Unbalanced it should be 2Zn+2Hcl=2ZnCl2+H2
500,000 g of baking soda is present in 1000 boxes of 500 g baking soda boxes.
Answer:
Option C.
Explanation:
As 500 g of baking soda is taken in each box of that company. The total weight of baking soda in all the boxes can be determined by adding the weights of each box. This is possible only when the number of boxes is less. But if the number of boxes are large, then we can determine the total weight of baking soda by multiplying the number of boxes with the weight in each box.
So in this case, 1000 boxes are present and in that 500 g of baking soda are present in each box.
So total grams of baking soda will be 1000 * 500 = 5,00,000 g.
Thus, 500,000 g of baking soda is present in 1000 boxes of 500 g baking soda boxes.
I think the answer is d not real fir sure I'll find out in the morning sorry
