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ss7ja [257]
3 years ago
5

Which bond is broken when energy is released?

Chemistry
2 answers:
Tanya [424]3 years ago
7 0
Energy can be released or absorbed when various chemical bonds form or are broken. Exothermic reactions release energy in the form of heat whereas endothermic reactions absorb energy.
blsea [12.9K]3 years ago
4 0
Energy is released when a new bonds form.
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Which of the following is a possible set of quantum numbers for an electron (n,l,m0,ms) ?
Mashutka [201]

Answer is: (3, 2, 0, -1/2).

The principal quantum number (n) is one of four quantum numbers which are assigned to each electron in an atom to describe that electron's state.

For principal quantum number n=3:  

1) azimuthal quantum number (l) can be l = 0...n-1:  

l = 0, 1, 2.  

The azimuthal quantum number determines its orbital angular momentum and describes the shape of the orbital.  

2) magnetic quantum number (ml) can be ml = -l...+l.  

ml = -2, -1, 0, +1, +2.

Magnetic quantum number specify orientation of electrons in magnetic field and number of electron states (orbitals) in subshells.  

3) the spin quantum number (ms), is the spin of the electron.  

ms = +1/2, -1/2.  

(1, 1, 0, +1/2)  is not correct because orbital quantum number cannot be l = 1 for n = 1.

(2, 1, 2, +1/2)  is not correct because magnetic quantum number cannot be ml = 2 for orbital quantum number l = 1.

(3, -2, 1, -1/2) is not correct because orbital quantum number cannot be l = -2 for principal quantum number n = 3.

5 0
3 years ago
Astrology, a belief in a relationship between constellations of stars and life events,
yanalaym [24]

Answer:

Pseudoscience

Explanation:

"Astrology has not demonstrated its effectiveness in controlled studies and has no scientific validity, thus regarded as pseudoscience."

3 0
3 years ago
How does a magnet separate mixtures like sand and iron filings?​
algol13

Answer:

because iron is magnetic and sand is not

Explanation:

6 0
3 years ago
Read 2 more answers
Which of the following statements is TRUE? (A) An ionic bond is much stronger than most covalent bonds.(B) Once dissolved in wat
Ainat [17]

Answer:

A)  An ionic bond is much stronger than most covalent bonds.

Explanation:

D) Ionic compounds have high melting points causing them to be solid at room temperature, and conduct electricity when dissolved in water. Covalent compounds have low melting points and many are liquids or gases at room temperature.

C) An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions. Ionic bonds are formed between a cation, which is usually a metal, and an anion, which is usually a nonmetal. A covalent bond involves a pair of electrons being shared between atoms.

A) Covalent bonds are stronger if you compare with ionic molecules, because their molecular orbital overlap is bigger. However, ionic molecules form lattices, thus the energy to break this lattice bond is stronger hence the ionic bond is stronger.

4 0
3 years ago
A 50.0 mL solution of 0.129 M KOH is titrated with 0.258 M HCl. Calculate the pH of the solution after the addition of each of t
kobusy [5.1K]

Answer:

A- pH = 13.12

B- pH = 12.91

C- pH = 12.71

D- pH = 12.43

E- pH = 11.55

F- pH = 7

G- pH = 2.46

H- pH = 1.88

Explanation:

This is a titration of a strong base with a strong acid. The neutralization reaction is: KOH (aq) + HCl (aq) →  H₂O(l) + KCl(aq)

Our pH at the equivalence point is 7, because we have made a neutral salt.

To determine the volume at that point we state the formula for titration:

mmoles of base = mmoles of acid

Volume of base  . M of base = Volume of acid . M of acid

50mL . 0.129M = 0.258 M . Volume of acid

Volume of acid = (50mL . 0.129M) / 0.258 M →  25 mL (Point <u>F</u>)

When we add 25 mL of HCl, our pH will be 7.

A- At 0 mL of acid, we only have base.

KOH → K⁺ + OH⁻

[OH⁻] = 0.129 M

To make more easy the operations we will use, mmol.

mol . 1000 = mmoles → mmoles / mL = M

- log 0.129 = 0.889

14 - 0.889 = 13.12

B-  In this case we are adding, (7 mL . 0.258M) = 1.81 mmoles of H⁺

Initially we have  0.129 M . 50 mL = 6.45 mmoles of OH⁻

1.81 mmoles of H⁺ will neutralize, the 6.45 mmoles of OH⁻ so:

6.45 mmol - 1.81 = 4.64 mmoles of OH⁻

This mmoles of OH⁻ are not at 50 mL anymore, because our volume has changed. (Now, we have 50 mL of base + 7 mL of acid) = 57 mL of total volume.

[OH⁻] = 4.64 mmoles / 57 mL = 0.0815 M

- log 0.0815 M = 1.09 → pOH

pH = 14 - pOH → 14 - 1.09 = 12.91

C- In this case we add (12.5 mL . 0.258M) = 3.22 mmoles of H⁺

<em>Our initial mmoles of OH⁻ would not change through all the titration. </em>

Then 6.45 mmoles of OH⁻ are neutralized by 3.22 mmoles of H⁺.

6.45 mmoles of OH⁻ - 3.22 mmoles of H⁺ = 3.23 mmoles of OH⁻

Total volume is: 50 mL of base + 12.5 mL = 62.5 mL

[OH⁻] = 3.23 mmol / 62.5 mL = 0.0517 M

- log  0.0517 = 1.29 → pOH

14 - 1.11 = 12.71

D- We add (18 mL . 0.258M) = 4.64 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 4.64 mmoles of H⁺.

6.45 mmoles of OH⁻ - 4.64 mmoles of H⁺ = 1.81 mmoles of OH⁻

Total volume is: 50 mL of base + 18 mL = 68 mL

[OH⁻] = 1.81 mmol / 68 mL = 0.0265 M

- log  0.0265 = 1.57 → pOH

14 - 1.57 = 12.43

E- We add (24 mL . 0.258M) = 6.19 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 6.19 mmoles of H⁺.

6.45 mmoles of OH⁻ - 6.19 mmoles of H⁺ = 0.26 mmoles of OH⁻

Total volume is: 50 mL of base + 24 mL = 74 mL

[OH⁻] = 0.26 mmol / 74 mL = 3.51×10⁻³ M

- log  3.51×10⁻³  = 2.45 → pOH

14 - 2.45 = 11.55

F- This the equivalence point.

mmoles of OH⁻ = mmoles of H⁺

We add (25 mL . 0.258M) = 6.45 mmoles of H⁺

All the OH⁻ are neutralized.

OH⁻  +  H⁺  ⇄   H₂O              Kw

[OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷  →  pOH = 7

pH → 14 - 7 = 7

G- In this case we have an excess of H⁻

We add (26 mL . 0.258M ) = 6.71 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

6.71 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 0.26 mmoles of H⁺

[H⁺] = 0.26 mmol / Total volume

Total volume is: 50 mL + 26 mL → 76 mL

[H⁺] = 0.26 mmol / 76 mL → 3.42×10⁻³ M

- log 3.42×10⁻³ = 2.46 → pH

H- Now we add (29 mL . 0.258M) = 7.48 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

7.48 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 1.03 mmoles of protons

Total volume is 50 mL + 29 mL = 79 mL

[H⁺] = 1.03 mmol / 79 mL → 0.0130 M

- log 0.0130 = 1.88 → pH

After equivalence point, pH will be totally acid, because we always have an excess of protons. Before the equivalence point, pH is basic, because we still have OH⁻ and these hydroxides, will be neutralized through the titration, as we add acid.

5 0
3 years ago
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