It is important to determine the qcalorimeter before determining the qmetal because the qcalorimeter is a constant and it's value is used to find the value of qmetal.
Answer: Increasing the volume of the gas would be the correct answer. Charles law of gas states this: 1) If the volume of a container is increased, the temperature increases.
2) If the volume of a container is decreased, the temperature decreases.
Hope this helps
Explanation:
If Thomson’s model of the atom were correct, Rutherford should have seen a complete reflection of those particles when fired at the foil. Those particles would have hit a solid object (Thomson model) and would have been reflected back at the emitter. However, Rutherford discovered that most of those particles passed through the sheet, either continuing straight through or having the angle change a small amount (bouncing off the nucleus). Only a small number of the particles were reflected back, having hit the nucleus straight on. This led him to believe that there was a small, dense portion of the atom that resided within a larger, emptier space.
<span>kinetic energy and gravity. Hope this helps! Good luck!</span>
Answer:
The mass percent of carbon in C₁₄H₁₉NO₂ is 72.1%.
Explanation:
To calculate the mass percent of carbon in C₁₄H₁₉NO₂ we have to follow two simple steps:
1st) Look for the atomic weight of each element in the Periodic Table and then multiply each one by its subscript in the molecule and finally sum all of them to find the molar weight of the molecule:
Carbon atomic weight . Carbon subscript = 12 . 14 = 168g
Hydrogen atomic weight . Hydrogen subscript = 1 . 19 = 19g
Nitrogen atomic weight . Nitrogen subscript = 14 . 1 = 14g
Oxygen atomic weight . Oxygen subscript = 16 . 2 = 32g
The molar weight of C₁₄H₁₉NO₂ is: 168g + 19g + 14g +32g = 233g.
2nd) Find the mass percent of carbon with a Rule of three:
If 233g represents the 100% of mass, the 168g of carbon in the molecule will represents the 72.1% of mass.
The Rule of three is:
233g ------------- 100%
168g ------------- = (168 . 100)/233 = 72.1% mass of carbon.
