Assume that 33.0 mL of a 0.10 M solution of a weak base B that accepts one proton is titrated with a 0.10 M solution of the monoprotic strong acid HX.How many moles of have been added at the equivalence point?n = ? mol

Answer:

The reaction given is:

In (g) + H₂ (g) ⇔ InH₂ (g), the Kp is 1.48 at 973 K.

The partial pressures measured in the reaction vessel is Partial pressure of In is 0.0540 atm, partial pressure of H₂ is 0.0250 atm, and the partial pressure of InH₂ is 0.0780 atm. By using the table given in the attachment below, the value of PInH₂ is (0.078-x), PIn is (0.054 + x), and the value of PH2 is (0.025 + x).

Kp = PInH₂/PIn × PH₂ = (0.078 - x) / (0.054 +x) (0.025 + x)

1.48 = (0.078 - x) / (0.054 +x) (0.025 + x)

x = 0.06689

Now the partial pressures of In, H₂ and InH₂ will be,

PH₂ = 0.025 + x = 0.025 + 0.0668 = 0.0918 atm

PIn = 0.054 + 0.0668 = 0.1208 atm

PInH₂ = 0.078 - 0.0668 = 0.0112 atm

Now the Qp or the reaction quotient will be,

Qp = (0.078) / (0.054) (0.025) = 57.78.

Answer:

See the answer below

Explanation:

The complete question can be seen in the attached image.

Phenolphthalein is an indicator that is often utilized in an acid-base reaction to indicate the endpoints of such reactions due to its ability to change color from pink/colorless to colorless/pink depending on if the final solution is acidic or basic.

Phenolphthalein is usually colorless in acidic solutions and appears pink in basic solutions. The more basic or alkaline a solution is, the stronger the pink color of phenolphthalein. Hence;

1. Ammonia with a pH of 11 is basic, phenolphthalein will turn pink.

2. Battery acid with a pH of 1 is acidic, it will remain colorless.

3. Lime juice with a pH of 2 is acidic, it will remain colorless.

4. Mashed avocado with a pH of 6.5 is acidic, it will remain colorless.

5. Seawater with a pH of 8.5 is basic, it will turn pink.

6. Tap water with a pH of 7 is neutral, it will remain colorless

Phenolphthalein is a chemical compound with the formula[tex]C_{20}H_{14}O_4[/tex]. Phenolphthalein is often used as an indicator in acid-base titrations. For this application, it turns colorless in acidic solutions and pink in basic solutions

Phenolphthalein works as in:-

According to the question, There are 5 solutions having different ph and the indication only turns basic solution to pink

The indicator only turn the basic solution pink and these solutions are as follows,

Hence, these are the answer.

For more information, refer to the link:-

Answer:

[tex]pH=2.28[/tex]

Explanation:

Hello,

In this case, for the acid dissociation of formic acid (HCOOH) we have:

[tex]HCOOH(aq)\rightarrow H^+(aq)+HCOO^-(aq)[/tex]

Whose equilibrium expression is:

[tex]Ka=\frac{[H^+][HCOO^-]}{[HCOOH]}[/tex]

That in terms of the reaction extent is:

[tex]1.8x10^{-4}=\frac{x*x}{0.16-x}[/tex]

Thus, solving for [tex]x[/tex] which is also equal to the concentration of hydrogen ions we obtain:

[tex]x=0.00528M[/tex]

[tex][H^+]=0.00528M[/tex]

Then, as the pH is computed as:

[tex]pH=-log([H^+])[/tex]

The pH turns out:

[tex]pH=-log(0.00528M)\\\\pH=2.28[/tex]

Regards.

Answer:

A

Explanation:

Absorbing moisture to protect goods from damage. Hence, option A is correct.

Silica gel is a desiccant, or drying agent, that manufacturers often place in little packets to keep moisture from damaging certain food and commercial products.

Silica Gel is a good drying agent for preventing corrosion, contamination, spoilage, and mould growth in many commodities and products due to its physical properties.

