When an aqueous solution of AgNO3 is electrolyzed, a gas is observed to form at the anode. The gas is

Answer:

E. HCl

Explanation:

Cl atom does not have enough electronegativity to make enough positive charge on H.

HCl is the compound which doesn't have hydrogen bonds. This is because of

the higher size of the chlorine atom.

There is no hydrogen bond because of the high size of the chlorine.

Chlorine have electrons with a very low density. It is also very

electronegative which explains why the formation of hydrogen bonds in the

compound HCl is not possible.

Instead, HCl has covalent bonds in which electron is shared between the

hydrogen and  chlorine to achieve a stable configuration.

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Explanation:

Half life = 12.3years

Time =  30 years

Basically half life is the amount of time taken for the intial concentration to be reduced to half.

First half life = 12.3 years = 10/2 = 5 Kg left

Second half life = 24.6  years= 5/2 = 2.5 Kg left

Third half life = 36.9 years = 2.5 / 2 = 1.25 Kg left

This means that after 30 years, the amount of tritium left would e betweem 1.25kg to 2.5 kg.

The number of tritium that will be left after 30 years is  1.844.

The isotope, tritium, has a half-life of 12.3 years. Assume we have 10 kg of the substance Also the number of years should be 30 years

So,

= 10*2^(-30/12.3)

= 1.844

Therefore, we can conclude that The number of tritium that will be left after 30 years is  1.844.

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Answer:

114 ppm

Explanation:

Data obtained from the question include:

Conc. of compound in mol/L = 1.5×10¯⁴ mol/L

Molar mass of compound = 760 g/mol

Conc. in ppm =..?

Next, we shall determine the concentration of the compound in grams per litre (g/L) . This is illustrated below:

Conc. in mol/L = conc. in g/L / Molar mass

1.5×10¯⁴ = conc. In g/L / 760

Cross multiply

Conc. in g/L = 1.5×10¯⁴ x 760

Conc. in g/L = 0.114 g/L

Next, we shall convert 0.114 g/L to milligrams per litre (mg/L). This is illustrated below:

1 g/L = 1000 mg/L

Therefore, 0.114 g/L = 0.114 x 1000 = 114 mg/L

Finally, we shall convert 114 mg/L to parts per million (ppm). This is illustrated below:

1 mg/L = 1 ppm

Therefore, 114 mg/L = 114 ppm

From the calculations made above,

1.5×10¯⁴ mol/L Is equivalent to 114 ppm.

Answer:

C2H4O3

Explanation:

We would have to do some preparations between before solving it the normal way. The main goal is to get the masses of the Individual elements. So here goes;

We can get the mass of C from CO2 using the following steps:

1 mole of CO2 has a mass of 44g (Molar mass) and contains 12g of C.

How did we know the molar mass of CO2 is 44g?

Easy. 1 mole of C = 12, 1 mole of O = 16

But we have two O’s so the total mass of O = (2 * 6) = 32

Total mass of CO2 = mass of C + Mass of O = 12 + 32 = 44

So if 44g of CO2 contains 12g of C, how much of C would be present in 35.21g CO2.

12 = 44

X = 35.21

X = (35.21 * 12) / 44 = 9.603g

We can also get the mass of H from H2O. 1 mole of H2O has a mass of 18g and contains 2g of H.

How did we know the molar mass of H2O is 18g?

Easy. 1 mole of H = 1, 1 mole of O = 16

But we have two H’s so the total mass of H = (2 * 1) = 2

Total mass of H2O = mass of H + Mass of O = 2 + 16 = 18

So how much of H would be present in 14.42g of H2O?

2 = 18

X =14.42

X = (14.42 * 2 ) / 18 = 1.602g

Now we have the masses of C and H. But the question says the compound contains the C, H and O.

