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jok3333 [9.3K]
3 years ago
6

For this question, the "entropy term" refers to "-TΔS". Addition reactions are generally favorable at low temperatures because _

_______.
Chemistry
1 answer:
nata0808 [166]3 years ago
8 0

Answer:

Lowering the temperature typically reduces the significance of the decrease in entropy. That makes the Gibbs Free energy of the reaction more negative. As a result, the reaction becomes more favorable overall.  

Explanation:

In an addition reaction there's a decrease in the number of particles. Consider the hydrogenation of ethene as an example.

\rm H_2C\text{=}CH_2\; (g) + H_2\; (g) \stackrel{\text{Ni}^\ast}{\to} H_3C\text{-}CH_3\; (g).

When \rm H_2 is added to \rm H_2C\text{=}CH_2 (ethene) under heat and with the presence of a catalyst, \rm H_3C\text{-}CH3 (ethane) would be produced.

Note that on the left-hand side of the equation, there are two gaseous molecules. However, on the right-hand side there's only one gaseous molecule. That's a significant decrease in entropy. In other words, \Delta S < 0.

The equation for the change in Gibbs Free Energy for a particular reaction is:

\Delta G = \Delta H + (\underbrace{- T \, \Delta S}_{\text{entropy}\atop \text{term}}).

For a particular reaction, the more negative \Delta G is, the more spontaneous ("favorable") the reaction would be.

Since typically \Delta S < 0 for addition reactions, the "entropy term" of it would be positive. That's not very helpful if the reaction needs to be favorable.

T (absolute temperature) is always nonnegative. However, lowering the temperature could help bring the value of

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How do you write 3.40 x 10 to the eighth power in standard form?
solong [7]

Answer:

272

Explanation:

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3 years ago
1.42 mol sample of neon gas at a temperature of 13.0 °C is found to occupy a volume of 25.5 liters. The pressure of this gas sam
Dima020 [189]

Answer: 996 mmHg

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

According to the ideal gas equation:

PV=nRT

P = Pressure of the gas = ?

V= Volume of the gas = 25.5 L

T= Temperature of the gas = 13°C = (273+13) K  = 286K

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= 1.42

P=\frac{nRT}{V}=\frac{1.42\times 0.0821\times 286}{25.5}=1.31atm=996mmHg      (760mmHg=1atm)

Thus pressure of this gas sample is 996 mm Hg.

3 0
3 years ago
51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

where;

r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0
3 years ago
How many atoms are in 175.8 grams of Hg?
dybincka [34]
5.22*22^3 should be the answer
5 0
3 years ago
A saline solution with a mass of 400 g has 30 g of NaCl dissolved in it. What is the mass/mass percent concentration of the solu
maria [59]
(g solute/g solution)*100 = % mass/mass

30 g / 400 * 100 

0,075 * 100

= 7,5% w/w

hope this helps!

3 0
3 years ago
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