Answer:
By stopping the synthesis of proteins.
Explanation:
Streptomycin combines with the bacteria's ribosomes causing them to lose their function, the cells die and actually, this antibiotic destroys all the bacteria by stopping the synthesis of protein bacteria.
This antibiotic is blocking the development of bacteria. Streptomycin was used to fight tuberculosis and it's been made from the micro-organism <em>Actinobacterium Streptomyces griseus. </em>
The infant's heart rate decreases as they grow and their blood pressure increases. The normal heart rate of infant is at 70 to 190 but as they reaches toddller age it is decreased to 60 to 140. However blood pressure increased as the child grows as blood pressure is quite dependent on size and the weight of the child.
A
CO₂ is the waste product of respiration of cells. Oxygen, on the other hand, is required in cellular respiration. Therefore at the alveolar in the lungs of an organism, CO₂ will be expelled while O₂ will be taken in.
Explanation:
This occurs passively due to the diffusion gradient, CO₂ usually has a higher concentration in the blood and lower concentration in alveoli hence this gas diffuses from the bloodstream into the lungs. On the other hand, oxygen is higher in concentration in the alveoli than in the blood capillaries hence difuses into the blood and taken to the cell for respiration. The CO₂ in the lungs is then expelled through exhalation of the lungs.
Learn More:
For more on gaseous exchange in the lungs check out;
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Answer:
120 gramos de CO₂
Explanation:
La ecuación química balanceada para la reacción de combustión del hidrocarburo C₃H₈ (propano) es la siguiente:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Por lo tanto, 1 mol de C₃H₈ produce 3 moles de CO₂. Ahora debemos convertir las moles a gramos utilizando el peso molecular (PM) de cada compuesto:
PM(C₃H₈)= (3 x 12 g/mol) + (8 x 1 g/mol) = 44 g/mol
1 mol C₃H₈ = 1 mol x 44 g/mol = 44 g
PM(CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol
3 mol CO₂ = 3 mol x 44 g/mol = 132 g
Por lo tanto, se producen 132 gramos de CO₂ a partir de 44 gramos de C₃H₈ y la relación estequiométrica es:
3 mol CO₂/1 mol C₃H₈ = 132 g CO₂/44 g C₃H₈
Finalmente, para calcular cuántos gramos de CO₂ se producen al quemar 40 g de C₃H₈, multiplicamos la relación estequiométrica por la masa a quemar:
masa CO₂ producida = 40 g C₃H₈ x 132 g CO₂/44 g C₃H₈= 120 g
