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kumpel [21]
3 years ago
15

Calculate the standard entropy change for the following reaction cu(s) + 1/2 O2(g) —> cuo(s)

Chemistry
2 answers:
Evgen [1.6K]3 years ago
8 0

3433Explanation:

the awser sucks and its not RIGHT

IceJOKER [234]3 years ago
8 0

Answer:

The standard entropy change for the reaction -93.12 J/K.mol

Explanation:

The Standard entropy change:

\Delta S^{0} = \Sigma nS^{0}(products) - \Sigma mS^{0}(reactants)

ΔS° = [42.59] - [33.17 + (205.07/2)] J/K.mol

ΔS° = [42.59] - [33.17 + 102.535] J/K.mol

ΔS° = [42.59] - [135.705] J/K.mol

ΔS° = - 93.12 J/K.mol

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One of the intermediates in the synthesis of glycine from ammonia, carbon dioxide, and methane is aminoacetonitrile, C2H4N2. The
sleet_krkn [62]

Answer:

Mass of C₂H₄N₂ produced = 3.64 g

Explanation:

The balanced chemical equation for the reaction is given below:

3CH₄ (g) + 5CO₂ (g) + 8NH₃ (g) → 4C₂H₄N₂ (g) + 10H₂O (g)

From the equation, 3 moles of CH₄ reacts with 5 moles of CO₂ and 8 moles of NH₃ to produce 4 moles of C₂H₄N₂ and 10 moles of H₂O

Molar masses of the compounds are given below below:

CH₄ = 16 g/mol; CO₂ = 44 g/mol; NH3 = 17 g/mol; C₂H₄N₂ = 56 g/mol; H₂O g/mol

Comparing the mole ratios of the reacting masses;

CH₄ = 1.65/16 = 0.103

CO₂ = 13.5/44 = 0.307

NH₃ = 2.21/17 = 0.130

converting to whole number ratios by dividing with the smallest ratio

CH₄ = 0.103/0.103 = 1

CO₂ = 0.307/0.103 = 3

NH₃ = 0.130/0.103 = 1.3

Multiplying through with 5

CH₄ = 1 × 5 = 5

CO₂ = 3 × 5 = 15

NH₃ = 1.3 × 5 = 6.5

Therefore, the limiting reactant is NH₃

8 × 17 g (136 g) of NH₃ reacts to produce 4 × 56 g (224 g) of C₂H₄N₂

Therefore, 2.21 g of NH₃ will produce (2.21 × 224)/136 g of C₂H₄N₂ = 3.64 g of C₂H₄N₂

Mass of C₂H₄N₂ produced = 3.64 g

7 0
3 years ago
A sample of bleach was analyzed as in this procedure. The only procedural difference is that the student weighed out the bleach
Bogdan [553]

Answer:

% = 5.69%

Explanation:

To do this, we need to write the equations taking place here. First, this is a REDOX reaction where the hypoclorite and thiosulfate solution reacts. The balanced equations are:

ClO⁻ + 2I⁻ + 2H⁺ -------> Cl⁻ +  I₂ + H₂O

I₂ + 2S₂O₃²⁻ -----------> 2I⁻ + S₄O₆²⁻

We already have the required volume and concentration of the thiosulfate solution, so we can calculate the moles of thiosulfate. With this moles, we can calculate the moles of hypochlorite, then the mass and finally the %.

The moles of thiosulfate would be:

moles S₂O₃²⁻ = V * M

moles S₂O₃²⁻ = 0.01324 * 0.0732 = 9.69x10⁻⁴ moles

Now according to the above reactions, we can see that

moles I₂ = moles ClO⁻

and

moles I₂ / moles S₂O₃²⁻ = 1/2

Therefore, let's calculate the moles of ClO⁻:

moles ClO⁻ = 9.69x10⁻⁴ / 2 = 4.845x10⁻⁴ moles

Now, we can calculate the mass of these moles, using the molar mass of sodium hypochlorite which is 74.44 g/mol:

m = 74.44 * 4.845x10⁻⁴

m = 0.036 g

Finally the % of this, in the bleach sample would be:

% = 0.036 / 0.634 * 100

<h2>% = 5.69%</h2>
6 0
3 years ago
Pis
loris [4]

Answer:

5.00×10-19 J

Explanation:

E = hv

but v = c ÷ wavelength

E = (6.626×10-34) × ((3×10^8)÷(398×10-9))

= 4.9945×10-19

≈5.00×10-19

Hope this helps

8 0
2 years ago
A fish may have hundreds of offspring at a time and only a small number can survive. which characteristics of fish might allow t
mafiozo [28]
Well depends on what type of fish the parent is but i guess you can say:
what they eat
teeth
instincts
color
venom
blending in,etc
8 0
3 years ago
In a reaction, gaseous reactants form a liquid product. The heat absorbed by the surroundings is 1.1 MJ, and the work done on th
ratelena [41]

Answer: \Delta E is 1155 kJ

Explanation:

According to first law of thermodynamics:

\Delta E=q+w

\Delta E=Change in internal energy

q = heat absorbed or released

w = work done or by the system

w = work done by the system=-P\Delta V  {Work done on the system is positive as the final volume is lesser than initial volume}

w =13.2kcal=55.2kJ   (1kcal=4.184 kJ)

q = +1.1 MJ = 1100 kJ  (1MJ=1000kJ)   {Heat absorbed by the system is positive}

\Delta E=1100kJ+(55.2)kJ=1155kJ

Thus \Delta E is 1155 kJ

4 0
3 years ago
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