Answer:
Explanation:
Unclear question.
I infer you want a clear rendering, which reads;
A 258.4 g sample of ethanol (C2H5OH) was burned in a calorimetric pump using a Dewar glass. As a consequence, the water temperature rose to 4.20 ° C.
If the heat capacity of the water and the surrounding glass was 10.4 kJ / ° C, calculate the heat of combustion of one mole of ethanol.
Answer:
Plants, algae, and a group of bacteria called cyanobacteria are the only organisms capable of performing photosynthesis
Answer:
54g of water
Explanation:
Based on the reaction, 1 mole of methane produce 2 moles of water.
To solve this question we must find the molar mass of methane in order to find the moles of methane added. With the moles of methane and the chemical equation we can find the moles of water produced and its mass:
<em>Molar mass CH₄:</em>
1C = 12g/mol*1
4H = 1g/mol*4
12g/mol + 4g/mol = 16g/mol
<em>Moles methane: </em>
24g CH₄ * (1mol / 16g) = 1.5 moles methane
<em>Moles water:</em>
1.5moles CH₄ * (2mol H₂O / 1mol CH₄) = 3.0moles H₂O
<em>Molar mass water:</em>
2H = 1g/mol*2
1O = 16g/mol*1
2g/mol + 16g/mol = 18g/mol
<em>Mass water:</em>
3.0moles H₂O * (18g / mol) =
<h3>54g of water</h3>
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 />
Answer:
35.8 g
Explanation:
Step 1: Given data
Mass of water: 63.5 g
Step 2: Calculate how many grams of KCl can be dissolved in 63.5. g of water at 80 °C
Solubility is the maximum amount of solute that can be dissolved in 100 g of solute at a specified temperature. The solubility of KCl at 80 °C is 56.3 g%g, that is, we can dissolve up to 56.3 g of KCl in 100 g of water.
63.5 g Water × 56.3 g KCl/100 g Water = 35.8 g KCl
