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Salsk061 [2.6K]
2 years ago
13

Atom , classify ithelppoppppoeowiie​

Chemistry
1 answer:
timama [110]2 years ago
5 0

Answer:

A atom is a small unit to which matter can be divided without releasing electrically charged particles

Explanation:

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What is the difference between ionization energy and ionization potential​
rjkz [21]

Answer:

They are similarly

Explanation:

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Deval drew the models of particles in a substance shown below. Which model best represents the particles in solid
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it would be the second one thank me later

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2 years ago
Identify the weak diprotic acid. identify the weak diprotic acid. h2so4 hcooh
slega [8]
Among formic acid (HCOOH ) and sulfuric acid (H₂SO₄), formic acid is the weak acid. Acidic strength of any acid is the tendency of that acid to loose proton. Among these two acids formic acid has a pKa value of 3.74 greater than that of sulfuric acid i.e. -10. Remember! Greater the pKa value of acid weaker is that acid and vice versa. Below I have drawn the Ionization of both acids to corresponding conjugate bases and protons. The structures below with charges are drawn in order to explain the reason for strength. As it is seen in charged structure of formic acid, there is one positive charge on carbon next to oxygen carrying proton. The electron density is shifted toward carbon as it is electron deficient and demands more electron hence, attracting electron density from oxygen and making the oxygen hydrogen bond more polar. While, in case of sulfuric acid it is depicted that Sulfur attached to oxygen containing proton has 2+ charge, means more electron deficient as compared to carbon of formic acid, hence, more electron demanding and strongly attracting electrons from oxygen and making the oxygen hydrogen bond very polar and highly ionizable.

7 0
3 years ago
What is the identity of the atom shown?
Charra [1.4K]
The answer is a , as aluminium has 13 protons and electrons .
6 0
3 years ago
Read 2 more answers
(b) Use the first law of thermodynamic to calculate AU for the following situations: (i) A coiled spring unwinds producing 153 J
il63 [147K]

Answer:

(i) ΔU = 116 J

(ii) ΔU = 289 J

(iii) ΔU = 1 KJ

(iv) ΔU = 0 J

(v) ΔU = 3.25 KJ

Explanation:

first law:

  • ΔU = Q + W

(i) W = 153 J;  Q = - 37 J ( Q ( - ), losing friction )

⇒ ΔU = 153 - 37 = 116 J

(ii) W = 289 J; Q = 0 ( insulated)

⇒ ΔU = W = 289 J

(iii) Q = 1 KJ , W = 0 ( isovolumetric process)

⇒ ΔU = Q = 1 KJ

(iv) isothermal ( constant temperature )

  • ΔU = Cv * ΔT

∴ ΔT = 0° ( isothermal )

⇒ ΔU = 0 J

(v) isobaric ( constant pressure )

⇒ ΔU = Q + W

∴ Q = 15.6 KJ

∴ W = - ∫ P dV = - P ΔV;  W (-) the system performs a job and the volume increases

.

∴ P = 950 KPa * ( 1000 Pa / KPa ) = 950000 Pa = 950000 J/m³

∴ ΔV = 18 - 5 = 13 L * ( m³ / 1000 L ) = 0.013 m³

⇒ W = - ( 950000 J/m³) * ( 0.013 m³ ) = - 12350 J ( - 12.35 KJ )

⇒ ΔU = 15.6 KJ + ( - 12.35 KJ )

⇒ ΔU = 3.25 KJ

7 0
3 years ago
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