Learn more about silica gel here:

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Answer:

B. 4.52 X10-9 M

Explanation:

Our goal for this question is to calculate the concentration of hydronium ions [tex]H^+[/tex] produced by water in a vessel with a concentration of hydroxide ions of [tex]2.21X10^-^6~M[/tex]. So, our first approach can be the ionization reaction of water:

[tex]H_2O_(_l_)~->~H^+~_(_a_q_)~+~OH^-~_(_a_q_)[/tex]

If we write the Keq expression for this reaction we will have:

[tex]Keq=[H^+][OH^-][/tex]

Now, water is the universal solvent, so, Keq has a special name. In the equilibrium problems for water we have to use "Kw" instead of "Keq":

[tex]Kw=[H^+][OH^-][/tex]

From this equation, we know the Kw value () and the concentration of the hydroxide ions ([2.21X10^-^6~M]). If we replace these values into the equation we can solve for [tex][H^+][/tex]:

[tex]1.0X10^-^1^4=[H^+][2.21X10^-^6~M][/tex]

[tex][H^+]=\frac{1.0X10^-^1^4}{2.21X10^-^6}=4.52X^-^9[/tex]

I hope it helps!

Answer:

B. 4.52 * 10^-9M

Explanation:

did the test

Answer:

14, 508J/K

ΔHrxn =q/n

where q = heat absorbed and n = moles

Explanation:

m = mass of substance (g) = 0.1184g

1 mole of Mg - 24g

n moles - 0.1184g

n = 0.0049 moles.

Also, q = m × c × ΔT

Heat Capacity, C of MgCl2 = 71.09 J/(mol K)

∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)

= 14, 508 J/K/kg

ΔT=  (final - initial) temp = 38.3 - 27.2

= 11.1 °C.

mass of MgCl2 = 95.211 × 0.1184 = 11.27

⇒ q = 11.27g × 11.1 °C × 14, 508 j/K/kg

= 1,7117.7472 J °C-1 g-1

∴ ΔHrxn = q/n

=1,7117.7472  ÷ 0.1184

= 14, 508J/K

Answer:

A. Saturated acids  -  1. myristic acid, 4. palmitic acid

B. Monounsaturated acids  - 2. oleic acid

C. Polyunsaturated acids -  3. linoleic acid

Explanation:

Saturated fatty acids are the fats combined together with a single bond and no double or triple bond while unsaturated fatty acids are the fats having a double or triple bond between them.

Monosaturated acids have only one double or triple bond while polyunsaturated acids have more than one double or triple bond.

So, an appropriate match of given acids are:

A. Saturated acids  - Myristic acid (CH3(CH2)12COOH) and Palmitic acid (CH3(CH2)14COOH) dont have any double or triple bonds.

B. Monounsaturated acids  - Oleic acid (CH3(CH2)7CH=CH(CH 2)7COOH) have only one double bond.

C. Polyunsaturated acids - Linoleic acid (HOC–7CH=CH–CH 2–CH=CH–3H) has two double bonds.

Hence, the correct answer is:

A. Saturated acids  -  1. myristic acid, 4. palmitic acid

B. Monounsaturated acids  - 2. oleic acid

C. Polyunsaturated acids -  3. linoleic acid

Answer:

Major product does not undergo oxidation since it is a tertiary alcohol whereas minor product undergoes oxidation to ketone as it is  secondary alcohol.

Explanation:

Hello,

In this case, given the attached picture, the hydration of the 1 methylcyclohexene yields to alcohols; 1-methylcyclohexan-1-ol and 1-methylcyclohexan-2-ol. Thus, since the OH in the 1-methylcyclohexan-1-ol (major product) is bonded to a tertiary carbon (bonded with other three carbon atoms) it is not able to increase the number of oxygen bonds (oxidation) as it already attained the octet whereas the 1-methylcyclohexan-2-ol (minor product) is able to undergo oxidation to ketone as the carbon bonded to it is secondary (bonded with other two carbon atoms), so one extra bond the oxygen is allowed to be formed to carbonyl.