So we still have to calculate the mass of Oxygen. We obtain this from;

Mass of Compound = Mass of Carbon + Mass of Oxygen + Mass of Hydrogen

Mass of Oxygen = Mass of compound – (Mass of Carbon + Mass of Hydrogen)

Mass of Oxygen = 30.42 – (9.603 + 1.602)

Mass of Oxygen = 30.42 - 11.205  = 19.215

Now we have all the masses so we are good too go. Let’s have our table.

Elements Carbon (C) Hydrogen (H) Oxygen (O)

Mass        9.603             1.602           19.215

                0.800            1.602           1.2001 (Divide by molar mass)

                1                    2                   1.5       (Divide by lowest number)

                2                    4                   3        (Convert to simple integers by * 2)

The Empirical formula of the compound is C2H4O3

Answer:

yes!you are right a cloudy formation will be formed when they will react.its because if nitrogen.

Answer: The concentration of [tex]CaSO_4[/tex]  in a saturated solution is [tex]3.0\times 10^{-3}M[/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 for the ionization of [tex]CaSO_4[/tex]  is given as:

[tex]K_{sp}[/tex] of [tex]CaSO_4[/tex]  = [tex]9.0\times 10^{-6}[/tex]

By stoichiometry of the reaction:

1 mole of  [tex]CaSO_4[/tex] gives 1 mole of [tex]Ca^{2+}[/tex] and 1 mole of [tex]SO_4^{2-}[/tex]

When the solubility of [tex]CaSO_4[/tex] is S moles/liter, then the solubility of [tex]Ca^{2+}[/tex] will be S moles\liter and solubility of [tex]SO_4^{2-}[/tex] will be S moles/liter.

[tex]K_{sp}=[Ca^{2+}][SO_4^{2-}][/tex]

[tex]9.0\times 10^{-6}=[s][s][/tex]

[tex]9.0\times 10^{-6}=s^2[/tex]

[tex]s=3.0\times 10^{-3}M[/tex]

Thus concentration of [tex]CaSO_4[/tex]  in a saturated solution is [tex]3.0\times 10^{-3}M[/tex]

Answer:

a. Acid B

b. Acid B

c. lower

Hope this helps you

Answer:

Energy

Explanation:

Paper is oxidized in the presence of oxygen. This reaction however proceeds very slowly until energy is supplied to the system. This implies that the reaction is not spontaneous at room temperature. A spontaneous reaction takes place without any need for external supply of energy.

The need for supply of external energy must be as a result of the high activation energy required for the reaction to go to completion. If a chemical reaction has a high activation energy, it will require an external supply of energy in order for such reaction to occur.

Answer: The enthalpy of combustion of iso-octane in excess oxygen at 298 K is -5462.2kJ/mol

Explanation:

The balanced reaction for combustion of isooctane is:

[tex]C_8H_{18}(l)+\frac{25}{2}O_2(g)\rightarrow 8CO_2(g)+9H_2O(l)[/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{(CO_2(g))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(1\times \Delta H^o_f_{(C_8H_{18}(g))})+(\frac{25}{2}\times \Delta H^o_f_{(O_2(g))})][/tex]

We are given:

[tex]\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H^o_{C_8H_{18}(l)}=-258.07kJ/mol[/tex]

Putting values in above equation, we get:

[tex]\Delta H^o_{rxn}=[(8\times (-393.51))+(9\times (-285.8))]-[(1\times (-258.07))+(\frac{25}{2}\times (0))]\\\Delta H^o_{rxn}=-5462.2kJ/mol[/tex]

The enthalpy of combustion of iso-octane in excess oxygen at 298 K is -5462.2kJ/mol

Answer:

volume = 2128L

Explanation:

volume at stp = 22.4dm³ =22400L

CO2 = 12 + 2(16) = 44

4.20/44 = volume/22400L

0.095 = volume/22400L

volume = 0.095 x 22400

volume = 2128L

Answer: 28.02 g

Explanation:

The M stands for molarity. It is moles of solute/liters of solution. We can use the molarity to convert liters to mL, then make a proportion to find the grams.

