Nickel (II) ions form a complex ion in the presence of ammonia with a formation constant (Kf) of 2.0×10^8:Ni2+ + 6NH3 ⇌ [Ni(NH3)6]2+
Calculate the molar solubility of NiS in 3.1 M NH3. g

Answers

Answer 1
Answer:

Answer:

The molar solubility of NiS is 7.7 * 10⁻⁷ M

Explanation:

To answer this question, we need to keep in mind two equilibriums.

First, we have the solubilization of NiS:

NiS ⇄ Ni²⁺ + S²⁻   ksp= 3.0 * 10⁻²¹  (we know this from standard tables)

Second, we have the formation of the complex:

Ni²⁺ + 6NH₃ ⇄ [Ni(NH₃)₆]²⁺   kf=2.0 * 10⁻⁸

Combine the two equilibriums and we have

NiS + 6NH₃ ⇄ [Ni(NH₃)₆]²⁺ + S²⁻  K= ksp * kf =6.0* 10⁻¹³=([S^(2-)][Ni(NH3)6^(+2)])/([NH3]^(6))

The molar solubility s is equal to both [Ni(NH₃)₆²⁺] and [S²]

At equilibrium, [NH₃]= 3,1 M - 6s

Thus, if we replace those terms in the formula for K, we're left with:

(s^(2) )/((3,1-6s)^(2))=6*10^(-13)

Using an approximation we can ignore the denominator and we have

  • s²=6.0 * 10⁻¹³
  • s=7.7 * 10⁻⁷

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Transpiration is the loss of water from the leaves of plants. The stomata of leaves must open to allowcarbon dioxide to enter the leaf for photosynthesis, but when they are open, water vapor escapes into
the atmosphere.
HYPOTHESIS: As the intensity of light is increased, the rate of transpiration will increase, as
measured in by the loss of mass of the plant.
Independent variable:
Dependent variable:

Answers

Independent variable are the light intensity that what we change and dependent variable are the rate of transpiration that what we measure.

What is photosynthesis ?

Photosynthesis is the process that can be used by the plants and other animals to covert light energy into the chemical energy. Photosynthesis is the main source of food in the earth.

Photosynthesis is the process in which oxygen is released. For survival oxygen is very important, from this process we obtain sufficient amount of oxygen.

The dependent variable is the rate of transpiration and the independent variable is time.The dependent variable is the rate of transportation because it is depends on the environmental factor the plant is placed.

Thus,Independent variable are the light intensity that what we change and dependent variable are the rate of transpiration that what we measure.

To learn more about the photosynthesis, follow the link;

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Answer:

Independent variable: The light intensity (what YOU CHANGE)

Dependent variable: The rate of transpiration (what YOU MEASURE)

The [Fe(H₂O)₆]³⁺ complex requires a relatively small amount of energy to promote an electron from the t2g to the eg. Based on the UV data, predict the spin of this complex.

Answers

Answer:

The spin of the complex is 5.92 B.M

Explanation:

Please see the attachments below

Final answer:

The [Fe(H₂O)₆]³⁺ complex is a high-spin complex due to the relatively small energy required to promote an electron from the t2g to the eg orbital. As such, it is reasonable to predict that it has a high-spin state with five unpaired electrons.

Explanation:

The [Fe(H₂O)₆]³⁺ complex is a type of coordination complex in which the central metal atom, Fe³⁺, is surrounded by six water molecules acting as ligands. The spin state of such a complex can be determined based on the energy required to promote an electron from the t2g to the eg.

In [Fe(H₂O)₆]³⁺, the field produced by the water ligands is relatively weak, resulting in a small crystal field splitting (Aoct <P). Given that it requires less energy for the electrons to occupy the eg orbitals than to pair up, there will be an electron in each of the five 3d orbitals before any pairing occurs. Hence, for the six d electrons on the Fe³⁺ ion in [Fe(H₂O)₆]³⁺, there should be one pair (two electrons) and four unpaired electrons.

High-spin complexes are those in which the electrons tend not to pair up because the crystal field splitting is not large enough to make it energetically favorable for them to do so. Given that the [Fe(H₂O)₆]³⁺ complex falls under the categories of high-spin complexes, it is reasonable to predict that it exhibits a high-spin state with five unpaired electrons.

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Classify each type of matter as an element, a compound, a heterogeneous mixture, or a homogeneous mixture. drag each item to the appropriate bin. sugar air hot tea salt water

Answers

Answer :

Sugar : Compound

Air : Homogeneous mixture

Hot tea : Homogeneous mixture

Salt water : Homogeneous mixture

Explanation :

Element : It is a pure substance which is composed of atoms of similar elements.

Compound : It is a pure substance which is made from atoms of different elements combined together in a fixed ratio by mass.

Heterogeneous mixtures : It is a mixture that has non-uniform composition throughout the solution and the particle size or shapes are also different.

There is a physical boundary between the dispersed phase and dispersion medium.

Homogeneous mixtures : It is a mixture that has uniform composition throughout the solution and the particle size or shapes are not different.

There is no physical boundary between the dispersed phase and dispersion medium.

As per question,

Sugar : It is a compound that is made from atoms of different elements combined together in a fixed ratio by mass.

Air : It is a homogeneous mixture of the gaseous substances of nitrogen, oxygen, and smaller amounts of other substances.

Hot tea : It is a homogeneous mixture because its composition is the same throughout the solution.

