Enter your answer in the provided box. Calcium hydroxide may be used to neutralize (completely react with) aqueous hydrochloric acid. Calculate the number of g of hydrochloric acid that would be neutralized by 0.685 mol of solid calcium hydroxide. g HCI

Answer: The correct answer is Option C.

Explanation:

The chemical formula for trinitrotoluene is [tex]C_7H_5N_3O_6[/tex]

In 1 mole of TNT, 7 moles of carbon atom, 5 moles of hydrogen atom, 3 moles of nitrogen atom and 6 moles of oxygen atom are present.

We know that:

Mass of trinitrotoluene = 227.1 g/mol

Mass of carbon = 12.01 g/mol

We are given:

Mass of TNT = 57.6 grams

To calculate the mass of carbon in given amount of TNT, we apply unitary method:

In 227.1 grams of TNT, amount of carbon present is = [tex](7\times 12.01)=84.07g[/tex]

So, in 57.6 grams of TNT, the amount of carbon present is = [tex]\frac{84.07g}{227.1g}\times 57.6g=21.3g[/tex]

Hence, the correct answer is Option C.

Answer:

Point source pollution:

 If pollution comes from a fix source then is called point source pollution.It have a specific location where pollution comes.

Ex: Air pollution ,water pollution.

Non point source pollution:

 If pollution comes from number of sources then is called point source pollution.This pollution does not have a specific point of source.

Ex: Spills ,leaks ,Sewage over flow etc.

Answer:

Defined as under.

Explanation:

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

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:

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:

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.  

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].

Answer: Option (d) is the correct answer.

Explanation:

The given reaction will be as follows.

       [tex]Ba(OH)_{2}(aq) + H_{2}SO_{4}(aq) \rightarrow BaSO_{4}(s) + 2H_{2}O(l)[/tex]

In the ionic form, the equation will be as follows.

           [tex]Ba^{2+}(aq) + SO^{2-}_{4}(aq) \rightarrow BaSO_{4}(s)[/tex] ........ (1)

           [tex]H^{+}(aq) + OH^{-}(aq) \rightarrow H_{2}O(l)[/tex] ............ (2)

Hence, for the net ionic equation we need to add both equation (1) and (2). Therefore, the net ionic equation will be as follows.

     [tex]Ba^{2+}(aq) + SO^{2-}_{4}(aq) + H^{+}(aq) + OH^{-}(aq) \rightarrow BaSO_{4}(s) + H_{2}O(l)[/tex]        

Now, balancing the atoms on both the sides we get the net ionic equation as follows.

         [tex]Ba^{2+}(aq) + SO^{2-}_{4}(aq) + 2H^{+}(aq) + 2OH^{-}(aq) \rightarrow BaSO_{4}(s) + 2H_{2}O(l)[/tex]          

The correct net ionic reaction for the experiment is H+(aq) + OH-(aq) → H2O(l), an example of an acid-base neutralization reaction. The other options included the formation of a precipitate, which is not part of the net ionic reaction.

Based on the provided options, the correct net ionic reaction for the experiment seems to be H+(aq) + OH-(aq) → H2O(l). This reaction is a classic example of an acid-base neutralization reaction, where an acid (H+) and a base (OH-) react to form water. The other two reactions involve the formation of a precipitate (BaSO4), but the full reaction is simplified to leave out the precipitate ions on either side. This does not occur in the first reaction. Hence, the first reaction is the correct net ionic reaction for this experiment.

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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: 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:

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:

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: 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:

a) 5,3176x10⁻⁴ moles

b) 6,85x10⁻⁴ moles

c) The appropriate formula to calculate is Henderson-Hasselbalch.

d) pH = 4,86. Acidic solution but slighty

Explanation:

a) moles of acetic acid:

9,20x10⁻³L × 57,8x10⁻³M = 5,3176x10⁻⁴ moles

b) moles of sodium acetate:

56,2x10⁻³g ÷ 82,0 g/mole = 6,85x10⁻⁴ moles

c) The appropriate formula to calculate is Henderson-Hasselbalch:

pH= pka + log₁₀ [tex]\frac{[A^-]}{[HA]}[/tex]

d) pH= 4,75 + log₁₀ [tex]\frac{[6,85x10_{-4}]}{[5,3176x10_{-4}]}[/tex]

pH = 4,86

3 < pH < 7→ Acidic solution but slighty

I hope it helps!