Best regards.

Answer:

shape

volume

Hope this helps! （づ￣3￣）づ╭❤～

Explanation:

Answer:

Pb + 2H2O --> PbO2 + 2H2

Explanation:

Products:

Solid metal; PbO2

Hydrogen; H

Reactants:

Metal; Pb

Steam; H2O

Reactants --> Products

Pb + H2O --> PbO2 + H2

Upon balancing we have;

Pb + 2H2O --> PbO2 + 2H2

Answer:

89%.

Explanation:

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

2Li + 2H2O —> 2LiOH + H2

Next, we shall determine the masses of Li and H2O that reacted and the mass of LiOH produced from the balanced equation.

This is illustrated below:

Molar mass of Li = 7 g/mol

Mass of Li from the balanced equation = 2 x 7 = 14 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36 g

Molar mass of LiOH = 7 + 16 + 1 = 24 g/mol

Mass of LiOH from the balanced equation = 2 x 24 = 48 g

Summary:

From the balanced equation above,

14 g of Li reacted with 36 g of H2O to produce 48 g of LiOH.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

14 g of Li reacted with 36 g of H2O.

Therefore, 7.4 g of Li will react with = (7.4 x 36)/14 = 19.03 g of H2O.

From the calculation made above, we can see that it will take a higher amount i.e 19.03 g than what was given i.e 10.2 g of H2O to react completely with 7.4 g of Li.

Therefore, H2O is the limiting reactant and Li is the excess reactant.

Next, we shall determine the theoretical yield of LiOH.

In this case we shall use the limiting reactant.

The limiting reactant is H2O and the theoretical yield of LiOH can be obtained as follow:

From the balanced equation above,

36 g of H2O reacted to produce 48 g of LiOH.

Therefore, 10.2 g of H2O will react to produce = (10.2 x 48)/36 = 13.6 g of LiOH.

Therefore, the theoretical yield of LiOH is 13.6 g

Finally, we shall determine the percentage yield of LiOH. This can be obtained as follow:

Actual yield = 12.1 g

Theoretical yield = 13.6 g

Percentage yield =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 12.1/ 13.6 x 100

Percentage yield = 89%

Therefore, the percentage yield LiOH is 89%.

Answer:

C) formaldehyde, H2C=O.

Explanation:

Hello,

In this case, given that the hydrogen bondings are known as partial intermolecular interactions between a lone pair on an electron rich donor atom, particularly oxygen, and the antibonding molecular orbital of a bond between hydrogen and a more electronegative atom or group. Thus, among the options, C) formaldehyde, H2C=O, will exhibit hydrogen bonding since the lone pair of electrons of the oxygen at the carbonyl group, are able to interact with hydrogen (in the form of water).

Best regards.

Answer:

The correct answer is 0.25 atm.

Explanation:

As mentioned in the given question that the chamber comprises equal molar concentrations of He, Ne, Ar and Kr gas. So, let us assume that the moles of all the gases will be x then the total number of moles will be 4x.  

The formula for calculating mole fraction is,  

Mole fraction = mole of the substance/total moles  

The mole fraction of Kr = x/4x = 0.25

The total pressure given in the chamber is 1 atm. Therefore, the partial pressure will be,  

Partial pressure = mole fraction * Total pressure

Partial pressure = 0.25 * 1 = 0.25 atm.  

The partial pressure of Kr is 0.25 atm.

We are told that there is an equal molar amounts of He, Ne, Ar, and Kr. If we decide to say, let the molar amount of each gas be x, the total number of moles  of all the gases will now be; x + x+ x + x = 4x

Let us recall that partial pressure of a gas can be obtained form the formula;

Partial pressure = mole fraction × total pressure

Mole fraction of Kr = x/4x = 1/4

Total pressure = 1 atm

Partial pressure of Kr = 1/4 × 1 = 0.25 atm

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Answer:

Explanation:

Answer:

1, 1, 1, 1  

Explanation:

potassium hydrogen sulfate + potassium hydroxide ⟶ potassium sulfate + water(l)

                 KHSO₄                   +               KOH              ⟶            K₂SO₄   + H₂O

1. Put a 1 in front of the most complicated-looking formula (K₂SO₄?):

KHSO₄  + KOH ⟶ 1K₂SO₄ + H₂O

2. Balance S:

We have fixed 1 S on the right. We need 1 S on the left. Put a 1 in front of KHSO₄ to fix it.

