Answer:
The volume of water in water bath is 1,011 Liters.
Explanation:
Length of the water bath, L = 1.85 m
Width of the water bath, W= 0.810 m
Height of the water bath ,H= 0.740 m
Height of the water in water bath, h= 0.740 m - 2.57 inches
1 m = 39.37 inch
Volume of the water in bath = L × W × h
The volume of water in water bath is 1,011 Liters.
As you increase in elevation, there is less air above you thus the pressure decreases. As the pressure decreases, air molecules spread out further (i.e. air expands) and the temperature decreases. If the humidity is at 100 percent (because it's snowing), the temperature decreases more slowly with height.
M P₂O₅ = 31g×2 + 16g×5 = 142g/mol
1 mole of P₂O₅ contains 5 moles of atoms
5 moles of atoms = 30,01×10²³ a.
142g ----------- 30,01×10²³ a.
0,69g ---------- X
X = (0,69×30,01×10²³)/142
X = 0,14626×10²³ = 1,4626×10²² atoms
:)
Answer:
Option-4 (3:2) is the correct answer.
Explanation:
Following steps are taken to balance the given unbalanced chemical equation.
Step 1: Write the unbalanced chemical equation,
N₂ + H₂ → NH₃
Step 2: Balance Nitrogen Atoms;
There are 2 nitrogen atoms on left hand side and 1 nitrogen atoms on right hand site therefore, to balance them multiply NH₃ on right hand side by 2 i.e.
N₂ + H₂ → 2 NH₃
Step 3: Balance Hydrogen Atoms;
Now, there are 2 hydrogen atoms on left hand side and 6 hydrogen atom on right hand site therefore, to balance them multiply H₂ on left hand side by 3 i.e.
N₂ + 3 H₂ → 2 NH₃
Now, the equation is balanced.
Step 4: Finding out mole ratios:
From balanced chemical equation it can be concluded that 3 moles of H₂ are involved in producing 2 moles of NH₃ hence, the mole ratio of consumption of H₂ to production of NH₃ is 3:2.
Answer:
2H₂(g) + O₂(g) → 2H₂O(g), in presence of Pt as a catalyst.
Explanation:
The reaction:
<em>2H₂(g) + O₂(g) → 2H₂O(g), in presence of Pt as a catalyst.</em>
2.0 moles of hydrogen gas react with 1.0 mole of oxygen gas to produce 2.0 moles of water vapor in presence of Pt as a catalyst.
