You need 158.70 grams of Fe2O3 to produce 111 grams of Fe. This is calculated by using the molar masses and stoichiometric relationship of the two compounds.
Solution:
MM Fe = 55.845 g/mol
MM Fe2O3 = 159.69 g/mol
Fe: Fe2O3 = 2 mol:1 mol
11 g FE (1 mol Fe/55.845 g Fe) (1 mol Fe2O3/2 mol Fe) (159.69 g Fe2O3 / 1 mole Fe2O3) = 158.70 grams Fe2O3
Answer:
https://www.wolframalpha.com/
Explanation:
Im pretty sure that the current will increase so the answer is A . Hope that helps
First, in order to calculate the specific heat capacity of the metal in help in identifying it, we must find the heat absorbed by the calorimeter using:
Energy = mass * specific heat capacity * change in temperature
Q = 250 * 1.035 * (11.08 - 10)
Q = 279.45 cal/g
Next, we use the same formula for the metal as the heat absorbed by the calorimeter is equal to the heal released by the metal.
-279.45 = 50 * c * (11.08 - 45) [minus sign added as energy released]
c = 0.165
The specific heat capacity of the metal is 0.165 cal/gC
