Find the final equilibrium temperature when 15.0 g of milk at 13.0 degrees c is added to 148 g of coffee with a temperature of 8
1 answer:
According to zeroth law of thermodynamics, when two objects are kept in contact, heat (energy) is transferred from one to the other until they reach the same temperature (are in thermal equilibrium). When the objects are at the same temperature there is no heat transfer.
So, at equilibrium, =
,
+
q=m×c×T, where q = heat energy, m = mass of a substance, c = specific heat (units J/kg∙K), T is temperature
=(15X13X4.19)+(148X88.3X4.19)
= 81.37 ° C
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<u>Answer:</u> The theoretical yield of iron(III) sulfate is 26.6 grams
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of iron(III) phosphate = 20.00 g
Molar mass of iron(III) phosphate = 150.82 g/mol
Putting values in equation 1, we get:
The given chemical equation follows:
As, sodium sulfate is present in excess. So, it is considered as an excess reagent.
Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate
So, 0.133 moles of iron(III) phosphate will produce = of iron(III) sulfate
Now, calculating the mass of iron(III) sulfate from equation 1, we get:
Molar mass of iron(III) sulfate = 399.9 g/mol
Moles of iron(III) sulfate = 0.0665 moles
Putting values in equation 1, we get:
Hence, the theoretical yield of iron(III) sulfate is 26.6 grams