = 4,76

When this reaction is in equilibrium:

[CH₃COOH] = 0,100 -x

[CH₃COO⁻] = x

[H⁺] = x

Thus, equilibrium equation is:

1,74x10⁻⁵ = [tex]\frac{[x][x] }{[0,100-x]}[/tex]

The equation you will obtain is:

x² + 1,74x10⁻⁵x - 1,74x10⁻⁶ = 0

Solving:

x = -0,0013278193 ⇒ No physical sense. There are not negative concentrations

x = 0,0013104193

As x = [H⁺] and pH = - log [H⁺]

pH = 2,88

b) Here, it is possible to use:

CH₃COOH + KOH → CH₃COOK + H₂O

With adition of 5,0 mL of 0,200M KOH solution the initial moles are:

CH₃COOH = [tex]0,025 L.\frac{0,100 mol}{L} =[/tex] = 2,5x10⁻³ mol

KOH = [tex]0,005 L.\frac{0,200 mol}{L} =[/tex] = 1,0x10⁻³ mol

CH₃COOK = 0.

In equilibrium:

CH₃COOH = 2,5x10⁻³ mol - 1,0x10⁻³ mol = 1,5x10⁻³ mol

KOH = 0 mol

CH₃COOK = 1,0x10⁻³ mol

Now, you can use Henderson–Hasselbalch equation:

pH = 4,76 + log [tex]\frac{1,0x10^{-3} }{1,5x10^{-3} }[/tex]

pH = 4,58

c) With adition of 10,0 mL of 0,200M KOH solution the initial moles are:

CH₃COOH = [tex]0,025 L.\frac{0,100 mol}{L} =[/tex] = 2,5x10⁻³ mol

KOH = [tex]0,010 L.\frac{0,200 mol}{L} =[/tex] = 2,0x10⁻³ mol

CH₃COOK = 0.

In equilibrium:

CH₃COOH = 2,5x10⁻³ mol - 2,0x10⁻³ mol = 0,5x10⁻³ mol

KOH = 0 mol

CH₃COOK = 2,0x10⁻³ mol

Now, you can use Henderson–Hasselbalch equation:

pH = 4,76 + log [tex]\frac{2,0x10^{-3} }{0,5x10^{-3} }[/tex]

pH = 5,36

d) With adition of 12,5 mL of 0,200M KOH solution the initial moles are:

CH₃COOH = [tex]0,025 L.\frac{0,100 mol}{L} =[/tex] = 2,5x10⁻³ mol

KOH = [tex]0,0125 L.\frac{0,200 mol}{L} =[/tex] = 2,5x10⁻³ mol

CH₃COOK = 0.

Here we have the equivalence point of the titration, thus, the equilibrium is:

CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ kb = kw/ka where kw is equilibrium constant of water = 1,0x10⁻¹⁴; kb = 5,75x10⁻¹⁰

Concentrations is equilibrium are:

[CH₃COOH] = x

[CH₃COO⁻] = 0,06667-x

[OH⁻] = x

Thus, equilibrium equation is:

5,75x10⁻¹⁰ = [tex]\frac{[x][x] }{[0,06667-x]}[/tex]

The equation you will obtain is:

x² + 5,75x10⁻¹⁰x - 3,83x10⁻¹¹ = 0

Solving:

x = -0.000006188987⇒ No physical sense. There are not negative concentrations

x = 0.000006188

As x = [OH⁻] and pOH = - log [OH⁻]; pH = 14 - pOH

pOH = 5,21

pH = 8,79

e) The excess volume of KOH will determine pH:

With 12,5mL is equivalence point, the excess volume is 15,0 -12,5 = 2,5 mL

2,5x10⁻³ L × [tex]\frac{0,200 mol}{1L}[/tex] ÷ 0,040 L = 0,0125 = [OH⁻]

pOH = - log [OH⁻]; pH = 14 - pOH

pOH = 1,90

pH = 12,10

I hope it helps!

Answer:

The pH of solution on addition of 0.0, 5.0, 10.0 12.5 and 15 ml of KOH will be 2.87, 4.56, 5.34, 4.74,  and 12.09 respectively.

