<span>Let's assume
that the oxygen gas has ideal gas behavior.
Then we can use ideal gas formula,
PV = nRT</span>
Where, P is the pressure of the gas (Pa), V is the volume of the gas
(m³), n is the number of moles of gas (mol), R is the universal gas
constant ( 8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin.
<span>
P = 2.2 atm = 222915 Pa
V = 21 L = 21 x 10</span>⁻³ m³
n = ?
R = 8.314 J mol⁻¹ K⁻¹
<span>
T = 87 °C = 360 K
By substitution,
</span>222915 Pa x 21 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻<span>¹ x 360 K
n
= 1.56</span><span> mol</span>
<span>
Hence, 1.56 moles of the oxygen gas are </span><span>
left for you to breath.</span><span>
</span>
Answer:
a. 113 min
Explanation:
Considering the equilibrium:-
2N₂O₅ ⇔ 4NO₂ + O₂
At t = 0 125 kPa
At t = teq 125 - 2x 4x x
Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x
125 - 3x = 176 kPa
x = 17 kPa
Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given that:
The rate constant, k = 
min⁻¹
Initial concentration = 125 kPa
Final concentration = 91 kPa
Time = ?
Applying in the above equation, we get that:-
Table Giving Answer
Element Atomic mass % Amount
Mg_24 24 79 18.96
Mg_25 25 10 2.5
Mg_26 26 11 2.86
Total 24.32
Discussion
The method of calculation for this table, which was done in Excel (a spread sheet) is shown below. Assume that there is 100 grams of material of "pure" magnesium. What is it's mass?
<em><u>Sample Calculation</u></em>
The the sample atomic mass = 24
Mass = % * sample atomic mass
Mass = 79% * 24
Mass = (79/100) * 24
Mass = 18.96
<em><u>Note</u></em>
The other two elements are found exactly the same as the sample calculation.
Then all you do is add the 3 masses together.
Answer
The mass of Mg to 1 decimal place is 24.3 <<<< Answer.
Answer:
Explanation:
Initial burette reading = 1.81 mL
final burette reading = 39.7 mL
volume of NaOH used = 39.7 - 1.81 = 37.89 mL .
37.89 mL of .1029 M NaOH is used to neutralise triprotic acid
No of moles contained by 37.89 mL of .1029 M NaOH
= .03789 x .1029 moles
= 3.89 x 10⁻³ moles
Since acid is triprotic , its equivalent weight = molecular weight / 3
No of moles of triprotic acid = 3.89 x 10⁻³ / 3
= 1.30 x 10⁻³ moles .
Answer:
electron (-) and proton (+)
