Answer: C. He didn't measure the weight of the log before it was burned.
Answer:
A. Final Temp = 36.428C and
47.9g ice will melt
Explanation:
Given the following data:
Mass of water (M1) = 45.0g = 0.045kg
Temperature (T1) = 85C = 358k
Mass of ice (M2) = 105g = 0.105kg
Temperature (T2) = 0c = 273k
Specific heat of water (C) = 4.18j/gC = 0.00418kj/kgc
Molar heat of fusion of water = 6.01kj/mol
Therefore, heat required (q) = MCT
M1C(T1-T2) = M2C∆T
By putting the data we have
0.045×0.00418×(358-273) = 0.105×0.00418×∆T
∆t = 0.045×0.00418×85/0.105×0.00418
∆t = 36.428C
Gram of ice that would melt would be 47.9g
Answer:
12.9 g O₂
Explanation:
To find the mass of oxygen gas produced, you need to (1) convert grams KClO₃ to moles KClO₃ (via molar mass from periodic table values), then (2) convert moles KClO₃ to moles O₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles O₂ to grams O₂ (via molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs to match the given value (33.0 g).
Molar Mass (KClO₃): 39.098 g/mol + 35.45 g/mol + 3(15.998 g/mol)
Molar Mass (KClO₃): 122.542 g/mol
2 KClO₃ ---> 2 KCl + 3 O₂
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
33.0 g KClO₃ 1 mole 3 moles O₂ 31.996 g
-------------------- x ------------------- x ----------------------- x ------------------ =
122.542 g 2 moles KClO₃ 1 mole
= 12.9 g O₂