Answer: The mass of decane produced is [tex]1.743\times 10^2g[/tex]

Explanation:

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]       ......(1)

Mass of hydrogen gas = 2.45 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1:, we get:

[tex]\text{Moles of }H_2=\frac{2.45g}{2g/mol}=1.225mol[/tex]

The chemical equation for the hydrogenation of decene follows:

[tex]C_{10}H_{20}(l)+H_2(g)\rightarrow C_{10}H_{22}(s)[/tex]

As, decene is present in excess. So, it is considered as an excess reagent.

Thus, hydrogen gas is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of hydrogen gas produces 1 mole of decane.

So, 1.225 moles of hydrogen gas will produce = [tex]\frac{1}{1}\times 1.225=1.225mol[/tex] of decane

Now, calculating the mass of decane by using equation 1, we get:

Moles of decane = 1.225 mol

Molar mass of decane = 142.30 g/mol

Putting values in equation 1, we get:

[tex]1.225mol=\frac{\text{Mass of decane}}{142.30g/mol}\\\\\text{Mass of carbon dioxide}=(1.225mol\times 142.30g/mol)=174.3g=1.743\times 10^2g[/tex]

Hence, the mass of decane produced is [tex]1.743\times 10^2g[/tex]

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).

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:

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.

For more information, refer the link:

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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.

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: 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:

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:

Q=7200 W

q=7200/72=100 [tex]W/m^2[/tex]

Explanation:

Given that

ΔT=15° C

Thickness ,t=15 cm

Thermal conductivity ,K=1 W/m.°C

Height,h=6 m

Length ,L=12 m

As we know that heat conduction through wall given as

[tex]Q=\dfrac{KA}{t}\Delta T[/tex]

Now by putting the values

A= 6 x 12 =72 [tex]m^2[/tex]

[tex]Q=\dfrac{KA}{t}\Delta T[/tex]

[tex]Q=\dfrac{1\times 72}{0.15}\times 15\ W[/tex]

Q=7200 W

Q is the total heat transfer.

Heat flux q

 q=Q/A [tex]W/m^2[/tex]

q=7200/72=100 [tex]W/m^2[/tex]

q is the heat flux.

As w know that heat flux(q) is the heat transfer rate from per unit area and on the other hand heat transfer(Q) is the total heat transfer from the surface.

Heat flux q=Q/A

That is why these both are different.

Answer:

5.65

Explanation:

Given that:

[tex]pK_{b}\ of\ CH_3NH_2=3.44[/tex]

[tex]K_{b}\ of\ CH_3NH_2=10^{-3.44}=3.6308\times 10^{-4}[/tex]

[tex]K_a\ of\ CH_3NH_3^+Cl^-=\frac {K_w}{K_b}=\frac {10^{-14}}{3.6308\times 10^{-4}}=2.7542\times 10^{-11}[/tex]

Concentration = 0.18 M

Consider the ICE take for the dissociation as:

                              [tex]CH_3NH_3^+[/tex]   ⇄     H⁺ +  [tex]CH_3NH_2[/tex]

At t=0                           0.18                 -                            -

At t =equilibrium        (0.18-x)                x                         x            

The expression for dissociation constant of acetic acid is:

[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [CH_3NH_2 \right ]}{[CH_3NH_3^+]}[/tex]

[tex]2.7542\times 10^{-11}=\frac {x^2}{0.18-x}[/tex]

x is very small, so (0.18 - x) ≅ 0.18

Solving for x, we get:

x = 0.2227×10⁻⁵  M

pH = -log[H⁺] = -log(0.2227×10⁻⁵) = 5.65

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: 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:

[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:

b)boiling water

Explanation: