Temperature changes with an increase or decrease of altitude. This change is known as the "lapse rate" and it varies depending on the amount of moisture in the particular mass of air. The "dry adiabatic lapse rate" (for dry air masses) is a temperature decrease of about 3 degrees C per thousand feet of altitude, while the "wet adiabatic lapse rate" (for moist air masses) is a temperature decrease of about 1.66 degrees C per thousand feet of altitude.
For average conditions, a figure of 3.5 degrees F (2 degrees C) per 1000 feet is commonly used.
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Answer:
0.7g of HCl
Explanation:
First, let us write a balanced equation for the reaction between HCl and Al(OH)3.
This is illustrated below:
Al(OH)3 + 3HCl —> AlCl3 + 3H2O
Next, let us obtain the masses of Al(OH)3 and HCl that reacted together according to the equation. This can be achieved as shown below:
Molar Mass of Al(OH)3 = 27 + 3(16+1)
= 27 + 3(17) = 27 + 51 = 78g/mol.
Molar Mass of HCl = 1 + 35.5 = 36.5g/mol
Mass of HCl from the balanced equation = 3 x 36.5 = 109.5g
Now we can obtain the mass of HCl that would react with 0.5g of Al(OH)3. This can be achieved as follow:
Al(OH)3 + 3HCl —> AlCl3 + 3H2O
From the equation above,
78g of Al(OH)3 reacted with 109.5g of HCl.
Therefore, 0.5g of Al(OH)3 will react with = (0.5 x 109.5)/78 = 0.7g of HCl
Answer:
219.95 °C
Explanation:
Given data:
Volume of gas = 9.71 L
Initial pressure = 209 torr (209/760 = 0.275 atm)
Initial temperature = 10.1 °C (10.1 +273 = 283.1 K)
Final temperature = ?
Final pressure = 364 torr (364/760 =0.479 atm)
Solution:
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
0.275 atm / 283.1 K = 0.479 atm/T₂
T₂ = 0.479 atm × 283.1 K/ 0.275 atm
T₂ = 135.6 atm. K /0.275 atm
T₂ = 493.1 K
Kelvin to °C:
493.1 K - 273.15 = 219.95 °C