S=50.05/32 O=49.95/16, S=1.5640625/1.5640625 O=3.121875/1.5640625, S=1 O=1.99, S=1 O=2 Empirical Formula= SO2
Answer:
the positively charged particle found in atomic nuclel
Explanation:
protons carry positive charge and attracts negatively charged particles called electron
Answer : The compound contains the highest percentage of lead (by mass) is, PbS.
Explanation :
To calculate the percentage of lead in sample, we use the equation:
<u>For :</u>
Mass of = 239.3 g
Mass of Pb = 207.2 g
Putting values in above equation, we get:
The percentage of lead in the is 86.58 %.
<u>For :</u>
Mass of = 267.2 g
Mass of Pb = 207.2 g
Putting values in above equation, we get:
The percentage of lead in the is 77.55 %.
<u>For :</u>
Mass of = 259.7 g
Mass of Pb = 207.2 g
Putting values in above equation, we get:
The percentage of lead in the is 79.78 %.
<u>For :</u>
Mass of = 545.3 g
Mass of Pb = 207.2 g
Putting values in above equation, we get:
The percentage of lead in the is 37.99 %.
Hence, from this we conclude that, the compound PbS contains the highest percentage of lead (by mass).
Answer:
Explanation:
6CO₂ + 6 H₂O ⇄ C₆H₁₂0₆ + 6O₂
This is the chemical equation given .
1. The equation shows a __Chemical equation_______the breaking and forming of chemical bonds that leads to a change in the composition of matter.
2. In the equation, CO₂ is a___reactant_____.
3. In the equation, C₆H₁₂0₆ is a ___product________.
4. In O₂, the type of bond that holds the two oxygen atoms together is a_nonpolar_covalent bond_________.
5. In H₂O, the type of bond that holds one of the hydrogen atoms to the oxygen atom is a__polar_hydrogen bond____.
6. The number of oxygen atoms on the left side of the equation is__equal to_________ the number of oxygen atoms on the right side.
Answer:
15 oxygens
Explanation:
Given the partially balanced reaction:
The subscripts (small number to the right of each element symbol) are the number of atoms of the element within each compound/molecule, and the coefficients (numbers in front of each compound) represent the number of that molecule involved in one full reaction (if the equation were balanced).
The product side of the reaction is on the right of the arrow.
To determine the total number of Oxygens on the product side, we need to identify how many Oxygens are in each molecule (the subscript on the Oxygen), and then multiply times the number of that molecule that would be involved (Coefficient of the compound containing Oxygen). There are multiple compounds on the right side of the equation that contain Oxygen, so we'll need to add together the number of Oxygens each part contributes.