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ira [324]
3 years ago
10

The most common form of elemental sulfur is S8, in which eight sulfur atoms linked in a ring of single bonds. At high temperatur

e, in the gas phase, S8 can break apart to give S2, the sulfur analog of molecular oxygen:
S8(g) 4S2(g) ΔH = +239 kJ at T = 800 K
Using the appropriate data in Table 3-2 S-S 240 calculate the S=S double-bond energy (in kJ/mol) in S2(g).
Chemistry
1 answer:
aliina [53]3 years ago
8 0

Answer:

Explanation:

In S₈ , there are 8 single bonds which breaks up first . Energy absorbed

= 8 x 240 = 1920 kJ

In S₈ four double bonds of S₂ are formed . Let bond energy be x . In this process energy will be released . energy released in four S₂ molecules formed = 4 x

Given

1920 + 4x = 239

4x = 239 - 1920

x = - 420.25 kJ .

So bond energy of S₂ = 420.25 kJ .

You might be interested in
A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
irga5000 [103]

Answer:

1. pH = 1.23.

2. H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the pKa is:

pKa=-log(Ka)=-log(5.90x10^{-2})=1.23

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

Best regards!

4 0
3 years ago
A 250 mL sample of gas is collected over water at 35°C and at a total pressure of 735 mm Hg. If the vapor pressure of water at 3
ELEN [110]

Answer:

The volume of the gas sample at standard pressure is <u>819.5ml</u>

Explanation:

Solution Given:

let volume be V and temperature be T and pressure be P.

V_1=250ml

V_2=?

P_{total}=735 mmhg

1 torr= 1 mmhg

42.2 torr=42.2 mmhg

so,

P_{water}=42.2mmhg

T_1=35°C=35+273=308 K

Now

firstly we need to find the pressure due to gas along by subtracting the vapor pressure of water.

P_{gas}=P_{total}-P_{water}

=735-42.2=692.8 mmhg

Now

By using combined gas law equation:

\frac{P_1*V_1}{T_1} =\frac{P_2*V_2}{T_2}

V_2=\frac{P_1*}{P_2}*\frac{T_2}{T_1} *V_1

V_2=\frac{P_gas}{P_2}*\frac{T_2}{T_1} *V_1

Here P_2 \:and\: T_2 are standard pressure and temperature respectively.

we have

P_2=750mmhg \:and\: T_2=273K

Substituting value, we get

V_2=\frac{692.8}{750}*\frac{273}{308} *250

V_2= 819.51 ml

4 0
2 years ago
When 3.0 kg of water is warmed from 10 °C to 80 °C, how much heat energy is needed?
Naddik [55]

Answer:

THE HEAT NEEDED TO CHANGE 3KG OF WATER FROM 10 C TO 80 C IS 877.8kJ OR 877,800 J.

Explanation:

Mass = 3.0 kg = 3 * 1000 = 3000 g

Initial temperature = 10 C

Final temperature = 80 C

Change in temperature = 80 - 10 = 70 C

Specific heat of water = 4.18 J/g C

Heat needed = unknown

Heat is the amount of energy in joules needed to change a gram of water by 1 C.

Heat = mass * specific heat * change in temperature

Heat = 3000 g * 4.18 J/g C * 70 C

Heat = 877 800 Joules

Heat = 877.8 kJ.

The heat needed to change 3 kg mass of water from 10 C to 80 C is 877,800 J or 877.8 kJ.

6 0
3 years ago
What does AuCl, Gold (I) Chloride, decompose into?
Alexxx [7]

Answer:

Gold

Explanation:

Gold(I) chloride is a compound of gold and chlorine with the chemical formula AuCl

6 0
3 years ago
A 1.59 mol sample of Kr has a volume of 641 mL. How many moles of Kr are in a 4.41 L sample at the same temperature and pressure
Marina86 [1]

Answer:

The correct answer is 10.939 mol ≅ 10.94 mol

Explanation:

According to Avogadro's gases law, the number of moles of an ideal gas (n) at constant pressure and temperature, is directly proportional to the volume (V).

For the initial gas (1), we have:

n₁= 1.59 mol

V₁= 641 mL= 0.641 L

For the final gas (2), we have:

V₂: 4.41 L

The relation between 1 and 2 is given by:

n₁/V₁ = n₂/V₂

We calculate n₂ as follows:

n₂= (n₁/V₁) x V₂ = (1.59 mol/0.641 L) x 4.41 L = 10.939 mol ≅ 10.94 mol

5 0
2 years ago
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