Answer:
8.55 × 10³ cal
Explanation:
Step 1: Given and required data
- Specific heat of water (c): 1 cal/g.°C
- Initial temperature: 22.7 °C
- Final temperature: 38.8 °C
Step 2: Calculate the temperature change (ΔT)
ΔT = Final temperature - Initial temperature = 38.8 °C - 22.7 °C = 16.1 °C
Step 3: Calculate the heat required (Q)
We will use the following expression.
Q = c × m × ΔT
Q = 1 cal/g.°C × 531 g × 16.1 °C = 8.55 × 10³ cal
Answer: centrifugation, boiling/heating
, and long standing
Explanation:
Answer:
5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O=
Answer: 9.04 g of H2O
Explanation:
First set up equation: C4H10 (g)+ O2(g) -> CO2(g) + H2O(g)
Next balance it: 2C4H10 (g)+ 13O2(g) -> 8CO2(g) + 10H2O (g)
Use equation to get moles and plug given
5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O
