Using the Henderson-Hasselbalch equation, calculate the percentage of an ammonia solution that would be in the ionic form at pH 9.40, assuming a pKa for the ammonium ion of 9.25.

Answer:

The answer is in the explanation

Explanation:

Acetic acid, CH₃COOH, is a weak acid that will produce a buffer when its conjugate base, CH₃COO⁻, acetate ion, is added to the solution.

That is because a buffer is the mixture of a weak acid and its conjugate base or vice versa.

When an acid (HX) is added to the solution, the acetate ion will react producing acetic acid, thus:

CH₃COO⁻ + HX → CH₃COOH + X⁻

For this reason, the pH doesn't change abruptly because H⁺ ions are not produced.

Now, if a  base (BOH) is added to the buffer, CH₃COOH will react producing acetate ion and water, thus:

CH₃COOH + BOH → CH₃COO⁻ + H₂O + B⁺.

In the same way, there are not produced free OH⁻ and the pH doesn't change significantly.

Answer:

Option C. Keq = [SO2]² [O2] /[SO3]²

Explanation:

The equilibrium constant keq for a reaction is simply the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.

Now, let us determine the equilibrium constant for the reaction given in the question.

This is illustrated below:

2SO3(g) <==> 2SO2(g) + O2(g)

Reactant => SO3

Product => SO2, O2

Keq = concentration of products /concentration of reactants

Keq = [SO2]² [O2] /[SO3]²

Answer:

Evolution of gas.

Formation of a precipitate.

Change in color.

Explanation:

Complete question:

On a hot summer day, the density of air at atmospheric pressure at 35.5°C is 1.1970 kg/m3. (a) What is the number of moles contained in 1.00 m3 of an ideal gas at this temperature and pressure.

Answer:

The  number of moles contained by an ideal gas at this temperature and pressure is 41.32 moles.

Explanation:

Given;

density of dry air, ρ = 1.1970 kg/m³

temperature of the air, T = 35.5°C  = 273 + 35.5 = 308.5 K

air volume, V = 1 m³

Apply ideal gas law for dry to calculate the air pressure;

[tex]P = \rho R_dT[/tex]

where;

P is the air pressure

ρ is the air density

Rd is gas constant for dry air = 287 J/kg/K

P = 1.197 x 287 x 308.5 = 105,981.78 Pa

(a) Now, determine the number of moles contained by an ideal gas at this temperature and pressure, by applying ideal gas law;

PV = nRT

where;

P is the pressure of the gas (Pa)

V is the volume of the gas (m³)

n is number of gas moles

R is gas constant = 8.314 m³.Pa / mol.K

T is temperature (K)

n = (PV) / (RT)

n = (105,981.78 x 1) / (8.314 x 308.5)

n = 41.32 moles

Therefore, the  number of moles contained by an ideal gas at this temperature and pressure is 41.32 moles.

The number of moles of an ideal gas at this temperature and pressure is 41.5 moles.

Given that;

Density of dry air = 1.1970 kg/m3

Pressure of dry air = ?

Temperature of dry air = 35.5°C + 273 = 308.5 K

Hence;

P = Density × gas constant of dry air × Temperature

P = 1.1970 kg/m3 × 287.1 J/Kg/K × 308.5 K

P = 106019 Pa or 1.05 atm

Using the ideal gas equation;

PV = nRT

n = PV/RT

n = 1.05 atm × 1000 L/0.082 atmL/K.mol × 308.5 K

n = 41.5 moles

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Answer:

997.65cmH2O

Explanation:

Barometric pressure = 739 mmHg

density of Hg = 13.5 g/ml

density of water (H2O) = 1.00 g/ml

Calculate Barometric pressure in centimetres of water ( cmH20)

equate the barometric pressure of Hg and water

739 * 13.5 * 9.8 = x * 1 * 9.81

x ( barometric pressure of water in mmH2O ) = 739 *13.5 / 1 = 9976.5mmH2O

in cmH2O = 997.65cmH2O

Answer:

63.518

Explanation:

The following data were obtained from the question:

Mass of Isotope A = 62.9 amu

Abundance of isotope A (A%) = 69.1%

Mass of isotope B = 64.9 amu

Abundance of isotope B (B%) = 30.9%

Atomic weight of the element =..?