[tex]\frac{2.75 mol}{L} *\frac{1L}{1000mL} =\frac{2.75 mol}{1000mL}[/tex]

Now that we have molarity in moles and mL, we can use the 107mL to get moles.

[tex]\frac{2.75moles}{1000mL} *107mL=0.29425mol[/tex]

We would multiply moles by molar mass to get grams. The molar mass of magnesium chloride is 95.211 g/mol.

[tex]0.29425mol*\frac{95.211g}{mol} =28.02g[/tex]

Answer:

Choice A. Ammonium chloride.

Explanation:

Consider the bonds in each of the four compounds.

Ammonium chloride [tex]\rm NH_4Cl[/tex] is an ionic compound. Each

The [tex]\rm {NH_4}^{+}[/tex] and [tex]\rm Cl^{-}[/tex] ions in [tex]\rm NH_4Cl[/tex] are connected with ionic bonds.

What make [tex]\rm NH_4Cl[/tex] special is that its cation [tex]\rm {NH_4}^{+}[/tex] is polyatomic. In other words, each [tex]\rm {NH_4}^{+}[/tex] ion contains more than one atoms. These atoms (one [tex]\rm N[/tex] atom and four [tex]\rm H[/tex] atoms) are connected with covalent bonds. Therefore, [tex]\rm NH_4Cl[/tex] has both ionic and covalent bonds.

Carbon dioxide [tex]\rm CO_2[/tex] is a covalent compound. Each [tex]\rm CO_2[/tex] molecule contains two [tex]\rm C=O[/tex] double bonds in total. [tex]\rm CO_2[/tex] molecules have no ionic bond.

The name "ethyl ethanoate" might sound like the name of a salt (think about sodium ethanoate.) However, in reality, ethyl ethanoate [tex]\rm CH_3COOCH_2CH3[/tex] is an ester. The "ethyl" here refers to the [tex]\rm -OCH_2CH3[/tex] part, originating from ethanol. On the other hand, "ethanoate" refers to the [tex]\rm CH_3C(O)-[/tex] part, which can be obtained from ethanoic acid.

These two parts are connected with a covalent [tex]\rm C-O[/tex] single bond. (The [tex]\rm C[/tex] in ethanoic acid is connected to the [tex]\rm O[/tex] in ethanol.) As a result, there's no ionic bond in ethyl ethanoate, either.

Sodium chloride [tex]\rm NaCl[/tex] is an ionic compound. Both the [tex]\rm Na^{+}[/tex] ion and the [tex]\rm Cl^{-}[/tex] are monoatomic. While the [tex]\rm Na^{+}[/tex] and [tex]\rm Cl^{-}[/tex] in sodium chloride are connected with ionic bonds, neither [tex]\rm Na^{+}[/tex] nor [tex]\rm Cl^{-}[/tex] contains covalent bond.

Answer:

HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)

NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)

Explanation:

A buffer is a solution that resists changes in acidity or alkalinity. A buffer is able to neutralize a little amount of acid or base thereby maintaining the pH of the system at a steady value.

A buffer may be an aqueous solution of a weak acid and its conjugate base or a weak base and its conjugate acid.

The equations for the neutralizations that occurred upon addition of HCl or NaOH are;

HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)

NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)

Answer: 8368 Joules

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]Q=m\times c\times \Delta T[/tex]

Q = Heat absorbed or released =?

c = specific heat capacity of water = [tex]4.184J/g^0C[/tex]

Initial temperature of water = [tex]T_i[/tex] = [tex]20^0C[/tex]

Final temperature of water = [tex]T_f[/tex]  = [tex]60^0C[/tex]

Change in temperature ,[tex]\Delta T=T_f-T_i=(60-20)^0C=40^0C[/tex]

Putting in the values, we get:

[tex]Q=50.0g\times 4.184J/g^0C\times 40^0C=8368J[/tex]

Thus energy in Joules required is 8368.