1KHSO₄  + KOH ⟶ 1K₂SO₄ + H₂O

3. Balance K:

We have fixed 2 K on the right and 1 K on the left. We need 1 more K on the left. Put a 1 in front of KOH.

1KHSO₄  + 1KOH ⟶ 1K₂SO₄ + H₂O

4. Balance O

We have fixed 4 O on the right and 5 O on the left. We need 1 more O on the right. Put a 1 in front of H₂O.

1KHSO₄  + 1KOH ⟶ 1K₂SO₄ + 1H₂O

Every formula has a coefficient. The equation should be balanced.

5. Check that atoms balance:

[tex]\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{K} & 2 &2\\\text{H} & 2 & 2\\\text{S} & 1 & 1\\\text{O}&5&5\\\end{array}[/tex]

It checks.

The coefficients are 1, 1, 1, 1.

Answer:

[tex]V_{HCl}=0.208L=208mL[/tex]

Explanation:

Hello,

In this case, since the chemical reaction is:

[tex]2HCl+Mg(OH)_2\rightarrow MgCl_2+2H_2O[/tex]

We can see that hydrochloric acid and magnesium hydroxide are in a 2:1 mole ratio, which means that the neutralization point, we can write:

[tex]n_{HCl}=2*n_{Mg(OH)_2}[/tex]

In such a way, the moles of magnesium hydroxide (molar mass 58.3 g/mol) in 500 mg are:

[tex]n_{Mg(OH)_2}=500mg*\frac{1g}{1000mg}*\frac{1mol}{58.3g} =0.00858mol[/tex]

Next, since the pH of hydrochloric acid is 1.25, the concentration of H⁺ as well as the acid (strong acid) is:

[tex][H^+]=[HCl]=10^{-pH}=10^{-1.25}=0.0562M[/tex]

Then, since the concentration and the volume define the moles, we can write:

[tex][HCl]*V_{HCl}=2*n_{Mg(OH)_2}[/tex]

Therefore, the neutralized volume turns out:

[tex]V_{HCl}=\frac{2*0.00858mol}{0.0562\frac{mol}{L} }\\ \\V_{HCl}=0.208L=208mL[/tex]

Best regards.

Answer:

4.84g of FeO is the theoretical yield

Explanation:

The Iron, Fe(s), reacts with oxygen, O₂(g), producing Iron (II) oxide, as follows:

2Fe(s) + O₂(g) → 2FeO

Theoretical yield is the yield of a reaction in which you assume the 100% of reactants is converted in products.

To find theoretical yield we need to find moles of Iron, and, knowing 2 moles of Fe produce 2 moles of FeO (Ratio 1:1), we can find theoretical yield of FeO as follows:

Moles Fe (Molar mass: 55.845g/mol)

Using the molar mass of the compound we can convert grams to moles, thus:

3.76g Fe × (1mol / 55.845g) = 0.0673 moles of Fe

Moles and mass of FeO

As there are in reaction 0.0673 moles Fe, assuming a theoretical yield (And as ratio of the reaction is 1:1), you will obtain 0.0673 moles of FeO.