The atomic weight of the element can be obtained as follow:

Atomic weight = [(Mass of A x A%)/100] + [(Mass of B x B%) /100]

Atomic weight = [(62.9 x 69.1)/100] + [(64.9 x 30.9)/100]

Atomic weight = 43.4639 + 20.0541

Atomic weight = 63.518

Therefore, the atomic weight of the element is 63.518.

Answer:

When copper is heated, it decomposes to form copper oxide and carbon dioxide. It is an endothermic reaction, which means that it absorbs heat. When heated, copper is easily bent or molded into shapes.

Explanation:

Answer:

Total numbe of protons and neutrons in a single atom of that element

Explanation:

Hello,

I'll answer the question by filling in the blank spaces

"The atomic mass of an element is equal to the total number of proton and neutron in a particular atom of the element. The atomic mass of an element is equal to the atomic weight. Its mass number one-twelfth of the mass of carbon-12 atom a weighted mass of all naturally occurring isotopes of the elements. Its atomic mass is the average mass of all the naturally occurring isotopes of the element."

The atomic mass of an element is the total number of protons and neutrons in a single atom of that element.

The atomic mass of an element is equal to a weighted average mass of all of the naturally occurring isotopes of the element. The correct answer is option 2.

Isotopes are elements with the same number of protons (atomic number) but differing numbers of neutrons (mass number).

Most elements exist in nature as a mixture of isotopes, each with a different mass number and abundance. The atomic mass of an element is computed by adding the masses of all isotopes, multiplying by their relative abundance, and dividing by the total abundance of all isotopes.

This gives a weighted average mass that corresponds to the normal mass of an element's atom in nature.

Therefore, the correct answer is option 2. to a weighted average mass of all of the naturally occurring isotopes of the element.

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Answer:

The answer is A

Explanation:

Answer:

C(s) + 3 H₂(g) + 1/2 O₂(g) ⇒C₂H₅OH(l)

Explanation:

Ethanol (C₂H₅OH) is an alcohol and it is formed by carbon (C), H (hydrogen) and O (oxygen) atoms. These elements in their standard states are:

C: C(s), it is solid, could be graphite, diamond, among others.

H: H₂(g), it is a diatomic gas.

O: O₂(g), it is a diatomic gas.

So, we can write the equation for the formation of  C₂H₅OH from C(s), H₂ and O₂ as follows:

C(s) + H₂(g) + O₂(g) ⇒C₂H₅OH(l)

Finally, we have to balance the equation by adding the estequiometrical coefficients:

C(s) + 3 H₂(g) + 1/2 O₂(g) ⇒C₂H₅OH(l)

2C(s)+3[tex]H_{2} [/tex](g)+[tex]\frac{1}{2} [/tex][tex]O_{2} [/tex](g)→[tex]C_{2} [/tex][tex]H_{5} [/tex]OH(l)

Explanation:

Standard state of carbon: C(s)

Standard state of oxygen: [tex]O_{2} [/tex](g)

Standard State of Hydrogen: [tex]H_{2} [/tex](g)

Then balance the equation C2H5OH(l) to get 2C(s)+3[tex]H_{2} [/tex](g)+[tex]\frac{1}{2} [/tex][tex]O_{2} [/tex](g)→[tex]C_{2} [/tex][tex]H_{5} [/tex]OH(l).

Answer:Mass of CO2 = 0.60g

Explanation:

Given the chemical rection

Li2CO3(s) + 2HCl(aq) --> 2LiCl(aq) + H2O(l) + CO2(g

No of moles = mass / molar mass

molar mass Li2CO3 = Molecular mass  calculation: 6.941 x 2 + 12.0107 + 15.9994 x 3 =  

= 73.8909 g/mol

therefore Number of moles Li2CO3 = 1.0g / 73.89 g/mol

= 0.0135 moles Li2CO3

From our given Balanced equation,  shows that  

Li2CO3(s) + 2HCl(aq) --> 2LiCl(aq) + H2O(l) + CO2(g

1 mole Li2CO3 produces 1 mole CO2

therefore 0.0135 mol Li2CO3 will produce  0.0135 moles of CO2

Also

No of moles = mass / molar mass

Mass = No of moles x molar mass

molar mass of CO2=12.0107 + 15.9994 x 2=44.0095 g/mol

Mass of CO2= 0.0135 X 44.0095 g/mol =0.594≈0.60g

Answer:

D. Hund's rule

Explanation:

Not sure, but I would go with Hund's since it talks about filing electrons in each orbital before you can pair them up. The reason sulfur has lower ionization is because it has one set of paired electrons which makes the orbital unstable whereas phosphorus has 3 unpaired e's which means it is more stable. Thus it is easier to remove electron from sulfur hence lower ionization energy.

