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3 years ago

A) Select the strongest acid:H2Te, GeH4, AsH3, H2S, SiH3

Chemistry

1 answer:

sineoko [7]3 years ago

4 0

Answer:

(A). H2S.

(B). BrO-.

(C). H2SbO3 .

(D). HClO4.

(E). CH3NH2.

(F). HClO > HBrO > HIO.

Explanation:

(A). The strongest acid is H2S because it has the lowest atomic size. If we move from the left side of the periodic table to the right side of the periodic table the atomic size decreases. As the atomic size decreases, it will be easier for H^+ ion to be released. Therefore, the ease of releasing Hydrogen ion makes H2S more acidic than the others in this group.

(B). BrO- is more basic because of bond pair - lone pair repulsion. BrO2- , BrO4-, and ClO2- exhibit resonance and therefore, they do not possess good basic features.

(C). H2SbO3 has the lowest pKa because the atomic size of Sb is greater than As and P , As is greater than P. This means that It is more difficult to remove Hydrogen ion from Sb.

(D). HClO4 is the Acid with the highest pKa because of high resonance.

(E). CH3NH2 is more basic because it has the highest electron Density. In BrNH2, the bromine ion reduces the electron Density.

(F). In halogens, as one goes down the group in the periodic table the acidic properties decreases, hence, HClO > HBrO > HIO.

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Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:

Lapatulllka [165]

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

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What is the net force on this object?

blondinia [14]

The answer is 3N up.

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Carbon (C): 1s(H)2s(I)2p(J) What is H, I, and J equal?

Artemon [7]

Answer:

H =2; I = 2; J = 2

Explanation:

Carbon is element 6 in the Periodic Table.

Start at element 1 (H) and count from left to right until you reach element 6 (C).

You get the electron configuration

C: 1s² 2s²2p².

Thus,

H =2; I = 2; J = 2

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3 years ago

List the numbers of protons, neutrons, and electrons in Chlorine-18.

OLEGan [10]

17 protons 17 electrons 18 neutrons

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3 years ago

The area surrounding the Pacific Ocean is often referred to as the ring of Firee due to the concentration of geologic events. De

viva [34]

Answer:

The amount of movement of tectonic plates in the area

Explanation:

The population density vary across the countries

Japan(127m), Philippines ( 103m), Indonesia (267m).

The volcanoes in Indonesia are among the most active of the pacific Ring of Fire along the north east island

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3 years ago