Answer:
just write all the things you learned in chemistry and then double space it
Answer:
The given sample is aluminum.
Explanation:
Given data:
Mass of unknown sample = 54 g
Volume of given sample = 20cm³
Is sample is Al = ?
Solution:
In order to solve this problem we must know the density of Al from literature.
density of Al = 2.7 g/cm³
Now we will calculate the density of given unknown sample.
d = m/v
d = 54 g/ 20cm³
d = 2.7 g/cm³
Thus given sample is aluminum.
When hydrogen atoms are energized by electricity, an emission spectrum of specific colors occurs because of the different wavelengths.
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
Well the answer is 207.2. You take the lowest sig figs that you are using which would be 2500= 2 sig figs. 207.2 rounded to 2 sig figs would be 210