Answer:

Explanation:

Cu(NO₃)₂ + 4NH₃ = Cu(NH₃)₄²⁺  + 2 NO₃⁻

187.5 gm      4M           1 M

187.5 gm reacts with 4 M ammonia

18.8 g     reacts with  .4 M ammonia

ammonia remaining left after reaction

= .8 M - .4 M = .4 M .

187.5 gm reacts with 4 M ammonia   to form 1 M Cu(NH₃)₄²⁺

18.8 g reacts with .4 M ammonia  to form 0.1 M Cu(NH₃)₄²⁺  

At equilibrium , the concentration of Cu²⁺ will be zero .

concentration of ammonia will be .4 M

concentration of  Cu(NH₃)₄²⁺ formed will be 0.1 M

Answer:

900g of POCl₃

Explanation:

Hello,

To solve this question, we'll require the equation of reaction.

P₄O₁₀ + 6PCl₅ → 10POCl₃

Molar mass of P₄O₁₀ = 283.886 g/mol

Molar mass of PCl₅ = 208.24 g/mol

Molar mass of POCl₃ = 153.33 g/mol

But Number of moles = mass / molar mass

Mass = molar mass × number of moles

Mass of POCl₃ = 153.33 × 10 = 1533.3g

Mass of PCl₅ = 208.24 × 6 = 1249.44g

Mass of P₄O₁₀ = 283.886 × 1 = 283.886g

From the equation of reaction,

283.886g of P₄O₁₀ + 1249.44g of PCl₅ produces 1533.33g of POCl₃

I.e 1533.33g of reactants produces 1533.33g of product (law of conservation of mass)

Therefore, (225g of P₄O₁₀ + 675g of PCl₅) = 900g will give x g of POCl₃.

1533.33g of reactants = 1533.33g of products

900g of reactants = x g of products

x = (900 × 1533.33) / 1533.33

x = 900g of POCl₃

Answer:

the atomic mass of any elemet contains avogardo numberof atoms

In case of Gallium,

69.72 gram is atomic mass and it cotnains around 6.023*10^23 atoms of Gallium

but, 2000 punds = 907184.7 grams

907184.7 gram of gallium contains= 6.023*10^23* 907184/69.72

                                                          = 79 *10^26 atoms

Explanation:

Answer:

combustion

Explanation:

The enthalpy change for the complete burning of one mole of a substance

is the enthalpy of __combustion_____ .

Answer:

p block.

Explanation:

jus took the test

Answer:

c p block

Explanation:

Answer:

Explanation:

C₉H₇O₄⁻ = weakest base

C₆H₅COO⁻ = strongest base

HCOO⁻ = intermediate base

HCOOH = not a Bronsted-Lowry base

HC₉H₇O₄ = not a Bronsted-Lowry base

C₆H₅COOH = not a Bronsted-Lowry base

Answer:

it's a two step elimination reaction

Explanation:

it follows a carbocationic pathway. When carbocation is stable, the equation is favourable, that is, double bond is formed by expelling hydrogen atom.

Answer:

The molar mass of the unknown compound is 153.3 g/mol

Explanation:

Step 1: Data given

Mass of an unknown molecular compound = 27.6 grams

Mass of benzene =  0.250 kg

Kf of benzene = 5.12 °C/m

freezing point depression of 3.69 °C

Step 2:  Calculate molality

ΔT = i*Kf*m

⇒with ΔT = reezing point depression of 3.69 °C

⇒with i = the van't Hoff factor of Benzene = 1

⇒with Kf = 5.12 °C/m

⇒ with m = molality = moles unknown compound / mass of benzene

3.69 = 1 * 5.12 * m

m = 0.72 molal

Step 3: Calculate moles of the unknown compound

molality = moles / mass benzene

0.72 molal = moles / 0.250 kg

Moles = 0.72 m * 0.250 kg

Moles = 0.18 moles

Step 4: Calculate molar mass of the unknown compound

molar mass = mass / moles

Molar mass = 27.6 grams / 0.18 moles

Molar mass = 153.3 g/mol

The molar mass of the unknown compound is 153.3 g/mol

Molar mass is the mass of the one mole of substance. The molar mass of the given unknown compound is 153.3 g/mol.

