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3 years ago

Wuy Piolcululugu Give the names of two organs in the chest. 1. ............... 2 m

Chemistry

1 answer:

Arada [10]3 years ago

8 0

Answer:

The organs present inside the chest are :

1. The lungs

2. The heart

Explanation:

The chest cavity is also called as the thoracic cavity. It is the second largest hollow space of the body.In the bottom , it is enclosed by the diaphragm.

This cavity actually contain three space each round with mesothelium , pleural cavity and precardial cavity.

This contain the lungs , the tracheobronchial tree , the heart , the blood vessels which transport the blood between the heart and the lungs.

It also contain the esophagus .

Esophagus is the path through which the food passes from the mouth to the stomach.

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Consider the following reaction: 2CH3OH ---> 2CH4+O2 delta H= +252.8kJ. Calculate the amount of heat transferred when 22.0g o

nydimaria [60]

To calculate the amount of heat transferred when an amount of reactant is decomposed, we must look at the balanced reaction and its corresponding heat of reaction. In this case, we can see that 252.8 kJ of heat is transferred per 2 moles of CH3OH used. When 22 g of CH3OH is used, 86.9 kJ is absorbed.

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2 years ago

4. Are all prokaryotes bad?<br> Provide an example to support<br> your answer.

Rina8888 [55]

Answer:

Although they receive a bad rap from the media and pharmaceuticals, the majority of prokaryotes are either harmless or actually help eukaryotes, such as animals and plants, to survive and only a small number of species are responsible for serious illnesses.

Explanation:

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2 years ago

For the reaction C2H4(g) + H2O(g) --> CH3CH2OH(g)

Dominik [7]

Answer : The value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

Explanation :

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy = -45.6 kJ = -45600 J

$\Delta S^o$ = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

$\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)$

$\Delta G^o=-12666.6J=-12.7kJ$

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln k$

where,

$\Delta G^o$ = standard Gibbs free energy = -12666.6 J

R = gas constant = 8.314 J/K.mol

T = temperature = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k$

$k=3.35\times 10^{2}$

Therefore, the value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

3 0

3 years ago

The K w for water at 0 ∘ C is 0.12 × 10 − 14 M 2 . Calculate the pH of a neutral aqueous solution at 0 ∘ C . p H = Is a pH = 7.2

Karo-lina-s [1.5K]

Answer:

pH → 7.46

Explanation:

We begin with the autoionization of water. This equilibrium reaction is:

2H₂O ⇄ H₃O⁺ + OH⁻ Kw = 1×10⁻¹⁴ at 25°C

Kw = [H₃O⁺] . [OH⁻]

We do not consider [H₂O] in the expression for the constant.

[H₃O⁺] = [OH⁻] = √1×10⁻¹⁴ → 1×10⁻⁷ M

Kw depends on the temperature

0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻] → [H₃O⁺] = [OH⁻] at 0°C

√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M

- log [H₃O⁺] = pH

pH = - log 3.46×10⁻⁸ → 7.46

8 0

3 years ago

When two hydrogen atoms bond with one oxygen Atom to form water what happens to the electrons

Likurg_2 [28]

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3 years ago