Answer:

The correct answer is (d) 24.8 Torr

Explanation:

When a solute is added to a solvent, the water pressure of the solution is lower than the vapor pressure of the pure solvent. This is called vapor pressure lowering and it is given by the following expression:

Psolution= Xsolvent x Pºsolvent

We have to calculate Xsolvent (mole fraction of solvent) which is given by the number of moles of solute divided into the total number of moles.

First, we calculate the number of moles of solute and solvent. The solute is glucose (C₆H₁₂O₆), and its number of moles is calculated from the mass and the molecular weight (MM):

MM (C₆H₁₂O₆)= (12 g/mol x 6) + (1 g/mol x 12) + (16 g/mol x 6) = 180 g/mol

moles of glucose= mass/MM= (72.9 g)/)(180 g/mol)= 0.405 moles

The solvent is water (H₂O) and again we calculate the number of moles as follows:

MM(H₂O)= (1 g/mol x 2) + 16 g/mol = 18 g/mol

moles of water= mass/MM= (115 g)/(18 g/mol)= 6.389 moles

Now, we calculate the total number of moles (nt):

nt= moles of glucose + moles of water= 0.405 moles + 6.389 moles= 6.794 moles

The mole fraction of water (Xsolvent) is given by:

Xsolvent= moles of water/nt= 6.389 moles/6.794 moles= 0.940

Finally, the vapor pressure of water over the solution will be the following:

Psolvent= Xsolvent x Pºsolvent= 0.940 x 26.4 Torr= 24.8 Torr

The number of radioactive atoms present in the sample is 1.01 × 10⁷ atoms.

Emission of Beta Particles takes place when there is a numerous amount of neutron to the proton in the nucleus. As a result, the neutron is transformed to proton (which is retained in the nucleus) and electron which leaves the nucleus under high energy.

Given that:

[tex]\mathbf{^{130} I \ \ \ \to ^{- \beta } \ \ \ ^{130}Xe }[/tex]

The rate of disintegration of the radioactive sample per minute is expressed by using the formula:

A = kN

where;

∴

[tex]\mathbf{N = \dfrac{A}{k}}[/tex]

[tex]\mathbf{N = \dfrac{9487}{0.00094}}[/tex]

N = 10092553.19 atoms

N = 1.01 × 10⁷ atoms  

Therefore, we can conclude that the number of radioactive atoms present in the sample is 1.01 × 10⁷ atoms.

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Answer:

54

Explanation:

The mass number of the atom is 54

Answer:

Ksp = 7.4x10⁻⁷

Explanation:

Molar solubility of a substance is defined as the amount of moles of that can be dissolved per liter of solution.

Ksp of Zn(OH)₂ is:

Zn(OH)₂(s) ⇄ Zn²⁺ + 2OH⁻

Ksp = [Zn²⁺] [OH⁻]²

And the molar solubility, X, is:

Zn(OH)₂(s) ⇄ Zn²⁺ + 2OH⁻

                 ⇄ X + 2X

Because X are moles of substance dissolved.

Ksp = [X] [2X]²

Ksp = 4X³

As molar solubility, X, is 5.7x10⁻³mol/L:

Ksp = 4X³

Ksp = 4 (5.7x10⁻³mol/L)³

Answer:

The Coulomb Barrier U is 25.91 MeV

Explanation:

Given that:

Atomic Mass of lead nucleus A = 208

atomic mass of an alpha particle A = 4

Radius of an alpha particle [tex]R_\alpha = R_o A^{^{\dfrac{1}{3}}[/tex]

where;

[tex]R_\alpha = 1.2 \times 10 ^{-15} \ m[/tex]

[tex]R_\alpha = R_o A^{^{\dfrac{1}{3}}[/tex]

[tex]R_\alpha = 1.2 \times 10 ^{-15} \ m \times (4) ^{^{\dfrac{1}{3}}[/tex]

[tex]R_\alpha = 1.905 \times 10^{-15} \ m[/tex]

Radius of the Gold nucleus

[tex]R_{Au}= R_o A^{^{\dfrac{1}{3}}[/tex]