Molality of the compound can be calculated using

ΔT = i Kf m

Where,

ΔT = freezing point depression = 3.69 °C

i =  Van't Hoff factor of Benzene = 1

Kf =  constant of freezing = 5.12 °C/m

m = molality = ?

Put the values in the equation,

3.69 = 1 x 5.12 x m

m = 0.72 molal

Number of moles of the compound,

[tex]\bold {molality =\dfrac { moles} { mass\ benzene}}\\\\\bold {0.72\ molal = \dfrac {moles }{0.250\ kg}}\\\\\bold {Moles = 0.72\ m \times 0.250\ kg}\\\\\bold {Moles = 0.18}[/tex]

So, molar mass of the unknown compound,

[tex]\bold {Molar\ mass =\dfrac { mass}{ moles}}\\\\\bold {Molar\ mass = \dfrac {27.6\ grams }{0.18\ moles}}\\\\\bold {Molar\ mass = 153.3 g/mol}[/tex]

The molar mass of the given unknown compound is 153.3 g/mol.

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Answer:

315mL

Explanation:

Data obtained from the question include the following:

Molarity of stock solution (M1) = 0.135 M

Volume of stock solution needed (V1) =?

Molarity of diluted solution (M2) = 0.0851 M

Volume of diluted solution (V2) = 500mL

The volume of the stock solution needed can be obtain as follow:

M1V1 = M2V2

0.135 x V1 = 0.0851 x 500

Divide both side by 0.135

V1 = (0.0851 x 500) / 0.135

V1 = 315mL

Therefore, the volume of the stock solution needed is 315mL

Answer:

0.129 M

Explanation:

0.100 M HCl = 0.100 mol/L solution HCl

5.00 mL = 0.00500 L solution HCl

0.100 mol/L HCl * 0.00500 L = 0.000500 mol HCl

                             HCl ------> H+ + Cl-

                           1 mol                   1 mol

                    0.000500 mol           0.000500 mol

0.200 M NaCl = 0.200 mol/L solution NaCl

2.00 mL = 0.00200 L solution NaCl

0.200 mol/L NaCl*0.00200 L = 0.000400 mol NaCl

                              NaCl ------> Na+ + Cl-

                            1 mol                        1 mol

                     0.000400 mol               0.000400 mol

Chloride ion altogether (0.000500 mol + 0.000400 mol) =0.000900 mol

Solution altogether (0.00500 L+0.00200 L) = 0.00700L

Molarity (Cl-)= solute/solution = 0.000900 mol/0.00700L = 0.129 mol/L=

= 0.129 M

Answer: their amounts vary throughout the atmosphere

Explanation:

There is very little that travels over the atmosphere

Vary=very little

Hope that helps

Answer:

a. The rate of the reaction is not proportional to the concentration of the reactant.

Explanation:

The rate expression for a zero order reaction is given as;

A → Product

Rate = k[A]⁰

[A]⁰ = 1

Rate = K

GGoing through the options;

a) This is correct because in the final form of the rate expression, the rate is independent of the concentration.

b) This option is wrong

c) This option is also wrong

d) Like options b and c this is also wrong becaus ethere is no relationship between either the concentration or t.

Answer:

Explanation:

17. it goes from solid copper to aqueous copper:

Cu(s) --> Cu₂(aq) + 2e⁻

18. complete ionic:

Cu(s) --> Cu₂(aq) + 2e⁻

19. net ionic, must include only reacting species, so

Cu(s) --> Cu₂(aq) + 2e⁻

20. this type of reaction is dissolution reaction(redox reaction)

copper reduced from Cu²⁺ to Cu.

Answer:

See the explanation

Explanation:

In this case, we have to keep in mind that in the monosubstituted product we only have to replace 1 hydrogen with another group. In this case, we are going to use the methyl group [tex]CH_3[/tex].

