Answer : The final temperature of the mixture is 
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
where,
= specific heat of iron = 
= specific heat of water = 
= mass of iron = 39.9 g
= mass of water = 
= final temperature of mixture = ?
= initial temperature of iron = 
= initial temperature of water = 
Now put all the given values in the above formula, we get
Therefore, the final temperature of the mixture is
Answer:
Percent Yield Fe = 82.5%
Explanation:
The actual yield is the value produced after an experiment is conducted. The theoretical yield is the value calculated using the balanced chemical equation and atomic/molar masses.
To find the percent yield of iron (Fe), you need to (1) convert grams Al to moles Al (via atomic mass), then (2) convert moles Al to moles Fe (via mole-to-mole ratio from equation coefficients), then (3) convert moles Fe to grams Fe (via atomic mass), and then (4) calculate the percent yield. It is important to arrange the ratios in a way that allows for the cancellation of units. The final answer should have 3 sig figs to reflect the sig figs of the given values.
Atomic Mass (Mg): 24.305 g/mol
Atomic Mass (Fe): 55.845 g/mol
3 Mg + 2 FeCl₃ -----> 2 Fe + 3 MgCl₂
20.5 g Mg 1 mole 2 moles Fe 55.845 g
----------------- x ----------------- x ---------------------- x ----------------- =
24.305 g 3 moles Mg 1 mole
= 31.4 g Fe
Actual Yield
Percent Yield = ---------------------------------- x 100%
Theoretical Yield
25.9 g Fe
Percent Yield = -------------------- x 100%
31.4 g Fe
Percent Yield = 82.5%
The answer to this item is TRUE. This can be explained through the Graham's law. This law states that the rate at which gases diffuse is inversely proportional to the square root of their densities which is also related to their molecular masses.
Answer:
See below
Explanation:
propane mole weight = 44 gm / mole
100 gm / 44 gm / mole = 2.27 moles
From the equation, 5 times as many moles of OXYGEN (O2)are required
= 11.36 moles of oxygen
at <u>STP</u> this is 254.55 liters of O2 (because 22.4 L = one mole) and
Using oxygen as 21 percent of air means that
.21 x = 254.55 = x = <u>1212.12 liters of air required </u>