Theoretical yield is given in grams, As molar mass of FeO is 71.844g/mol, theoretical yield of the reaction is:

0.0673 moles FeO × (71.844g / mol) =

Answer:

H₂O

Explanation:

The empirical formular of the compound is obtained using the following steps;

Step 1: Divide the percentage composition by the atomic mass

Hydrogen = 11.21 / 1 = 11.21

Oxygen = 88.79  / 16 = 5.55

Step 2: Divide by the lowest number

Hydrogen  = 11.21 / 5.55 = 2.02 ≈ 2

Oxygen = 5.55 / 5.55 = 1

This means the ratio of the elements is 2 : 1

The empirical formular (simplest formular of a compound) of the compound is;

H₂O

Answer:Empirical formula ======== H₂O    

Explanation:The empirical formula of a compound shows the whole number ratio for  each atom in a compound.

To find empirical formula. we follow the below steps

The total mass of the compound here  is 100 grams, that is (11.21% of hydrogen + 88.79% of oxygen) we can then  assume 11.21 grams  of hydrogen and 88.79grams of oxygen

                                          Hydrogen                   Oxygen

1.composition by mass    11.21                              88.79

molecular weight              1.007g/mol               15.990g/mol

2.Divide composition by mass  11.21/1.007            88.79/15.99    

by each molecular weight to get 11.13                            5.553

no of moles

3 Divide by the least number of moles

to get atomic ratio                       11.13/5.553          5.553/5.553

                                                         2.004                           1.00

4.Convert  to whole numbers             2                                 1

Empirical formula ======== H₂O    

Answer:

[tex]0.033g[/tex]

Explanation:

Hello,

In this case, since 11 mg per kilogram of body weight has the given lethality, the mg that turn out lethal for a chicken weighting 3 kg is computed by using a rule of three:

[tex]11mg\longrightarrow 1kg\\\\x\ \ \ \ \ \ \longrightarrow 3kg[/tex]

Thus, we obtain:

[tex]x=\frac{3kg*11mg}{1kg}\\ \\x=33mg[/tex]

That in grams is:

[tex]=33mg*\frac{1g}{1000mg} \\\\=0.033g[/tex]

Regards.

Answer:

D

Explanation:

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

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

It is a solution of a weak acid and salt.

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

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

Since

moles of HC2H3O2 = 1*0.250 = 0.250

moles of NaC2H3O2 = 1*0.050 = 0.050.

moles of HCl is added = 0.050

Now

The reaction between HCl and NaC2H3O2

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

Now

BCA table is

            NaC2H3O2  HCl       HC2H3O2

Before 0.050 0.050 0.250

Change -0.050 -0.050 +0.050

After 0 0 0.300

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

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

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Answer:

In 5 US liquid gallons, there are 80 cups.

Explanation:

To get from gallons to cups, just multiply the amount of gallons you have by 16.

Answer:

THE MOLAR HEAT CAPACITY OF HEXANE IS 290.027 J/ C

Explanation:

1604 J of heat is added to 48.9 g of hexane

To calculate the molar heat capacity of hexane, it is important to note that the molar heat capacity of a substance is the measure of the amount of heat needed to raise 1 mole of a substance by 1 K.

Since 1604 J of heat = 48.9 g of hexane

Molar mass of hexane = 86 g/mol = 1 mole

then;

1604 J = 48.9 g

x = 86 g

x = 1604 * 86 / 48.9

x = 4205.4 J

Hence, 4205.4 J of heat will be added to 1 mole or 86 g of hexane to raise the temperature by 14.5 C.

In other words,

heat = molar heat capacity * temperature change

molar heat capacity = heat/ temperature change

Molar heat capacity = 4205.4 J / 14.5 C

Molar heat capacity = 290.027 J/C

The molar heat capacity of hexane is 290.027 J/ C

Answer:

b

Explanation:

Provided the experiment is about the transfer of heat from one medium to another, any residual hot water on the metal would have added to the heat transferred from the metal to the cold water system.

The residual hot water on the metal posses its own heat and when such is transferred along with the metal into the cold water, the heat from the residual hot water will interfere with the measurement of the actual heat transferred to the cold water by the metal. Hence, the accuracy of the result would be impacted.

The correct option is b.

