Question 3 (1 point)

An ionic bond can be formed when one or more electrons are 1) equally shared by two atoms 2) unequally shared by two atoms 3) tr

Lyrx [107]

Answer:

4) transferred from the valence shell of one atom to the valence shell of another atom

Explanation:

Electrons are located outside of the nucleus which contains the protons and the neutrons.

For bonds to form, valence electrons located in the outermost shell electrons are involved. These are the valence electrons. These outer shell electrons can be shared or transferred between two combining atoms to form stable atoms.

In ionic bonds, the electrons are transferred from one specie to another. The atom that loses the electrons becomes positively charged and the receiving atom becomes negatively charged. This is the crux of ionic bonds.

6 0

3 years ago

What are the different acid bases

Alex

Answer:

different acids include acetic acid,lactic acid,tartaric acids DNA(deoxyribonucliec acid) .

different bases include calcium oxide Potassium hydroxide ,Sodium hydroxide and Barium hydroxide .

5 0

3 years ago

Which of the following is an essential condition for a redox reaction?

rusak2 [61]

The oxidation of at least two atoms should change

8 0

3 years ago

Read 2 more answers

What is the pH of a 0.640 M solution of C₅H₅NHBr (Kb of C₅H₅N is 1.7 × 10⁻⁹)?

Elden [556K]

The pH of a 0.64 M solution of pyridine (C₅H₅N) is 9.52.

A figure expressing the acidity or alkalinity of a solution on a logarithmic scale on which 7 is neutral, lower values are more acid and higher values more alkaline.

The equation for the protonation of the base pyridine is the following:

C₅H₅N + H₂O ⇄ C₅H₅NH⁺ + OH⁻ (1)

Kb = 1.7 × 10⁻⁹ (Given)

To calculate the pH of the solution we need to use the following equation:

pH + pOH = 14

=14 - [-log[OH⁻]]

= 14 + log[OH⁻]

Now, we need to find the concentration of the OH⁻ ions. Since pyridine is a weak base, at the equilibrium we have (eq 1):

C₅H₅N + H₂O ⇄ C₅H₅NH⁺ + OH⁻

0.64 - x x x

After entering the values of [C₅H₅N] = 0.64-x, [C₅H₅NH⁺] = x, and [OH⁻] = x, into equation (2) we can find the concentration of OH⁻:

1.7 × 10⁻⁹ =[C₅H₅NH⁺] [OH⁻] / [C₅H₅N]

= x . x / 0.64-x

1.7 × 10⁻⁹ (0.64-x) - x² = 0

Solving the above quadratic equation for x, we have :

x₁ = -3.32 x 10⁻⁵
x₂ = 3.32 x 10⁻⁵

Now, We can calculate the pH, after taking the positive value, x₂, (concentrations cannot be negative) and entering into above equation :

<em />

<em>pH = </em>14 + log[OH⁻]

= 14 + log (3.32 x 10⁻⁵)

= 9.52

Therefore, the pH of the solution of pyridine is 9.52.

Find more about pH here:

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3 0

2 years ago

How many moles of sodium carbonate are contained by 57.3g of sodium carbonate

Lady_Fox [76]

Answer:

$\boxed {\boxed {\sf 0.541 \ mol \ Na_2CO_3}}$

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

Na: 22.9897693 g/mol
C: 12.011 g/mol
O: 15.999 g/mol

Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

Na₂ = 22.9897693 * 2= 45.9795386 g/mol
O₃ = 15.999 * 3= 47.997 g/mol
Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol

We will convert using dimensional analysis. Set up a ratio using the molar mass.

$\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

We are converting 57.3 grams to moles, so we multiply by this value.

$57.3 \ g \ Na_2CO_3} *\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

Flip the ratio so the units of grams of sodium carbonate cancel.

$57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386 \ g \ Na_2CO_3}$

$57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }$

$\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3$

$0.5406295944 \ mol \ Na_2CO_3$

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

$0.541 \ mol \ Na_2CO_3$

There are approximately <u>0.541 moles of sodium carbonate</u> in 57.3 grams.

6 0

2 years ago