Is it easier to form a positive ion with an element that has a high ionization energy or element that has a low ionization energy? Explain.

Answers

Answer 1

Answer: It's easier when an atom or molecule has a low ionization energy. Because ionization energy is, the minimum energy needed to remove an electorn from an atom or a molecule in the gas phase. So if an atom has a lower ionization energy, you can remove an electron and form an ion easily.

Answer 2

Answer:

Final answer:

Atoms with low ionization energy form positive ions or cations more easily than atoms with high ionization energy due to less energy being required to remove an electron from the atom's valence shell.

Explanation:

The ease at which an atom forms a positive ion, also known as a cation, depends on its ionization energy. Ionization energy is the minimum amount of energy required to remove an electron from an atom in its ground state.

An atom with a low ionization energy will form a cation more readily than an atom with a high ionization energy. This is because it takes less energy to remove an electron from the atom's valence shell. On the other hand, atoms with a high ionization energy hold their electrons more tightly, thus it requires more energy to form a cation. This is seen in the trend across the periodic table where ionization energy increases across a period and decreases down a group due to the size and energy of the orbitals.

Sometimes, successive ionization energies or the energy necessary to remove subsequent electrons from an atom increase because core electrons, which are closer to the nucleus and thus experience a strong electrostatic attraction, are harder to remove than valence electrons.

Learn more about Ionization Energy here:

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What is the molecular mass of HCI

Answers

Answer:

36.46 g/mol

pls mark brainlest

A bobsled has a momentum of 4000 0 kg* m/s to the south. Friction on the track reduces its momentum to 500 kg* m/s to the south. What impulse is applied to the bobsled by the friction? A. 6500 kg* m/s north

OB. 1500 kg* m/s south O

C. 6500 kg* m/s south

O 1500 kg* m/s north

Answers

Answer: 1500 kg•m/s north

Explanation:

A PE X

The correct answer is c

N2(g) + 3 H2(g) ---> 2 HN3(g) (blanced)If 5.42 g of nitrogen are reacted with 5.42 g of hydrogen gas, which of the reactants is the limiting reactant?

Molar Mass of N2 = 28.02 g/mol
Molar Mass of H2 = 2.02 g/mol
Molar Mass of NH3 = 17.04 g/mol

Answers

Answer:

N₂

Explanation:

The limiting reactant is the one that gives the smaller amount of product.

Assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r: 28.02 2.02 17.04

N₂ + 3H₂ ⟶ 2NH₃

m/g: 5.42 5.42

1. Calculate the moles of each reactant

$\text{Moles of N}_(2) = \text{5.42 g} * \frac{\text{1 mol}}{\text{28.02 g}} = \text{0.1934 mol N}_(2)\n\n\text{Moles of H}_(2) = \text{5.42 g} * \frac{\text{1 mol}}{\text{2.02 g}} = \text{2.683 mol H}_(2)$

2. Identify the limiting reactant

Calculate the moles of NH₃ we can obtain from each reactant.

From N₂:

The molar ratio of NH₃:N₂ is 2:1.

$\text{Moles of NH}_(3) = \text{0.1934 mol N}_(2) * \frac{\text{2 mol NH}_(3)}{\text{1 mol N}_(2)} = \text{0.3867 mol NH}_(3)$

From H₂:

The molar ratio of NH₃:H₂ is 3:2.

$\text{Moles of NH}_(3) = \text{2.683 mol H}_(2) * \frac{\text{2 mol NH}_(3)}{\text{3 mol N}_(2)} = \text{1.789 mol NH}_(3)$

N₂ is the limiting reactant, because it gives the smaller amount of NH₃.

C,H,O,+ 3HCI → C,HCl, + 3H20
right 5 possible mole ratios

Answers

Answer:

Here's what I get

Explanation:

I think this may be the equation you intended to write:

C₃H₅(OH)₃ + 3HCl ⟶ C₃H₅Cl₃ + 3H₂O

The mole ratios are the ratios of the coefficients in the balanced equation.

Here are some of the possible molar ratios.

C₃H₅(OH)₃:HCl = 1:3
C₃H₅(OH)₃:C₃H₅Cl₃ = 1:1
C₃H₅(OH)₃:H₂O = 1:3
HCl:C₃H₅Cl₃ = 3:1
HCl:H₂O = 3:3
C₃H₅Cl₃:H₂O = 1:3

when an iron nail is placed in an aqueous solution of copper(ii)sulphate,a reaction take place.construct an ionic equation for reaction?

Answers

The net ionic equation for this reaction is $Fe(s) \ + \ Cu^(2+)(aq) \ \rightarrow \ Fe^(2+)(aq) \ + \ Cu(s)$ . In ionic equations the soluble compounds are written in their dissociated ion forms. The ionic equation can be determined by knowing the general solubility rules. In this problem there are sulfate compounds involved. It is important to know that sulfates are generally soluble except for sulfates of Ba, Pb, Ag, and Sr. Therefore, the ones dealt with in this problem, CuSO4 and FeSO4, are both soluble in water.

Further Explanation:

The complete equation for this reaction is:

$Fe(s) \ + \ CuSO_(4)(aq) \rightarrow \ FeSO_(4)(aq) \ + \ Cu(s)$

CuSO4 and FeSO4 are soluble in water which will lead to the ionic equation:

$Fe(s) \ + \ Cu^(2+)(aq) \ + \ SO_(4)^ {2-}(aq) \ \rightarrow \ Fe^(2+)(aq) \ + \ SO_(4)^(2-)(aq) \ + \ Cu(s)$

Since the sulfate ion appears on both sides, it is a spectator ion. Spectator ions are unchanged in the chemical reaction and are removed from the net ionic equation:

$\boxed {Fe(s) \ + \ Cu^(2+)(aq) \ \rightarrow \ Fe^(2+)(aq) \ + \ Cu(s)}$

Based on the net ionic equation it can be seen that this is a reduction-oxidation reaction. Iron was oxidized and copper was reduced.

Learn More

Learn more about solubility rules brainly.com/question/11617954
Learn more about redox reactions brainly.com/question/2388654
Learn more about chemical equations brainly.com/question/6976143

Keywords: ionic equation, solubility

What is the volume of a liquid with a density of 5.45g/ml and a mass of 65g

Answers

you have to use the fact that d=m/v and solve for v (v=m/d). with that equation you just divide the mass of the liquid by the density of the liquid. When done correctly you should get 11.93 mL of liquid.

I hope this helps.

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