Protons=47
nuetrons=61
electrons=47
35.9g x 1 mol/ 2.016g x 22.4 L/ 1 mol= 398.89 L
Answer:
The products are carbon dioxide and water
Explanation:
Step 1: Data given
Combustion = a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve O2 as one reactant.
Step 2: The complete combustion of C3H7OH:
For the combustion of 1-propanol, we need O2.
The products of this combustion are CO2 and H2O.
C3H7OH + O2→ CO2 + H2O
On the left side we have 3x C (in c3H7OH), on the right side we have 1x C (in CO2). To balance the amount of C, we have to multiply CO2 on the right side by 3
C3H7OH + O2→ 3CO2 + H2O
On the left side we have 8x H (in C3H7OH) and 2x on the right side (in H2O). To balance the amount of H, we have to multiply H2O, on the right side by 4.
C3H7OH + O2→ 3CO2 + 4H2O
On the left side we have 3x O (1x in C3H7OH and 2x in O2), on the right side we have 10x O (6x in CO2 and 4x in H2O).
To balance the amount of O on both sides, we have to multiply C3H7OH by 2, multiply O2 by 9. Then we have to multiply 3CO2 by 2 and 4H2O by 2. Now the equation is balanced.
2C3H7OH + 9O2→ 6CO2 + 8H2O
For 2 moles propanol, we need 9 moles of O2 to produce 6 moles of CO2 and 8 moles Of H2O
The products are carbon dioxide and water
Answer:
The particles in the neutral paper can shift, causing the paper to become polarized and attracted to the rod.
Explanation:
The neutral paper has an even distribution of its electrons throughout the paper. If a charged rod is brought near the neutral paper, this can cause the electrons in the paper to shift. If the rod is negative, the electrons will be repelled from the rod and cause the molecules in the paper to have a slight positive charge on the part of the paper closest to the rod. If the rod is positive, the electrons will be attracted to the rod and cause a slight negative charge on the side of the paper closest to the rod.
I think the answer you are looking for is D.
Hope this helps!!! :D
