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2 years ago

Δs is negative for the reaction __________.

1 answer:

6 0

ΔS =S(products) -S(reactants)

Where ΔS is the change of entropy in a reactions

a. ΔS = (2) - (2+1) = -1
b. ΔS = (1+1) -(1) = 1
c. ΔS = (1+2) - (1) = 2
d. ΔS = (2) - (2+1) = -1
e. ΔS = (1) - (1) = 0

ΔS is negative for reaction a. and d.

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PA to PB 100 pm to the left of the nucleus, along the -x axis.
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PAPB 100 pm in front of the nucleus, along the -y axis. 100 pm behind the nucleus, along the +y axis.

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2 years ago

Consider these generic half-reactions. Half-reaction E° (V) X+(aq)+e−⟶X(s) 1.52 Y2+(aq)+2e−⟶Y(s) −1.17 Z3+(aq)+3e−⟶Z(s) 0.84 Ide

olya-2409 [2.1K]

Answer:

strongest oxidizing agent: $X^{+}$

weakest oxidizing agent: $Y^{2+}$

strongest reducing agent: Y

weakest reducing agent: X

$X^{+}$ will oxidize Z

Explanation:

The higher the reduction potential of a species, higher will be the tendency to consume electrons from another species. Hence higher will be the oxidizing power of it's oxidized form and lower will be the reducing power of it's reduced form.

Alternatively, higher reduction potential value suggests that the oxidized form of the species acts as a stronger oxidizing agent and the reduced form of the species acts as a weaker reducing agent.

Order of reduction potential:

$E_{X^{+}\mid X}^{0}(1.52V)> E_{Z^{3+}\mid Z}^{0}(0.84V)> E_{Y^{2+}\mid Y}^{0}(-1.17V)$

So, strongest oxidizing agent: $X^{+}$

weakest oxidizing agent: $Y^{2+}$

strongest reducing agent: Y

weakest reducing agent: X

As reduction potential of the half cell $X^{+}\mid X$ is higher than the reduction potential of the half cell $Z^{3+}\mid Z$ therefore $X^{+}$ will oxidize Z into $Z^{3+}$ and itself gets converted into X.

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2 years ago

Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in

ira [324]

Explanation:

The given data is as follows.

m = 10.0 kg = 10,000 g (as 1 kg = 1000 g)

Initial temp. of block 1, $T_{1} = 100^{o}C$ = (100 + 273) K = 373 K

Initial temp. of block 2, $T_{2} = 0^{o}C$ = (0 + 273) K = 273 K

So, heat released by block 1 = heat gained by block 2

$mC \Delta T = mC \times \Delta T$

$10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C$

$T_{f} - 100^{o}C = 0^{o}C - T_{f}$

$2T_{f} = 100^{o}C$

$T_{f} = 50^{o}C$

Convert temperature into kelvin as (50 + 273) K = 323 K.

Also, we know that the relation between enthalpy and temperature change is as follows.

$\Delta H = mC \Delta T$

= $10000 g \times 0.385 J/K g \times 323 K$

= 1243550 J

or, = 1243.5 kJ

Now, calculate entropy change for block 1 as follows.

$\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{373}$

= $10000 g \times 0.385 J/K g \times -0.143$

= -554.12 J/K

Now, entropy change for block 2 is as follows.

$\Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{273}$

= $10000 g \times 0.385 J/K g \times 0.168$

= 647.49 J/K

Hence, total entropy will be sum of entropy change of both the blocks.

$\Delta S_{total} = \Delta S_{1} + \Delta S_{2}$

= -554.12 J/K + 647.49 J/K

= 93.37 J/K

Thus, we can conclude that for the given reaction $\Delta H$ is 1243.5 kJ and $\Delta S_{total}$ is 93.37 J/K.

6 0

2 years ago

To find the Ce4+ content in a solid sample, 4.3718 g of the solid sample were dissolved and treated with excess iodate to precip

gulaghasi [49]

Answer:

3.43 %

Explanation:

We need to calculate first the number of moles of CeO2 produced in the combustion. Given its formula we know how many moles of Ce atom are present. From there calculate the mass this number of moles this represent and then one can calculate the percentage.

0.1848 g CeO2 x 1 mol CeO2/172.114g = 0.00107 mol CeO2

0.00107 mol CeO2 x 1 mol Ce/ 1 mol CeO2 = 0.00107 mol Ce

.00107 mol Ce x 140.116 g Ce/ mol = 0.150 g Ce

0.150 g Ce/ 4.3718 g sample x 100 = 3.43 %

5 0

2 years ago

Read 2 more answers

Fe3+(aq)+6H2O(l)⇌Fe(H2O)63+(aq) : F e 3 + ( a q ) + 6 H 2 O ( l ) ⇌ F e ( H 2 O ) 6 3 + ( a q ) : blank is the Lewis acid and bl

Tasya [4]

Answer:

Lewis acid- Fe3+

Lewis base- water molecule

Explanation:

Acids and bases have been defined in diverse ways. There have been definitions put forward by Arrhenius, Brownstead and Lowry as well as Lewis. Each definition his useful in its own way.

Lewis acids are lone pair acceptors such as metal ions. This implies that in the particular instance of this reaction, Fe3+ is the lewis acid.

Similarly, a Lewis base is a lone pair donor, all ligands are lone pair donors since they donate one or more lone pairs of electrons to Lewis acids. In the particular instance of this reaction, the Lewis base is the water molecule.

6 0

2 years ago