Answer:
If you were to plug your nose, it would block out more than 80% of taste, causing all three items to have no taste- therefore making them taste the same.
Explanation:
False
Fact: Mammals and plants belong to the same domain, the Eukarya domain.
Evidence :All the organisms that possess a eukaryotic cell, plants, animals, protists, and fungi are in the Eukarya domain.
Answer: 0.4533mol/L
Explanation:
Molar Mass of CaCO3 = 40+12+(16x3) = 40+12+48 = 100g/mol
68g of CaCO3 dissolves in 1.5L of solution.
Xg of CaCO3 will dissolve in 1L i.e
Xg of CaCO3 = 68/1.5 = 45.33g/L
Molarity = Mass conc.(g/L) / molar Mass
Molarity = 45.33/100 = 0.4533mol/L
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:
18 grams of water
Explanation:
The Balance Chemical Reaction is as follow,
2 NH₄NO₃ → 2 N₂ + O₂ + 4 H₂O
According to Equation,
160 g (2 moles) NH₄NO₃ produces = 72 g (4 moles) of H₂O
So,
40 g of NH₄NO₃ will produce = X g of H₂O
Solving for X,
X = (40 g × 72 g) ÷ 160 g
X = 18 g of H₂O
<em>Hope This Helps!</em>
