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amm1812
3 years ago
11

Molecules can be constructed by combining atoms of the same element. True or false?

Chemistry
2 answers:
earnstyle [38]3 years ago
6 0
True

hope that helps :)
Vera_Pavlovna [14]3 years ago
5 0
By deffinition a molecule is a more or less stable grouping of two or more atoms held together by a chemical bond; they may be the same element.
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What is the change in internal energy for each of the following situations? 1. q-7.9 J out of the system and w 3.6 J done on the
adoni [48]

Answer:

A i. Internal energy ΔU = -4.3 J ii. Internal energy ΔU = -6.0 J B. The second system is lower in energy.

Explanation:

A. We know that the internal energy,ΔU = q + w where q = quantity of heat and w = work done on system.

1. In the above q = -7.9 J (the negative indicating heat loss by the system). w = 3.6 J (It is positive because work is done on the system). So, the internal energy for this system is ΔU₁ = q + w = -7.9J + 3.6J = -4.3 J

ii. From the question q = +1.5 J (the positive indicating heat into the system). w = -7.5 J (It is negative because work is done by the system). So, the internal energy for this system is ΔU₂ = q + w = +1.5J + (-7.5J) = +1.5J - 7.5J = - 6.0J

B. We know that ΔU = U₂ - U₁ where U₁ and U₂ are the initial and final internal energies of the system. Since for the systems above, the initial internal energies U₁ are the same, then we say U₁ = U. Let U₁ and U₂ now represent the final energies of both systems in A i and A ii above. So, we write ΔU₁ = U₁ - U and ΔU₂ = U₂ - U where ΔU₁ and ΔU₂ are the internal energy changes in A i and A ii respectively. Now from ΔU₁ = U₁ - U, U₁ = ΔU₁ + U and U₂ = ΔU₂ + U. Subtracting both equations U₁ - U₂ = ΔU₁ - ΔU₂

= -4.3J -(-6.0 J)= 1.7 J. Since U₁ - U₂ > 0 , U₂ < U₁ , so the second system's internal energy increase less and is lower in energy and is more stable.

8 0
3 years ago
Rank these acids according to their expected pKa values.ClCH2COOHClCH2CH2COOHCH3CH2COOHCl2CHCOOHIn order of highest pka to lowes
Ilia_Sergeevich [38]

Answer:

CH₃CH₂CH₂COOH > CH₃CH₂COOH > ClCH₂CH₂COOH  > ClCH₂COOH

Explanation:

Electron-withdrawing groups (EWGs) increase acidity by inductive removal of electrons from the carboxyl group.

Electron-donating groups (EDGs) decrease acidity by inductive donation of electrons to the carboxyl group.

  • The closer the substituent is to the carboxyl group, the greater is its effect.
  • The more substituents, the greater the effect.  
  • The effect tails off rapidly and is almost zero after about three C-C bonds.

CH₃CH₂-CH₂COOH —  EDG —                         weakest —  pKₐ = 4.82

      CH₃-CH₂COOH — reference —                                     pKₐ = 4.75

  ClCH₂-CH₂COOH — EWG on β-carbon— stronger —     pKₐ = 4.00

           ClCH₂COOH — EWG on α-carbon — strongest —  pKₐ = 2.87

6 0
3 years ago
Read 2 more answers
For which one of the following is ΔHfo zero?
Vlada [557]
The answer is most likely C
6 0
3 years ago
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When two your substances are combined so that each of the pure substances retains its own properties the result is a(n)?A. mixtu
goblinko [34]
Mixture is the answer 
8 0
3 years ago
Strike anywhere matches contain the compound tetraphosphorus trisulfide, which burns to form tetraphosphorus decaoxide and sulfu
erica [24]

Answer:

194.6 mL of SO₂

Explanation:

The reaction that takes place is:

P₄S₃ + 6O₂(g) → P₄O₁₀ + 3SO₂(g)

<u>To solve this problem we need to use PV=nRT</u>, so first let's convert the given units:

  • 23.8 °C → 23.8 + 273.15 = 296.95 K
  • 747 torr → 747/760 = 0.983 atm

We need to calculate V, so in order to do that we calculate n, using the mass of the reactant (P₄S₃):

0.576 g P₄S₃ * \frac{1molP_{4}S_{3}}{220gP_{4}S_{3}} *\frac{3molSO_{2}}{1molP_{4}S_{3}} = 7.85 * 10⁻³ mol SO₂ = n

  • Now we calculate V:

PV=nRT

0.983 atm * V =  7.85 * 10⁻³ mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.95 K

V = 0.1946 L

  • Finally we convert L into mL:

0.1946 * 1000 = 194.6 mL

8 0
3 years ago
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