Answer:

3-pentanol

Explanation:

In this case, we have alcohol as the main functional group (OH) with a molecular ion at 88. If the molecular ion is 88 the molar mass is also 88 g/mol therefore the formula for the unknown molecule is [tex]C_5H_1_2O[/tex].

Additionally, if the mass spectrum shows the molecular ion peak we can not have tertiary alcohols (tertiary alcohols often do not show M+ at  all). So, the structures only can be primary and secondary structures.

With this in mind, our options are:

-) 1-pentanol

-) 2-pentanol

-) 3-pentanol

Now we can analyze each structure:

-) 1-pentanol

The structure must explain all the fragments produced (73, 70, and 59). In this primary alcohol, we will have an alpha cleavage (the red bond would be broken). If this has to happen, we will have fragments at 31 and 57. These fragments dont fit with the reported ones, therefore this is not a possible structure (See figure 1).

-) 2-pentanol

On this structure, we will have also an alpha cleavage (red bond). In this rupture we will have fragments at 45 and 43, these m/z values dont fit with the reported ones, therefore this is not a possible structure (See figure 1).

-) 3-pentanol

In this structure, we have the "OH"  bonded to carbon three. So, we can analyze each fragment:

   -) m/z 59

This fragment, can be explained as an alpha cleavage. But, in this case we have two ruptures that can produce the same ion. The carbons on both sides of the C-OH bond.

   -) m/z 71

This fragment, can be explained as a loss of water (M-18) in which we have the production of a carbocation in the carbon where we previously have the C-OH bond.

   -) m/z 73

This fragment, can be explained as a beta cleavage. But, in this case, also we have two ruptures that can produce the same ion. The methyl groups on each end molecule.

See figure 2

I hope it helps!

Answer:

Explanation:

Number of moles of propane:

=Mass in grams ÷ Relative molecular Mass

= 100/((12*3) + (1*8))

= 100 ÷ 44

= 2.2727

Mole ratio propane:carbon (IV) oxide = 1:3(from the equation)

Number of moles of CO2 = 3*2.2727 = 6.8181

Mass in grams = Relative molecular Mass * Number of moles

= 44 * 6.8181

= 299.9964 grams

A number of moles of propane:

Mass in grams ÷ Relative molecular Mass

= 100/((12*3) + (1*8))

= 100 ÷ 44

= 2.2727

Mole ratio propane:carbon (IV) oxide = 1:3(from the equation)

Number of moles of CO2 = 3*2.2727 = 6.8181

Mass in grams = Relative molecular Mass * Number of moles

=44 * 6.8181

= 299.9964 grams

Carbon dioxide is used as a refrigerant, in fireplace extinguishers, for inflating lifestyles rafts and life jackets, blasting coal, foaming rubber and plastics, selling the increased vegetation in greenhouses, and immobilizing animals earlier than slaughter, and in carbonated liquids.

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Answer:

[tex]1.242 \times 10^{25}\text{ atoms H}[/tex]

Explanation:

You must convert the mass of B₄H₁₀ to moles of B₄H₁₀, then to molecules of B₄H₁₀, and finally to atoms of H.

1. Moles of B₄H₁₀

[tex]\text{Moles of B$_{4}$H}_{10} = \text{110.0 g B$_{4}$H}_{10} \times \dfrac{\text{1 mol B$_{4}$H}_{10}}{\text{53.32 g B$_{4}$H}_{10}} = \text{2.063 mol B$_{4}$H}_{10}[/tex]

2. Molecules of B₄H₁₀

[tex]\text{No. of molecules} = \text{2.063 mol B$_{4}$H}_{10} \times \dfrac{6.022 \times 10^{23}\text{ molecules B$_{4}$H}_{10}}{\text{1 mol B$_{4}$H}_{10}}\\\\=1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10}[/tex]

3. Atoms of H

[tex]\text{Atoms of H} = 1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10} \times \dfrac{\text{10 atoms H}}{\text{1 molecule B$_{4}$H}_{10}}\\\\= \mathbf{1.242 \times 10^{25}}\textbf{ atoms H}[/tex]

Answer:

1)Ni=1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d10 called full-filled

2)Cr=1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5 called half-filled

Answer:

5.94×10¯³

Explanation:

The following data were obtained from the question:

Concentration of N2, [N2] = 0.036 M

Concentration of H2, [H2] = 0.15 M

Equilibrium constant (Kc) = 0.29 M

Concentration of NH3, [NH3] =.…?