Explanation:

pH can be calculated by the evaluation of Hydronium ions in the solution.

[tex]\rm K_a[/tex] of [tex]\rm CH_3COOH[/tex] is 1.8 [tex]\rm \times10^-^5[/tex]

(a) Hydrogen ion concentration = [tex]\rm \sqrt{K_a\;\times\;CH_3COOH\;concentartion}[/tex]

Hydrogen ion concentration = [tex]\rm \sqrt{1.8\;\times\;10^-^5\;\times\;0.1}[/tex]

Hydrogen ion concentration = 1.34 [tex]\times\;10^-^3[/tex] M

pH of solution = log [Hydrogen ion concentration]

pH = log [[tex]\rm 1.34\;\times10^-^5[/tex]]

pH = 2.87

(b) On addition of 5 ml of KOH of 0.200 M, the moles of KOH added are:

moles of KOH = [tex]\rm \frac{volume\;(ml)}{1000}\;\times\;\frac{molarity}{L}[/tex]

moles of KOH = [tex]\rm \frac{5}{1000}\;\times\;\frac{0.200}{L}[/tex]

moles of KOH = [tex]\rm 1\;\times\;10^-^3[/tex] M

The initial moles of [tex]\rm CH_3COOH[/tex] are:

moles of [tex]\rm CH_3COOH[/tex] = [tex]\rm \frac{25}{1000}\;\times\;\frac{0.100}{L}[/tex]

moles of [tex]\rm CH_3COOH[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex]

At equilibrium, the concentration of [tex]\rm CH_3COOH[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex] - [tex]\rm 1\;\times\;10^-^3[/tex] mol

concentration of [tex]\rm CH_3COOH[/tex] = [tex]\rm 1.5\;\times\;10^-^3[/tex] moles.

The concentration of [tex]\rm CH_3COOK[/tex] = [tex]\rm 1\;\times\;10^-^3[/tex] moles

pH = [tex]\rm pK_a\;+\;log\;\frac{salt}{acid}[/tex]

pH = 4.76 + log [tex]\rm \frac{1\;\times\;10^-^3}{1.5\;\times\;10^-^3}[/tex]

pH = 4.58

(c) On addition of 10 ml of KOH,

moles of KOH = [tex]\rm \frac{10}{1000}\;\times\;\frac{0.200}{L}[/tex]

moles of KOH = [tex]\rm 2\;\times\;10^-^3[/tex] moles

moles of [tex]\rm CH_3COOH[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex] moles

moles of [tex]\rm CH_3COOK[/tex] = [tex]\rm 2\;\times\;10^-^3[/tex] moles

pH = 4.76 + log [tex]\rm \frac{2\;\times\;10^-^3}{0.5\;\times\;10^-^3}[/tex]

pH = 5.36

(d) On addition of 12.5 ml of KOH,

moles of KOH = [tex]\rm \frac{12.5}{1000}\;\times\;\frac{0.200}{L}[/tex]

moles of KOH = [tex]\rm 2.5\;\times\;10^-^3[/tex]

moles of  [tex]\rm CH_3COOH[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex] moles

moles of [tex]\rm CH_3COOK[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex] moles

pH = 4.76 + log [tex]\rm \frac{2\;\times\;10^-^3}{0}[/tex]

pH = 4.74

(e) On addition of 15.0 ml of KOH,

The excess point is reached after the addition of 12.5 ml of KOH. After further addition of KOH, the pOH will be there.

The OH concentration on addition of KOH = [tex]\rm \frac{(\frac{15}{1000}\;\times0.200\;moles)\;-\;(\frac{25}{1000}\;\times0.100\;moles) }{\frac{25}{1000}\;+\;\frac{15}{1000} }[/tex]

= 0.0125 M

pOH = -log [OH concentration]

pOH = -log [0.0125]

pOH = 1.903

pH = 14 - pOH

pH = 14 - 1.903

pH = 12.097

The pH of solution on addtion of KOH will be :

O ml KOH = 2.87

5 ml KOH= 4.56

10 ml KOH = 5.36

12.5 ml KOH = 4.74

15 ml KOH = 12.907.

