Carbon and fluorine have the same shielding effect True or False
1 answer:
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Answer:
The empirical formula of the compound =
Explanation:
Mass of carbon dioxide gas = 59.060 mg = 0.059060 g
1 mg = 0.001 g
Moles of carbon dioxide =
Moles of carbon in 0.0013 moles of carbon dioxide gas = 1 × 0.0013 mol = 0.0013 mol
Mass of 0.0013 moles of carbon =
Mass of water = 24.176 mg = 0.024176
Moles of water =
Moles of hydrogen in 0.0013 moles of water = 2 × 0.0013 mol = 0.0026 mol
Mass of 0.0013 moles of hydrogen=
Mass of sulfur dioxide = 20.326 mg = 0.020326 g
Moles of sulfur dioxide =
Moles of sulfur in 0.00032 moles of sulfur dioxide = 1 × 0.00032 mol = 0.00032 mol
Mass of 0.00032 moles of sulfur =
Mass of oxygen in the sample = x
Mass of sample = 33.153 mg = 0.033153 g
0.033153 g = 0.0156 g + 0.0013 g + 0.01024 g + x
x = 0.006013 g
Moles of oxygen =
For empirical formula divide the lowest number of moles of elemnt from all the moles of the all the elements:
Carbon :
Hydrogen:
Sulfur :
Oxygen :
The empirical formula of the compound =
Answer:
0.83Cal
Explanation:
Given parameters:
Amount of energy given;
35J;
Convert to Cal;
1 calorie = 42 joules
x calories = 35 joules
Cross multiply;
42x = 35
x = = 0.83Cal