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ivann1987 [24]
2 years ago
13

Which statement about spontaneous and nonspontaneous processes is correct? Spontaneous processes are favored by a decrease in H,

but nonspontaneous reactions are favored by an increase in H. Spontaneous processes are favored by a decrease in S, but nonspontaneous reactions are favored by an increase in S. An increase in S favors both spontaneous and nonspontaneous processes. A decrease in H favors both spontaneous and nonspontaneous processes.
Chemistry
2 answers:
Marina86 [1]2 years ago
7 0
I believe your answer would be the first one

hope this helps
Sav [38]2 years ago
5 0

Answer: Spontaneous processes are favored by a decrease in H, but nonspontaneous reactions are favored by an increase in H.

Explanation:

\Delta G=\Delta H-T\Delta S

\Delta G = gibbs free energy

\Delta H = change in enthalpy

\Delta S = change in entropy

For a reaction to be spontaneous, \Delta G= -ve. For this \Delta H  has to be negative i.e. there should be a decrease in enthalpy and \Delta S has to be positive i.e. there should be a increase in entropy.

\Delta G=(-ve)-T(+ve)=-ve-ve=-ve


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Iodine has a lower atomic weight than tellurium (126.90 for I, 127.60 for Te) even though it has a higher atomic number (53 for
andrew11 [14]

Answer and Explanation:

Iodine have lower atomic mass than tellurium even though the atomic number of iodine is more than the atomic number of tellurium

This is because the atomic weight of any element is the sum of number of proton and number of neutron, even though the number of proton in iodine is more so but the number of neutron is less as compared to tellurium which makes the tellurium of high atomic mass

6 0
2 years ago
Read 2 more answers
What is the molarity of a naoh solution if 11.9 ml of a 0.220 m h2so4 solution is required to neutralize a 25.0-ml sample of the
bulgar [2K]
0.355M x 0.0282L= 0.01 moles of H2SO4. Remember sulphuric acid is diprotic so it will release 2 from each molecule. 
<span>So moles of protons = 0.01 x 2 = 0.02 moles of H+ </span>
<span>For neutralization: moles H+ = moles OH- </span>
<span>Therefore moles of NaOH = 0.02 </span>
<span>conc = moles / volume </span>
<span>Conc NaOH = 0.02 / 0.025L = 0.8M </span>

4 0
2 years ago
Read 2 more answers
A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (
djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

Kb = \frac{[ HSO3-][0H-]}{SO3-2}

Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation Kb=\frac{[HSO3-][OH-]}{[SO3-2]}

We noe have:

1.485*10^-^7=\frac{x*x}{(0.056-x)}

x = [OH-] = 9.11*10^-^5

pOH = -log(OH) = -log(9.11*10^-^5)

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0
2 years ago
A 0.529-g sample of gas occupies 125 ml at 60. cm of hg and 25°c. what is the molar mass of the gas?
Llana [10]

<span>Let's </span>assume that the gas has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>

</span><span>Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹) and T is temperature in Kelvin.<span>
<span>
</span>P = 60 cm Hg = 79993.4 Pa
V = </span>125  mL = 125 x 10⁻⁶ m³

n = ?

<span> R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 25 °C = 298 K
<span>
By substitution,
</span></span>79993.4 Pa<span> x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 298 K<span>
                                          n = 4.0359 x 10</span>⁻³ mol

<span>
Hence, moles of the gas</span> = 4.0359 x 10⁻³ mol<span>

Moles = mass / molar mass

</span>Mass of the gas  = 0.529 g 

<span>Molar mass of the gas</span> = mass / number of moles<span>
                                    = </span>0.529 g / 4.0359 x 10⁻³ mol<span>
<span>                                    = </span>131.07 g mol</span>⁻¹<span>

Hence, the molar mass of the given gas is </span>131.07 g mol⁻¹

4 0
2 years ago
Hydrogen bonds are approximately _____% of the bond strength of covalent c-c or c-h bonds.
Lelu [443]
Hydrogen bonds are approximately 5% of the bond strength of covalent C-C or C-H bonds.
Hydrogen bonds strength in water is approximately 20 kJ/mol, strenght of carbon-carbon bond is approximately 350 kJ/mol and strengh of carbon-hydrogen bond is approximately 340 kJ/mol.
20 kJ/350 kJ = 0,057 = 5,7 %.
4 0
2 years ago
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