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Answer: The number of carbon and oxygen atoms in the given amount of carbon dioxide is [tex]4.72\times 10^{22}[/tex] and [tex]9.44\times 10^{22}[/tex] respectively

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Given mass of carbon dioxide gas = 3.45 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of carbon dioxide gas}=\frac{3.45g}{44g/mol}=0.0784mol[/tex]

1 mole of carbon dioxide gas contains 1 mole of carbon and 2 moles of oxygen atoms.

According to mole concept:

1 mole of a compound contains [tex]6.022\time 10^{23}[/tex] number of molecules

So, 0.0784 moles of carbon dioxide gas will contain [tex]1\times 0.0784\times 6.022\times 10^{23}=4.72\times 10^{22}[/tex] number of carbon atoms and [tex]2\times 0.0784\times 6.022\times 10^{23}=9.44\times 10^{22}[/tex] number of oxygen atoms

Hence, the number of carbon and oxygen atoms in the given amount of carbon dioxide is [tex]4.72\times 10^{22}[/tex] and [tex]9.44\times 10^{22}[/tex] respectively

Answer:

6298702 potassium atoms would have to be laid side by side to span a distance of 2,91 mm

Explanation:

If the radius of a potassium atom is 231 pm, then the diameter would be:

d = 2r = 2*(231) = 462 pm

So each potassium atom occupies a space of 462 pm, we can express this relationship as follows:

[tex]\frac{1 potassium atom}{462 pm}[/tex]

To solve this problem, we'll use the following conversion factors:

1 pm = 1×E−12 m

1000 mm = 1 m

We begin to accommodate all our relationships starting from that numerical expression that is not written as a relationship (usually the one in the question), and in such a way that the units are eliminated between them.

[tex]2, 91 mm * \frac{1 m}{1000 mm}*\frac{ 1 pm}{1*10^{-12}m }*\frac{1 potassium atom}{462 pm}= 6298701.3[/tex] potassium atoms

So, we'll need 6298702 potassium atoms to span a distance of 2,91 mm

Answer:

the correct answer is option 'b': More than

Explanation:

The 2 situations are represented in the attached figures below

When an object is placed in air it is acted upon by force of gravity of earth which is measured as weight of the object.

While as when any object is submerged partially or completely in any fluid the fluid exerts a force in upward direction and this force is known as force of buoyancy and it's magnitude is given by Archimedes law as equal to the weight of the fluid that the body displaces, hence the effective force in the downward direction direction thus the apparent weight of the object in water decreases.

Answer: The volume of liquid X is 246 mL

Explanation:

To calculate volume of a substance, we use the equation:

[tex]\text{Density of a substance}=\frac{\text{Mass of a substance}}{\text{Volume of a substance}}[/tex]

We are given:

Mass of liquid X = 205 g

Density of liquid X = 0.834 g/mL

Putting values in above equation, we get:

[tex]0.834g/mL=\frac{205g}{\text{Volume of liquid X}}\\\\\text{Volume of liquid X}=246mL[/tex]

Hence, the volume of liquid X is 246 mL

Answer: [tex]4.4\times 10^{2}[/tex]

Explanation:

The chemical reaction follows the equation:

     [tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex]

t = 0 [tex]5.000\times 10^{-3}[/tex] [tex]2.50\times 10^{-3}[/tex]    0

At eqm [tex](5.000\times 10^{-3}-2x)[/tex] [tex](2.50\times 10^{-3}-x)[/tex]   (2x)      

The expression for [tex]K_c[/tex] for the given reaction follows:

[tex]K_c=\frac{[SO_3]^2}{[SO_2]^2[O_2]}[/tex]

[tex]K_c=\frac{[2x]^2}{[5.000\times 10^{-3}-2x]^2[2.50\times 10^{-3}-x]}[/tex]

Given : [tex][SO_2]_{eqm}=2.80\times 10^{-3}[/tex]

