Can I+ (the iodine cation) becalled a Lewis base?
Fullyexlplain your answer.

Answers

Answer 1

Answer:

Answer: No I+ cannot be called a Lewis base.

Explanation:

According to Lewis Theory, it defines an acid as an electron-pair acceptor and a base as an electron-pair donor.

In terms of Lewis basicity, Iodide ion (anions) has the more readily available lone pair electrons for donation since iodide ion is less electronegative .

With the help of the net electronic structure one can understand the answer of the question, because we need to study the I+ ion (cation) structure.

Lewis acid is therefore any substance, that can accept a pair of nonbonding electrons.

From the picture below I+ is most likely ready to accept electrons not to give from it 5s orbital to become stable.

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Two hypothetical ionic compounds are discovered with the chemical formulas XCl2 and YCl2, where X and Y represent symbols of the imaginary elements. Chemical analysis of the two compounds reveals that 0.25 mol XCl2 has a mass of 100.0 g and 0.50 mol YCl2 has a mass of 125.0 g. (a) What are the molar masses of XCl2 and YCl2

1.81 g H2 is allowed to react with 10.2 g N2, producing 2.19 g NH3.What is the theoretical yield in grams for this reaction under the given conditions?3H2(g)+N2(g)→2NH3(g)

Answers

The theoretical yield : = 10.251 g

Further explanation

Given

Reaction

3H₂(g)+N₂(g)→2NH₃(g)

1.81 g H₂

10.2 g N₂

2.19 g NH₃

Required

The theoretical yield

Solution

Find limiting reactant :

H₂ : 1.81 g : 2 g/mol = 0.905 mol

N₂ : 10.2 g : 28 g/mol = 0.364 mol

mol : coefficient

H₂ = 0.905 : 3 = 0.302

N₂ = 0.364 : 1 = 0.364

H₂ as a limiting reactant(smaller ratio)

Moles NH₃ based on H₂, so mol NH₃ :

= 2/3 x mol H₂

= 2/3 x 0.905

=0.603

Mass NH₃ :

= mol x MW

=0.603 x 17 g/mol

= 10.251 g

Calculate the enthalpy for this reaction: 2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ Given the following thermochemical equations: C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(ℓ) ΔH° = -1,123 kJ C(s) + O2(g) ---> CO2(g) ΔH° = -340 kJ H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -211 kJ

Answers

Answer:

The enthalpy for given reaction is 232 kilo Joules.

Explanation:

$C_2H_2(g) + (5)/(2)O_2(g)\rightarrow 2CO_2(g) + H_2O(l), \Delta H^o_(1) = -1,123 kJ$ ...[1]

$C(s) + O_2(g)\rightarrow CO2(g), \Delta H^o_(2) = -340 kJ$ ..[2]

$H_2(g) + (1)/(2)O_2(g)\rightarrow H_2O(l) ,\Delta H^o_(3) = -211 kJ$ ..[3]

$2C(s) + H_2(g)\rightarrow C_2H_2(g),\Delta H^o_(4) =?$ ..[4]

2 × [2] + [3] - [1] ( Using Hess's law)

$\Delta H^o_(4)=2* \Delta H^o_(2)+\Delta H^o_(3) - \Delta H^o_(1)$

$\Delta H^o_(4)=2* (-340 kJ) + (-211 kJ) - (-1,123 kJ)$

$\Delta H^o_(4)=232 kJ$

The enthalpy for given reaction is 232 kilo Joules.

Write all of the allowed sét of quantum numbers for a 3p orbital

Answers

Answer:

Which of the following is a possible set of quantum numbers for a 3p orbital?

Since you are dealing with a 3p-orbital, your principal and angular momentum quantum numbers will be n=3 and l=1 .

How many moles of sulfur atoms are there in 5.0 g of sulfur?

Answers

Answer:

Number of moles = 0.153 mol

Explanation:

Given data:

Mass of sulfur = 5 g

Number of moles of sulfur atom = ?

Solution:

Formula:

Number of moles = mass/molar mass

Molar mass of sulfur is 32. 065g/mol.

By putting values,

Number of moles = 5 g/ 32.06 g/mol

Number of moles = 0.153 mol

Bobby created a dilution of 1/100 of a bacterial sample by adding 1 mL of sample to 99 mL of saline. Unfortunately, after Bobby completed the dilution, he knocked the container over spilling the majority of the diluted sample out. After cleaning up the mess, he found he had 19 mL of diluted sample remaining. Can he still completed the microbial count and if so, then write out the steps on how would he determine the original cell concentration of his total remaining samp

Answers

Answer:

There is no short answer.

Explanation:

In the given example Bobby is creating a solution for his bacteria count which consists of 1% bacterial sample.

Considering that the solution was mixed homogeneously, he can apply the procedure to the remaining sample and get the results he wants.

Or if the average number of bacteria in a 1 mL sample is known, he can apply that information proportionally to the 100 mL mixture and find the original cell concentration.

I hope this answer helps.

What is colloidal solutions

Answers

Explanation:

Colloidal solutions, or colloidal suspensions, are nothing but a mixture in which the substances are regularly suspended in a fluid. ... Colloidal systems can occur in any of the three key states of matter gas, liquid or solid. However, a colloidal solution usually refers to a liquid concoction.

Answer:

Colloidal solutions, or colloidal suspensions, are nothing but a mixture in which the substances are regularly suspended in a fluid.

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