Answer:
NO would form 65.7 g.
H₂O would form 59.13 g.
Explanation:
Given data:
Moles of NH₃ = 2.19
Moles of O₂ = 4.93
Mass of NO produced = ?
Mass of produced H₂O = ?
Solution:
First of all we will write the balance chemical equation,
4NH₃ + 5O₂ → 4NO + 6H₂O
Now we will compare the moles of NO and H₂O with ammonia from balanced chemical equation:
NH₃ : NO NH₃ : H₂O
4 : 4 4 : 6
2.19 : 2.19 2.19 : 6/4 × 2.19 = 3.285 mol
Now we will compare the moles of NO and H₂O with oxygen from balanced chemical equation:
O₂ : NO O₂ : H₂O
5 : 4 5 : 6
4.93 : 4/5×4.93 = 3.944 mol 4.93 : 6/5 × 4.93 = 5.916 mol
we can see that moles of water and nitrogen monoxide produced from the ammonia are less, so ammonia will be limiting reactant and will limit the product yield.
Mass of water = number of moles × molar mass
Mass of water = 3.285 mol × 18 g/mol
Mass of water = 59.13 g
Mass of nitrogen monoxide = number of moles × molar mass
Mass of nitrogen monoxide = 2.19 mol × 30 g/mol
Mass of nitrogen monoxide = 65.7 g
Answer:
height(h)=10 m
acceleration due to gravity (g)=10 m/s^2
mass (m)=?
potential energy=67
or,mgh= 67
or,m×10×10=67
or,m=67/100
or,m=0.67 kg
Therefore,m= 670 grams
Answer:
Metallic bonding may be described as the sharing of free electrons among a lattice of positively charged metal ions. The structure of metallic bonds is very different from that of covalent and ionic bonds. While ionic bonds join metals to nonmetals, and covalent bonds join nonmetals to nonmetals, metallic bonds are responsible for the bonding between metal atoms.
In metallic bonds, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “sea” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions. The electrons then move freely throughout the space between the atomic nuclei.
Explanation:
The answer is D. The rice cooker / pot.
hope this helps !
Explanation:
Student A would be introducing more error. While student B is subtracting the actual filter paper used in the experiment, Student A is using the mass of a different filter paper, which may have a slightly different mass than the filter paper that was used. To avoid having to aacount for this, or introducing such an error, follow Student B in weighing the filter paper before adding the sand.
I hope this helps.
