Respuesta :

51.86 grams would be in the container.

One ounce is an equivalent of 28.34 grams, so times that by 12.2
.
12.2 * 28.34 = 345.75.

Put the percentage into decimal form, so 15% would now be 0.15.

0.15 * 345.75 = 51.86.

Answer:

% of N present in 12.2 oz of fertlilizer = 51.9 g

Explanation:

Given:

% of N in fertilizer = 15

Mass of sample fertilizer = 12.2 oz

Conversion factor for Ounce to gram:

1 ounce = 28.3495 g

12.2 ounce (oz) = 12.2 × 28.3495

                         = 345.8639 g

% of N in fertlilizer = 345.8639 × 0.15

                               = 51.8795 g

                               = 51.9 g

% of N present in 12.2 oz of fertlilizer = 51.9 g

ACCESS MORE