[tex]R_{Au}= 1.2 \times 10 ^{-15} \ m \times (208) ^{^{\dfrac{1}{3}}[/tex]

[tex]R_{Au} = 7.11 \times 10^{-15} \ m[/tex]

[tex]R = R_\alpha + R_{Au}[/tex]

[tex]R = 1.905 \times 10^{-15} \ m + 7.11 \times 10^{-15} \ m[/tex]

[tex]R = 9.105 \times 10 ^{-15} \ m[/tex]

The electric potential energy of the Coulomb barrier [tex]U = \dfrac{Ke \ q_{\alpha} q_{Au}}{R}[/tex]

[tex]U = \dfrac{8.99 \times 10^9 \ N.m \ ^2/C ^2 \ \times 2 ( 82) \times \(1.60 \times 10^{-19} C \ \ e } {9.105 \times 10^{-15} \ m }[/tex]

U = 25908577.7eV

U = 25.908577 × 10⁶ eV

U =  25.91 MeV

The Coulomb Barrier U is 25.91 MeV

Answer:

160g

Explanation:

Mass in grams is equal to product of moles and molar mass of compound.

Answer:

a) Conjugate base F– b) Conjugate acid K+ c) Conjugate acid NH4+ d) Conjugate base NO2- e) Conjugate base HCOO– f) Conjugate acid CH4+

Explanation:

Acid will produce Conjugate base

Base will produce Conjugate acid.

Answer:

a. The acid HF: F-

b. The base KOH: H2O

c. The base NH3: NH4+

d. The acid HNO3: NO3-

e. The acid HCOOH: COOH-

f. The base CH3NH2: CH3NH3+

Explanation:

Answer:

[tex]Pb(ClO_{3})_{4}\\Pb(MnO_{4})_{4}\\Fe(ClO_{3})_{3}\\\Fe(MnO_{4})_{3}\\[/tex]

Explanation:

[tex]Pb^{4+}(ClO_{3}^{-})_{4}--->Pb(ClO_{3})_{4}\\Pb^{4+}(MnO_{4}^{-})_{4}--->Pb(MnO_{4})_{4}\\Fe^{3+}(ClO_{3}^{-})_{3}--->Fe(ClO_{3})_{3}\\\Fe^{3+}(MnO_{4}^{-})_{3}--->Fe(MnO_{4})_{3}\\[/tex]

Answer:

[tex]m_{CaCl_2}=39.96gCaCl_2[/tex]

Explanation:

Hello,

In this case, for the undergoing reaction, we can compute the grams of the formed calcium chloride by noticing the 1:1 molar ratio between calcium and it (stoichiometric coefficients) and using their molar mass of 40 g/mol and 111 g/mol by using the following stoichiometric process:

[tex]m_{CaCl_2}=14.4gCa*\frac{1molCa}{40gCa} *\frac{1molCaCl_2}{1molCa} *\frac{111gCaCl_2}{1molCaCl_2}\\ \\m_{CaCl_2}=39.96gCaCl_2[/tex]

Clearly, chlorine is not used since it is said there is enough for the reaction to go to completion.

Best regards.

Answer: H2O and OH^-  are added to balance the oxygen atoms.

Explanation:

Answer:

Renewable Resources: are considered unlimited, are replaced faster than used.

Nonrenewable Resources: are used more quickly than replaced, have fixed amounts, cannot be replaced in a short time.

Explanation:

Renewable resources are natural resources that are able to naturally regenerate themselves, hence, they are considered to be unlimited. They are usually replaced faster than they are used because they have a short regeneration time. A good example is the solar energy.

Nonrenewable resources are those natural resources that cannot naturally regenerate and when they do, it takes a very long time (usually millions of years). They are therefore used at a much faster rate than they are being replaced and their natural deposits are more or less fixed due to the long regeneration time. A good example is the crude oil deposit.

Hence:

Renewable Resources: are considered unlimited, are replaced faster than used.

Nonrenewable Resources: are used more quickly than replaced, have fixed amounts, cannot be replaced in a short time.

