Answer:
C) carbon dioxide and hydrigen
Answer:
41.3kJ of heat is absorbed
Explanation:
Based in the reaction:
Fe₃O₄(s) + 4H₂(g) → 3Fe(s) + 4H₂O(g) ΔH = 151kJ
<em>1 mole of Fe3O4 reacts with 4 moles of H₂, 151kJ are absorbed.</em>
63.4g of Fe₃O₄ (Molar mass: 231.533g/mol) are:
63.4g Fe₃O₄ × (1mol / 231.533g) = <em>0.274moles of Fe₃O₄</em>
These are the moles of Fe₃O₄ that react. As 1 mole of Fe₃O₄ in reaction absorb 151kJ, 0.274moles absorb:
0.274moles of Fe₃O₄ × (151kJ / 1 mole Fe₃O₄) =
<h3>41.3kJ of heat is absorbed</h3>
<em />
<span>Pre-1982 definition of STP: 37 g/mol
Post-1982 definition of STP: 38 g/mol
This problem is somewhat ambiguous because the definition of STP changed in 1982. Prior to 1982, the definition was 273.15 K at a pressure of 1 atmosphere (101325 Pascals). Since 1982, the definition is 273.15 K at a pressure of exactly 100000 Pascals). Because of those 2 different definitions, the volume of 1 mole of gas is either 22.414 Liters (pre 1982 definition), or 22.71098 liters (post 1982 definition). And finally, there's entirely too many text books out there that still use the 35 year obsolete definition. So let's solve this problem using both definitions and you need to pick the correct answer for the text book you're using.
First, determine how many moles of gas you have. Just simply divide the volume you have by the molar volume.
Pre-1982: 2.1 / 22.414 = 0.093691443 moles
Post-1982: 2.1 / 22.71098 = 0.092466287 moles
Now determine the molar mass. Simply divide the mass by the moles. So
Pre-1982: 3.5 g / 0.093691443 moles = 37.35666667 g/mol
Post-1982: 3.5 g / 0.092466287 moles = 37.85163333 g/mol
Finally, round to 2 significant figures. So
Pre-1982: 37 g/mol
Post-1982: 38 g/mol</span>
Answer:
119.85 grams Br or 120. grams Br (sig figs)
Explanation:
1.50 moles Br 79.90 g Br
--------------------- x ------------------------ = 119.85 grams Br or 120 grams Br (sig figs)
1 mole
Answer:
<span>5.1012⋅<span>10<span>−12</span></span>J t</span>he idea here is that you need to use the mass of a single proton and the mass of a single neutron to calculate the mass of a lithium-6 nucleus, then use the measured value to find its mass defect.
