Answer:
44.8 L
Explanation:
Using the ideal gas law equation:
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
At Standard temperature and pressure (STP);
P = 1 atm
T = 273K
Hence, when n = 2moles, the volume of the gas is:
Using PV = nRT
1 × V = 2 × 0.0821 × 273
V = 44.83
V = 44.8 L
Answer:
37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.
Explanation:
Equation for the reaction:
2 CO + 2 NO ------> N2 + 2 CO2
2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen
At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.
So therefore, we can say:
2 * 22.4 L of CO produces 22.4 L of N2
44.8 L of CO produces 22.4 L of N2
Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:
44.8 L of CO = 22.4 L of N
x L = 18.9 L
x L = 18.9 * 44.8 / 22.4
x L = 18.9 * 2
x = 37.8 L
The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L
The two forms of oxygen, O2 and O3 is "<span>They have different molecular structures and different properties."</span>
Answer:
After 5 second 25% C-15 will remain.
Explanation:
Given data:
Half life of C-15 = 2.5 sec
Original amount = 100%
Sample remain after 5 sec = ?
Solution:
Number of half lives = T elapsed / half life
Number of half lives = 5 sec / 2.5 sec
Number of half lives = 2
At time zero = 100%
At first half life = 100%/2 = 50%
At second half life = 50%/2 = 25%
Thus after 5 second 25% C-15 will remain.