Answer: !

Explanation:

Answer:

[H₂A] = 5.0409x10⁻⁷M

[HA⁻] = 0.001951M

[A²⁻] = 0.234

11.29 = pH

Explanation:

When Na₂A is in equilibrium with water, the reactions that occurs are:

2Na⁺ + A²⁻(aq) + H₂O(l) ⇄ HA⁻(aq) + 2Na⁺(aq) + OH⁻(aq)

As sodium ion doesn't react:

A²⁻(aq) + H₂O(l) ⇄ HA⁻(aq) + OH⁻(aq)

Kb1 = KwₓKa2 = 1x10⁻¹⁴/ 6.19x10⁻⁹ = 1.6155x10⁻⁶ = [HA⁻] [OH⁻] / [A²⁻]

And HA⁻ will be in equilibrium:

HA⁻(aq) + H₂O(l) ⇄ H₂A(aq) + OH⁻(aq)

Kb2 = KwₓKa1 = 1x10⁻¹⁴/ 7.68x10⁻⁵ = 1.3021x10⁻¹⁰ = [H₂A] [OH⁻] / [HA⁻]

In the reaction, you have 2 equilibriums, for the first reaction, concentrations in equilibrium are:

[HA⁻] = X

[OH⁻] = X

[A²⁻] = 0.236M - X

Replacing in Kb1:

1.6155x10⁻⁶ = [HA⁻] [OH⁻] / [A²⁻]

1.6155x10⁻⁶ = [X] [X] / [0.236-X]

3.8126x10⁻⁶ - 1.6155x10⁻⁶X = X²

3.8126x10⁻⁶ - 1.6155x10⁻⁶X - X² = 0

Solving for X

X = -0.00195 → False solution. There is no negative concentrations

X = 0.001952.

Replacing, concentrations for the first equilibrium are:

[HA⁻] = 0.001952

[OH⁻] = 0.001952

[A²⁻] = 0.234

Now, in the second equilibrium:

[HA⁻] = 0.001952 - X

[OH⁻] = X

[H₂A] = X

Replacing in Kb1:

1.3021x10⁻¹⁰ = [H₂A] [OH⁻] / [HA⁻]

1.3021x10⁻¹⁰ = [X] [X] / [0.001952 - X]

2.5417x10⁻¹³ - 1.3021x10⁻¹⁰X = X²

2.5417x10⁻¹³ - 1.3021x10⁻¹⁰X - X² = 0

Solving for X

X = -5.04x10⁻⁷ → False solution. There is no negative concentrations

X = 5.0409x10⁻⁷

Replacing, concentrations for the second equilibrium are:

[HA⁻] = 0.001951M

[OH⁻] = 5.0409x10⁻⁷M

[H₂A] =  5.0409x10⁻⁷M

Thus, you have concentrations of H2A, HA−, and A2−

Now, for pH, the sum of both productions of [OH⁻] is:

[OH⁻] = 0.0019525

pOH = -log[OH⁻] = 2.709

As 14 = pH+ pOH

11.29 = pH

As per the question the pH and the cons of the H2A, the HA−, and the A2−, at equilibrium for a 0.236 M.

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Answer:

because it covers a large area as there are more then 1 lakh compounds of organic chemistry.

Answer:

7 chlorine atoms

Explanation:

K=2.8.8.1

Cl=2.8.7

pottasium will give chlorine its I valence electron to form ions as follows

K=(2.8.8)+

Cl=(2.8.8)-

It will react with 1 chlorine atom.

An ionic bond is shaped by using the whole transfer of some electrons from one atom to every other. The atom losing one or more electrons becomes a cation—an undoubtedly charged ion. The atom gaining one or more electrons will become an anion—a negatively charged ion.

In those compounds carbon, nitrogen, oxygen, and chlorine atoms have four, three, and one bonds, respectively. The hydrogen atom and the halogen atoms form the most effective covalent bond to different atoms in maximum stable neutral compounds.

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