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3 years ago

When the temperature of a substance increases, do the intermolecular forces change? why or why not?

Chemistry

1 answer:

xeze [42]3 years ago

5 0

Answer:

If the temperature rises much more, the individual particles will have enough energy to overpower the intermolecular forces, causing the particles to disperse and the material to become a gas (assuming that their chemical bonds are not so weak that the compound decomposes from the high temperature).

Explanation:

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Answer:

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Explanation:

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Catalyst

Hoochie [10]

a. volume of NO : 41.785 L

b. mass of H2O : 18 g

c. volume of O2 : 9.52 L

<h3>Further explanation</h3>

Given

Reaction

4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Required

a. volume of NO

b. mass of H2O

c. volume of O2

Solution

Assume reactants at STP(0 C, 1 atm)

Products at 1000 C (1273 K)and 1 atm

a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

$\tt \dfrac{4}{5}\times 0.5=0.4$

volume NO at 1273 K and 1 atm

$\tt V=\dfrac{nRT}{P}=\dfrac{0.4\times 0.08206\times 1273}{1}=41.785~L$

b. 15 L NH3 at STP ( 1mol = 22.4 L)

$\tt \dfrac{15}{22.4}=0.67~mol$

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

$\tt \dfrac{6}{4}\times 0.67=1$

mass H2O(MW = 18 g/mol) :

$\tt mass=mol\times MW=1\times 18=18~g$

c. mol NO at 1273 K and 1 atm :

$\tt n=\dfrac{PV}{RT}=\dfrac{1\times 35.5}{0.08206\times 1273}=0.34$

mol ratio of NO : O2 = 4 : 5, so mol O2 :

$\tt \dfrac{5}{4}\times 0.34=0.425$

Volume O2 at STP :

$\tt 0.425\times 22.4=9.52~L$

5 0

3 years ago

For double-helix formation, DG can be measured to be 2 54 kJ mol 1 ( 2 13 kcal mol 1 ) at pH 7.0 in 1 M NaCl at 25 8C (298 K). T

levacccp [35]

Answer:

ΔS = -661.0J/mol is the entropy change for the system

ΔS = -842J/mol.K is the entropy change for the surroundings

Explanation:

From the relationship between ΔG, T, ΔH and ΔS,

Mathematically, ΔG = ΔH - TΔS

TΔS = ΔH - ΔS

ΔS = ΔH - ΔS / T

but ΔG = -54 kJ/mol, ΔH = -251 kJ/mol and T = 25 °C (298 K)

plugging into the equation,

ΔS = -251 kJ/mol - ( -54 kJ/mol) / 298

ΔS = -0.6610KJ/mol or in J.mol

ΔS = -661.0J/mol is the entropy change for the system

For entropy change for the surroundings = ΔS = ΔH/T

ΔS = -251 kJ/mol / 298K

ΔS = -0.84KJ/mol.K or -842J/mol.K is the entropy change for the surroundings

8 0

3 years ago