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.
Answer:
A.The sound becomes louder.
And
C.The sound waves get further.
Explanation:
Louder the sound it will travel long.
Answer:
51.2g of CO2
Explanation:
The first step is to balance the reaction equation as shown in the solution attached. Without balancing the reaction equation, one can never obtain the correct answer! Then obtain the masses of octane reacted and carbon dioxide produced from the stoichiometric equation. After that, we now compare it with what is given as shown in the image attached.
Answer:
310.53 g of Cu.
Explanation:
The balanced equation for the reaction is given below:
CuSO₄ + Zn —> ZnSO₄ + Cu
Next, we shall determine the mass of CuSO₄ that reacted and the mass Cu produced from the balanced equation. This can be obtained as follow:
Molar mass of CuSO₄ = 63.5 + 32 + (16×4)
= 63.5 + 32 + 64
= 159.5 g/mol
Mass of CuSO₄ from the balanced equation = 1 × 159.5 = 159.5 g
Molar mass of Cu = 63.5 g/mol
Mass of Cu from the balanced equation = 1 × 63.5 = 63.5 g
Summary:
From the balanced equation above,
159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.
Finally, we shall determine the mass of Cu produced by the reaction of 780 g of CuSO₄. This can be obtained as follow:
From the balanced equation above,
159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.
Therefore, 780 g of CuSO₄ will react to produce = (780 × 63.5)/159.5 = 310.53 g of Cu.
Thus, 310.53 g of Cu were obtained from the reaction.
Answer:
Molecular formula naphthalene → C₁₀H₈
Empirical formula naphthalene → C₅H₄
Explanation:
Centesimal composition means that in 100 g of compound we have x g of the element. Therefore in 100 g of naphthalene we have:
93.7 g of C
6.3 g of H
Let's make a rule of three:
In 100 g of naphthalene we have 93.7 g of C and 6.3 g of H
In 128 g of naphthalene we would have:
128 . 93.7 / 100 = 120 g of C
128. 6.3 / 100 = 8 g of H
We convert the mass to moles, by molar mass:
120 g . 1mol / 12 g = 10 moles C
8 g . 1mol/ 1g = 8 moles H
Molecular formula naphthalene → C₁₀H₈
Empirical formula naphthalene → C₅H₄
(The sub-index of each element is divided by the largest possible number)
