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.
Answer:
A concentrated solution is one that has a relatively large amount of dissolved solute. A dilute solution is one that has a relatively small amount of dissolved solute.
No, because boats and other mechanical vehicles can spill gas and oil into the freshwater. if the freshwater is scarce already, we should not contaminate it more by risking the gas and oil spills.
Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water
Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.
From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water
At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
= (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)-
= 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
= (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
= 268.356(T - 27.8)
Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C
Answer: 28.4 °C