[tex]5.000\times 10^{-3}-2x=2.80\times 10^{-3}[/tex]

[tex]x=1.1\times 10^{-3}[/tex]

Putting the values we get:

[tex]K_c=\frac{[2\times 1.1\times 10^{-3}]^2}{[5.000\times 10^{-3}-2\times 1.1\times 10^{-3}]^2[2.50\times 10^{-3}-1.1\times 10^{-3}]}[/tex]

[tex]K_c=4.4\times 10^2[/tex]

Therefore, the equilibrium concentration [tex]4.4\times 10^{2}[/tex]

Answer:

2 Pb(OH)2 + 2H2SO4 => 2 PbSO4 + 4 H20

Explanation:

Since there's no "?" shown in the equation, let's balance it and solve it entirely.

Pb(OH)2 + 2H2SO4 => PbSO4 + 2H20

1Pb + 10O + 6H + 2S ≠ 1Pb + 6O + 4H + 1S → it needs to be balanced.

To do this, let's start by looking at the elements that are only presnet once on each side:

On the products half, S is only present in PbSO4 → if we look at the reagents half, we can see it needs a "2" → then Pb is multiplied by 2 too → so Pb(OH)2 on the reagents half will also need a "2" → final count on O and H on the reagents side: 12O and 8H  → to balance it, you need 4 water molecules on the products side.

Answer : The volume of [tex]CO_2[/tex] gas is 1.00 L

Explanation :

First we have to determine the moles of [tex]C_3H_4PO_7^{3-}[/tex].

Molar mass of [tex]C_3H_4PO_7^{3-}[/tex] = 182.9 g/mole

[tex]\text{ Moles of }C_3H_4PO_7^{3-}=\frac{\text{ Mass of }C_3H_4PO_7^{3-}}{\text{ Molar mass of }C_3H_4PO_7^{3-}}=\frac{15.0g}{182.9g/mole}=0.0820moles[/tex]

Now we have to calculate the moles of [tex]CO_2[/tex].

The given balanced chemical reaction is:

[tex]C_5H_8P_2O_{11}^{4-}(aq)+H_2O(aq)+CO_2(g)\rightarrow 2C_3H_4PO_7^{3-}(aq)+2H^+(aq)[/tex]

From the reaction we conclude that,

As, 2 moles of [tex]C_3H_4PO_7^{3-}[/tex] produce from 1 mole of [tex]CO_2[/tex]

So, 0.0820 moles of [tex]C_3H_4PO_7^{3-}[/tex] produce from [tex]\frac{0.0820}{2}=0.041moles[/tex] of [tex]CO_2[/tex]

Now we have to calculate the volume of [tex]CO_2[/tex] gas.

Using ideal gas equation:

[tex]PV=nRT[/tex]

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = 298 K

n = number of moles of gas = 0.041 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get:

[tex](1.00atm)\times V=(0.041mole)\times (0.0821L.atmK^{-1}mol^{-1})\times (298K)[/tex]

[tex]V=1.00L[/tex]

Therefore, the volume of [tex]CO_2[/tex] gas is 1.00 L

Explanation:

The given data is as follows.

      Mass of antimony = 19.75 g

      Molar mass of Sb = 121.76 g/mol

Therefore, calculate number of moles of Sb as follows.

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

                                         = [tex]\frac{19.75 g}{121.76 g/mol}[/tex]

                                         = 0.162 mol

Mass of oxygen given is 6.5 g and molar mass of oxygen is 16 g/mol. Hence, moles of oxygen will be calculated as follows.

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

                                         = [tex]\frac{6.5 g}{16 g/mol}[/tex]

                                         = 0.406 mol

Hence, ratio of moles of Sb and O will be as follows

                          Sb : O

                      [tex]\frac{0.162}{0.162} : \frac{0.406}{0.162}[/tex]

                           1 : 2.5

We multiply both the ratio by 2 in order to get a whole number. Therefore, the ratio will be 2 : 5.

Thus, we can conclude that the empirical formula of the given oxide is [tex]Sb_{2}O_{5}[/tex].