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3 years ago

How much energy is released when 0.40 mol C6H6(g) completely reacts with oxygen?

Chemistry

2 answers:

podryga [215]3 years ago

6 0

Answer:

The right answer is -1268 kJ

Explanation:

Vsevolod [243]3 years ago

4 0

<span>This question asksyou to apply Hess's law.
You have to look for how to add up all the reaction so that you get the net equation as the combustion for benzene. The net reaction should look something like C6H6(l)+ O2 (g)-->CO2(g) +H2O(l). So, you need to add up the reaction in a way so that you can cancel H2 and C.
multiply 2 H2(g) + O2 (g) --> 2H2O(l) delta H= -572 kJ by 3
multiply C(s) + O2(g) --> CO2(g) delta H= -394 kJ by 12
multiply 6C(s) + 3 H2(g) --> C6H6(l) delta H= +49 kJ by 2 after reversing the equation.
Then,
6 H2(g) + 3O2 (g) --> 6H2O(l) delta H= -1716 kJ
12C(s) + 12O2(g) --> 12CO2(g) delta H= -4728 kJ
2C6H6(l) --> 12 C(s) + 6 H2(g) delta H= - 98 kJ
______________________________________...
2C6H6(l) + 16O2 (g)-->12CO2(g) + 6H2O(l) delta H= - 6542 kJ
I hope this helps and my answer is right.</span>

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Breaking larger molecules into smaller molecules and carbon

ella [17]

Answer: catabolism

Explanation:

Catabolism refers to the set of metabolic pathways which is necessary for the breaking down of molecules into smaller units. This is then oxidized for the release of energy or can be used to perform other anabolic reactions.

Catabolism is regarded as the opposite direction of anabolism which is simply the building-up of molecules. It should be noted that anabolism and catabolism work together in every living organisms and perform functions such as the production of energy and the repair of cells.

3 0

3 years ago

A solution of NaOH is titrated with H2SO4. It is found that 20.05 mL of 0.3564 M H2SO4 solution is equivalent to 43.42 mL of NaO

Darya [45]

Answer : The concentration of NaOH is, 0.336 M

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=0.3564M\\V_1=20.05mL\\n_2=1\\M_2=?\\V_2=43.42mL$

Putting values in above equation, we get:

$2\times 0.3564M\times 20.05mL=1\times M_2\times 43.42mL$

$M_2=0.336M$

Thus, the concentration of NaOH is, 0.336 M

3 0

4 years ago

How much work must be done on a 5-kg snowboard to increase its speed from 2 m/s to 4 m/s?

Alekssandra [29.7K]

Answer:

Option C = 30 j

Explanation:

Given data:

mass of snowboard = 5 Kg

Initial speed = 2 m/s

final speed = 4 m/s

work done = ΔE= ?

ΔE= change in kinetic energy

Solution:

Formula:

K.E (initial) = 1/2 × mv²

K.E (initial) = 1/2 × 5 Kg . (2m/s)²

K.E (initial) = 1/2 × 20 Kg.m²/s²

K.E (initial) = 10 Kg.m²/s² or 10 J

Kg.m²/s² = J

K.E (finial) = 1/2 × mv²

K.E (finial) = 1/2 × 5 Kg . (4m/s)²

K.E (finial) = 1/2 × 5 Kg . 16 m²/s²

K.E (finial) = 1/2 × 80 Kg.m²/s²

K.E (finial) = 40 Kg.m²/s² or 40 J

work done = ΔE = K.E (finial) - K.E (initial)

work done = ΔE = 40 J - 10 J

work done = ΔE = 30 J

6 0

3 years ago

What is a chemical property of soda ash

xz_007 [3.2K]

Soda ash which is also referred to as sodium carbonate with a chemical formula (NaCO3) has the following chemical formula
The melting point of 851 degrees
The molecular weight of 105.98
Its density at 25 degrees is =2.53 g/cm^3
Its specific heat capacity at 25 degrees = 0.249 cal/gm/ 0c

4 0

4 years ago

if 3.0 grams of aluminum and 6.0 grams of bromine react to form AlBr3, how many grams of product would theoretically be produced

Gelneren [198K]

1) Chemical reaction: 2Al + 3Br₂ → 2AlBr₃.
m(Al) = 3,0 g.
m(Br₂) = 6,0 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 3,0 g ÷ 27 g/mol.
n(Al) = 0,11 mol.
n(Br₂) = n(Br₂) ÷ m(Br₂).
n(Br₂) = 6 g ÷ 160 g/mol.
n(Br₂) = 0,0375 mol; limiting reagens.
n(Br₂) : n(AlBr₃) = 3 : 2.
n(AlBr₃) = 0,025 mol.
m(AlBr₃) = 0,025 mol · 266,7 g/mol.
m(AlBr₃) = 6,67 g.

2) m(Br₂) - all bromine reacts, so mass of bromine after reaction is zero grams (m(Br₂) = 0 g).
n(Al) = 0,11 mol - 0,025 mol = 0,085 mol.
m(Al) = 0,085 mol · 27 g/mol.
m(Al) = 2,295 g.
m(AlBr₃) = 6,67 g · 0,72 (yield of reaction).
m(AlBr₃) = 4,8 g.
n - amount of substance.
M - molar mass.

4 0

3 years ago