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nadezda [96]
3 years ago
14

Assuming constant pressure, rank these reactions from most energy released by the system to most energy absorbed by the system,

based on the following descriptions:
A. Surroundings get colder and the system decreases in volume.
B. Surroundings get hotter and the system expands in volume.
C.Surroundings get hotter and the system decreases in volume.
D. Surroundings get hotter and the system does not change in volume.
Also assume that the magnitude of the volume and temperature changes are similar among the reactions.
Rank from most energy released to most energy absorbed. To rank items as equivalent, overlap them.
Chemistry
1 answer:
gayaneshka [121]3 years ago
3 0

Answer:

B > D > C > A

Explanation:

For the first law of the thermodynamics, the total energy variation in a process is:

ΔU = Q - W

Where Q is the heat, and W the work. If the system loses heat, Q < 0, if it absorbs heat, Q>0. If work is done in the system (volume decreases), W < 0, if the system does the work (volume increases), W > 0.

A. If the surroundings get colder, the system is absorbing heat, so Q>0, and the system decreases in volume so W < 0 :

ΔU = +Q - (-W) = +Q + W (absorbs a higher energy)

B. If the surroundings ger hotter, the system is losing heat, so Q<0, and the system expands, so W>0:

ΔU = -Q -W (loses higher energy)

C. Surroundings get hotter, Q<0, and the system decreases in volume, W<0

ΔU = - Q + W = 0 (magnitude of heat and work is similar)

D. Surroundings get hotter, Q<0, and the system is not changing in volume, W = 0.

ΔU = -Q (loses energy)

For the most released (more negative) for the most absorbed (most positive):

B > D > C > A

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3 years ago
if 3.0 grams of aluminum and 6.0 grams of bromine react to form AlBr3, how many grams of product would theoretically be produced
Gelneren [198K]
1) Chemical reaction: 2Al + 3Br₂ → 2AlBr₃.
m(Al) = 3,0 g.
m(Br₂) = 6,0 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 3,0 g ÷ 27 g/mol.
n(Al) = 0,11 mol.
n(Br₂) = n(Br₂) ÷ m(Br₂).
n(Br₂) = 6 g ÷ 160 g/mol.
n(Br₂) = 0,0375 mol; limiting reagens.
n(Br₂) : n(AlBr₃) = 3 : 2.
n(AlBr₃) = 0,025 mol.
m(AlBr₃) = 0,025 mol · 266,7 g/mol.
m(AlBr₃) = 6,67 g.

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4 0
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Enter your answer in the provided box. what is the emf of a cell consisting of a pb2 / pb half-cell and a pt / h / h2 half-cell
sp2606 [1]

Answer:

The emf of the electrochemical cell has been calculated to be <u>3.364.</u>

in order to calculate the emf we need to apply Nernst equation.

The emf has been the potential of the cell in the reaction with the change in the electrons in the reaction. The emf of the cell has been given by the Nernst equation as -

                            emf= E⁰ cell - 0.059/n. log 1/concentration

Computation for the emf of the cell

  The given cell has the number of electrons transfer, n= 2

  The concentration of   Pb2= 0.57 M

  The concentration of   H= 0.090 M

The cell potential of the reaction has been= -0.126

Substituting the values for the emf of the cell:

                  emf= -0.126- 0.059/2 * log 1/ [0.57]x [0.090] ²

                  emf= 1.555* 2.337

                  emf=3.634 v

The emf of the electrochemical cell has been calculated to be 3.634 v.

Learn more about emf here-

brainly.com/question/9425530

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(carb weigh/total)×100
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