Answer:
d. 103.3
Explanation:
In the given question, the National Weather Service routinely supplies atmospheric pressure data to help pilots set their altimeters. And the units of atmospheric pressure used for reporting the atmospheric pressure data are inches of mercury. For a barometric pressure of 30.51 inches of mercury, we can calculate the pressure in kPa as follow:
In principle, 3.386 kPa is equivalent to the atmospheric pressure of 1 inch of mercury. Thus, 30.51 inches of mercury is equivalent to 30.51 in *(3.386 kPa/1 in) = 103.307 kPa.
Therefore, a barometric pressure of 30.51 inches of mercury corresponds to _____103.3_____ kPa.
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Answer:
The final temperature is 348.024°C.
Explanation:
Given data:
Specific heat of copper = 0.385 j/g.°C
Energy absorbed = 7.67 Kj (7.67×1000 = 7670 j)
Mass of copper = 62.0 g
Initial temperature T1 = 26.7°C
Final temperature T2 = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
Q = m.c. ΔT
7670 J = 62.0 g × 0.385 j/g °C ×( T2- 26.7 °C
)
7670 J = 23.87 j.°C ×( T2- 26.7 °C
)
7670 J / 23.87 j/°C = T2- 26.7 °C
T2- 26.7 °C = 321.324°C
T2 = 321.324°C + 26.7 °C
T2 = 348.024°C
The final temperature is 348.024°C.
Answer: A volume of 455 mL from 0.550 M KBr solution can be made from 100.0 mL of 2.50 M KBr.
Explanation:
Given: 
= ?, 
= 0.55 M
= 100.0 mL, 
= 2.50 M
Formula used to calculate the volume of KBr is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that a volume of 455 mL from 0.550 M KBr solution can be made from 100.0 mL of 2.50 M KBr.