Salt water : It is a homogeneous mixture because its composition is the same throughout the solution.

Sugar compound

Air Homogeneous mixture

Hot tea Homogeneous mixture

Salt water Homogeneous mixture

 

Element:

An element is a simplest type of a pure substance which is composed of only one type of matter. For examples; silver, gold platinum etc.

Pure substance:

A compound is a pure substance which is composed of two or more type of matter in definite proportion. For examples; water which is composed by Hydrogen and oxygen etc.

Mixture:

A mixture is made from two or more type of matter in any proportion. Mixture are generally two types:

1. Homogeneous mixture

This is also known as solution, in which the composition of the substances is uniform.

Lemon-flavored water is an example of homogeneous mixture because in this the composition of the substances is uniform.

2. Heterogeneous mixture

In this form of mixture the composition of substance are not uniform and the component of mixture also visible.

Stuffed mushrooms is an example of Heterogeneous mixture because in this form of mixture the composition of substance are not uniform and the component of mixture also visible.


To test Döbereiner’s idea, predict:(a) The boiling point of HBr from the boiling points of HCl (- 84.9°C) and HI (-35.4°C) (actual value = -67.0°C)
(b) The boiling point of AsH₃ from the boiling points of PH₃ (- 87.4°C) and SbH₃ (-17.1°C) (actual value = -55°C)

Answers

Answer:

a) Approximate boiling point of HBr = -60.15 °C

b) Approximate boiling point of AsH₃ = -52.25 °C

Explanation:

Döbereiner stated that some elements could be arranged in groups of 3 similar elements ( known as "triads) , and the element of the middle ( elements are ordered with respect to their atomic mass) would have properties between the other 2 ( the average value)

a) In the first case the triad would be the halogen triad ( Cl , Br and I ) . And according to Döbereiner , the boiling point of HBr should be the average of HCl and HI . Therefore

Approximate boiling point of HBr = [(- 84.9°C) + (-35.4°C)]/2 = -60.15 °C

b) Simmilarly for  AsH₃ , PH₃ and SbH₃ , the boiling point of AsH₃ would be

Approximate boiling point of AsH₃ = [(- 87.4°C) + (-17.1°C)]/2 = -52.25 °C

Be sure to answer all parts. Industrially, hydrogen gas can be prepared by combining propane gas (c3h8) with steam at about 400°c. The products are carbon monoxide (co) and hydrogen gas (h2). (a) write a balanced equation for the reaction. Include phase abbreviations. (b) how many kilograms of h2 can be obtained from 8.31 × 103 kg of propane

Answers

Balanced chemical reaction:

C₃H₈(g) + 3H₂O(g) → 3CO(g) + 7H₂(g).

M(C₃H₈) = 44.1 g/mol; molar mass of propane.

M(H₂) = 2 g/mol; molar mass of hydrogen.

From balanced chemical reaction: n(C₃H₈) : n(H₂) = 1 : 7.

7m(C₃H₈) : M(C₃H₈) = m(H₂) : M(H₂).

7·8310 kg : 44.1 g/mol = m(H₂) : 2 g/mol.

m(H₂) = 2638.09 kg; mass of hydrogen.

Answer: a) C_3H_8(g)+3H_2O(g)\rightarrow 3CO(g)+7H_2(g)

b) 2.64* 10^3kg

Explanation:

a) According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

C_3H_8(g)+3H_2O(g)\rightarrow 3CO(g)+7H_2(g)

b)\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

\text{Number of moles of propane}=(8.31* 10^6g)/(44.1g/mol)=0.188* 10^6moles

C_3H_8(g)+3H_2O(g)\rightarrow 3CO(g)+7H_2(g)

According to stoichiometry:

1 mole of C_3H_8 gives 7 moles of H_2

Thus 0.188* 10^6moles  moles of C_3H_8 will give =(7)/(1)* 0.188* 10^6=1.32* 10^6moles of H_2

Mass of H_2=moles* {\text {molar Mass}}=1.32* 10^6moles* 2g/mol=2.64* 10^6g=2.64* 10^3kg

Thus 2.64* 10^3kg of H_2 can be obtained from 8.31* 10^3 kg of propane

A solution contains 0.10 M Pb2+ and 0.10 M Cu2.. Which cation will precipitate first when a solution of NazS is slowly added to the mixture? Refer to the information sheet for solubility constants. P A) Pb2+ B) Cu2+ C) impossible to tell D) both cations

Answers

Answer:

b) Cu2+

Explanation:

  • information sheet for solubility constants:

Ksp PbS = 3.4 E-28

Ksp CuS = 6.0 E-37

  • PbS ↔ Pb2+  +  S2-

∴ Ksp = 3.4 E-28 = [ Pb2+ ] * [ S2- ]

∴ [ Pb2+ ] = 0.10 M

⇒ [ S2- ] = 3.4 E-28 / 0.10 = 3.4 E-27 M

  • CuS ↔ Cu2+  +  S2-

∴ Ksp = 6.0 E-37 = [ Cu2+ ] * [ S2- ]

∴ [ Cu2+ ] = 0.10 M

⇒ [ S2- ] = 6.0 E-37 / 0.10 = 6.0 E-36 M

we have:

(1) [ S2- ] PbS >> [ S2- ] CuS

(2) Ksp PbS >> Ksp CuS

from (1) and (2) it can determined, that separation can be carried out and also the cation that precipitates first is the Cu2+