Step 1: Change density from g/mL to g/L;
0.807 g/mL = 807 g/L
Step 2: Find Moles of N₂;
As,
Density = Mass / Volume
Or,
Mass = Density × Volume
Putting Values,
Mass = 807 g/L × 1 L
Mass = 807 g
Also,
Moles = Mass / M.mass
Putting values,
Moles = 807 g / 28 g.mol⁻¹
Moles = 28.82 moles
Step 3: Apply Ideal Gas Equation to Find Volume of gas occupied,
As,
P V = n R T
V = n R T / P
Putting Values, remember! don't forget to change temperatue into Kelvin (25 °C + 273 = 298 K)
V = (28.82 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 298 K) ÷ 1 atm
V = 704.76 L
Answer: low pressure
Explanation: Air at the equator is warmer and the air above expands, becomes less dense and rises
Answer:
1.5e+8 atoms of Bismuth.
Explanation:
We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):
For this, it is necessary to know the values in meters for any of these diameters:
Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.
<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>
1 atom of Bismuth = 320pm in diameter.
<h3>Diameter of a biscuit in meters</h3>
<h3>Resulting Ratio</h3>
How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:
In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.
The
reaction is<span>
C</span>₂H₄ +
O₂ → CO₂ + H₂O<span>
To balance the equation, both side have same
number of elements. Here,</span>
In left hand
side has
in right hand side has
4 H atoms
2 H
atoms
2 C atoms 1 C
atom
<span>
2 O atoms 3 O
atoms
First, we have to balance number of C atoms and number of H atoms in both side.
To balance C atoms, '2' should be added before CO</span>₂ and to balance H atoms, '2' should be added
before H₂<span>O.
Then number of oxygen atoms is </span>2 x 2 + 2 = 6 in right hand side. So, 3 should be
added before O₂<span> in left hand side.
After balancing the equation should be,</span>
C₂H₄<span> + 3O</span>₂<span> → 2CO</span>₂<span> + 2H</span>₂O