The equation for the reaction is given below:

N2(g) + 3H2(g) <=> 2NH3(g)

Thus, we can determine the concentration of NH3 by using the equilibrium expression for the reaction.

This is illustrated below:

The equilibrium constant for a reaction is simply defined as the ratio of the concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Kc = [NH3]² / [N2] [H2]³

Inputting the value of Kc, [N2], and [H2] the value of [NH3]can be obtained as follow:

Kc = [NH3]² / [N2] [H2]³

0.29 = [NH3]²/ 0.036 × 0.15³

Cross multiply

[NH3]² = 0.29 × 0.036 × 0.15³

[NH3]² = 3.5235×10¯⁵

Take the square root of both side

[NH3] = √(3.5235×10¯⁵)

[NH3] = 5.94×10¯³

Therefore, the concentration of NH3, [NH3] is 5.94×10¯³ M.

Answer:

The correct answer is 1.68 × 10⁻⁵ Ci

Explanation:

The activity of the uranium is determined by using the formula,  

R = 0.693 N/t1/2 -------------- (i)

The number of atoms is, N = nNA

Here, NA is the Avogadro number and n is the number of moles. The value of n is m/M, that is, mass/molecular mass. Now the value of N becomes,  

N = (m/M) NA

The m or mass of uranium given is 50.1 grams, and the molecular mass is 238 g/mol, now putting the values we get,  

N = (50 g/238 g) (6.023 × 10²³) = 1.26 × 10²³

The half-life of 238U from year to second is,  

t1/2 = (4.468 × 10⁸ year) (3.16 × 10⁷ s/ 1 year) = 1.412 × 10¹⁶ s

Substituting the values of t1/2 as 1.412 × 10¹⁶, and 1.26 × 10²³ for N in equation (i) we get,  

R = 0.639 (1.26 × 10²³) / 1.412 × 10¹⁶ s  

= 6.18 × 10⁶ Bq (2.7027 × 10⁻¹¹ Ci/1 Bq)

= 1.68 × 10⁻⁵ Ci

Hence, the activity of the plate is 1.68 × 10⁻⁵ Ci

Answer:

74.3g of methanol were introduced into the vessel

Explanation:

In the equilibrium:

CH₃OH(g) ⇄ CO(g) + 2H₂(g)

Kc is defined as the ratio between concentrations in equilibrium of :

Kc = 6.90x10⁻² = [CO] [H₂]² / [CH₃OH]

Some methanol added to the vessel will react producing H₂ and CO. And equilibrium concentrations must be:

[CH₃OH] = ? - X

[CO] = X

[H₂] = 2X

Where ? is the initial concentration of methanol

As [H₂] = 2X = 0.426M; X = 0.213M

[CH₃OH] = ? - 0.213M

[CO] = 0.213M

[H₂] = 0.426M

Replacing in Kc to solve equilibrium concentration of methanol:

6.90x10⁻² = [0.213] [0.426]² / [CH₃OH]

[CH₃OH] = 0.560

As:

[CH₃OH] = ? - 0.213M = 0.560M

? = 0.773M

0.7733M was the initial concentration of methanol. As volume of vessel is 3.00L, moles of methanol are:

3.00L * (0.773 mol / L) = 2.319 moles methanol.

Using molar mass of methanol (32.04g/mol), initial mass of methanol added was:

2.319 moles * (32.04g / mol) =

Answer:

Cu2 + 2Mg-> 2Cu+ Mg2

Explanation:

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

Hope it will be helpful

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

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

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

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

Learn more about balanced equations here:

brainly.com/question/15052184

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