In the axial position, we have a more steric hindrance because we have two hydrogens near to the [tex]CH_3[/tex] group. If we have more steric hindrance the molecule would be more unstable. In the equatorial positions, we don't any interactions because the [tex]CH_3[/tex] group is pointing out. If we don't have any steric hindrance the molecule will be more stable, that's why the molecule will the equatorial position.

See figure 1

I hope it helps!

Answer:

Molarity = 0.0428 M = 42.8 mM

Explanation:

Step 1: Data given

Mass of nickel(II) bromide = 1.87 grams

Molar mass of nickel(II) bromide = 218.53 g/mol

Volume = 200 mL = 0.200 L

Step 2: Calculate moles of nickel(II) bromide

Moles nickel (II) bromide = mass / molar mass

Moles nickel (II) bromide = 1.87 grams / 218.53 g/mol

Moles nickel (II) bromide = 0.00856 moles

Step 3: Calculate moles nickel (II) cation

For 1 mol NiBr2 we have 1 mol Ni^2+

For 0.00856 moles NiBr2 we have 0.00856 moles Ni^2+

Step 4: Calculate final molarity of Ni^2+

Molarity = moles / volume

Molarity = 0.00856 moles / 0.200 L

Molarity = 0.0428 M = 42.8 mM

Answer:

CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺

Explanation:

A buffer is defined as the mixture of a weak acid and its conjugate base or vice versa.

For the acetic acid buffer, CH₃CO₂H is the weak acid and its conjugate base is the ion without H⁺, that is CH₃CO₂⁻. The equilibrium equation in water knowing this is:

In the equilibrium, the acid is dissociated in the conjugate base and the hydronium ion.

The acetic acid/acetate buffer system is a common buffer used in the laboratory, the equilibrium equation for the acetic acid/acetate buffer system. The formula of acetic acid is CH3CO2H -

CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺

An acid buffer is a solution that contains roughly the same concentrations of a weak acid and its conjugate base.

The equation is:

CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺

where CH₃CO₂H - acetic acid

and, CH₃CO₂⁻ acetate ion

Thus, CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺ is the equilibrium equation for the acetic acid/acetate buffer system.

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Answer:

T = 48.39%

Explanation:

In this case we need to apply the Beer law which is the following:

A = CεL  (1)

Where:

A: Absorbance of solution

C: Concentration of solution

ε: Molar Absortivity (Constant)

L: Length of the cell

Now according to the given data, we have transmittance of 93% or 0.93. We can calculate absorbance using the following expression:

A = -logT (2)

Applying this expression, let's calculate the Absorbance:

A = -log(0.93)

A = 0.03152

Now that we have the absorbance, let's calculate the concentration of the solution, using expression (1).

A = CεL

C = A / εL

Replacing:

C = 0.03152 / 1 *ε   (3)

Now, we want to know the transmittance of the solution with a length of 10 cm. so:

A = CεL

Concentration and ε are constant, so:

A = (0.03152 / ε) * ε * 10

A = 0.3152

Now that we have the new absorbance, we can calculate the new transmittace:

T = 10^(-A)

T = 0.4839 ----> 48.39%

Answer:

Here's what I get  

Explanation:

1. POCl₃

(a) Lewis structure

Set P as the central atom, with O and Cl atoms directly attached to it.

Electrons available = P + O + 3Cl = 5 + 6 + 3×7 = 11 + 21 = 32

Arrange these electrons to give every atom an octet. Put a double bond between P and O.

You get the structure shown below.

(b) Geometry

There are four bond pairs and no lone pairs about the P atom.

Electron pair geometry — tetrahedral

    Molecular geometry — tetrahedral

(c) Ideal bond angles

Tetrahedral bond angle = 109.5°

2. AlCl₆³⁻

(a) Lewis structure

Set Al as the central atom, with the Cl atoms directly attached to it.

Electrons available = Al + 6Cl + 3(-) = 3 + 6×6 +3 = 6 + 36 = 42

Arrange these electrons to give every atom an octet. Assign formal charges.

You get the structure shown below.

(b) Geometry

There are six bond pairs and no lone pairs about the Al.

Electron pair geometry — octahedral

    Molecular geometry — octahedral

(c) Ideal bond angles

        Axial-equatorial =  90°

Equatorial-equatorial = 120°

                 Axial-axial = 180°