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) Pabs = 48960 KPa
b) T = 433.332 °C
Explanation:
∴ d = 1000 Kg/m³
∴ g = 9.8 m/s²
∴ h = 5000 m
∴ P gauge = - 40 KPa * ( 1000 Pa / KPa ) = - 40000 Pa; Pa≡Kg/m*s²
⇒ Pabs = - 40000 Kg/ms² + ( 1000 Kg/m³ * 9.8 m/s² * 5000 m )
⇒ Pabs = 48960000 Pa = 48960 KPa
a) at that height and pressure, we find the temperature at which the water boils by means of an almost-exponential graph which has the following equation:
P(T) = 0.61094 exp ( 17.625*T / ( T + 243.04 ))......P (KPa) ∧ T (°C)....from literature
∴ P = 48960 KPa
⇒ ( 48960 KPa / 0.61094 ) = exp ( 17.625T / (T+ 243.04))
⇒ 80138.803 = exp ( 17.625T / ( T + 243.04))
⇒ Ln ( 80138.803) = 17.625T / ( T + 243.04))
⇒ 11.292 * ( T + 243.04 ) = 17.625T
⇒ 11.292T + 2744.289 = 17.625T
⇒ 2744.289 = 17.625T - 11.292T
⇒ 2744.289 = 6.333T
⇒ T = 433.332 °C
Please refer to the attachment for a complete classification of your specified matter.
If iron (Fe) gained a proton it would become cobalt (CO).
