Answer:
7.78×10¯³ mole
Explanation:
From the question given above, the following data were obtained:
Volume (V) = 75 mL
Pressure (P) = 255 kPa
Temperature (T) = 22.5 °C
Number of mole (n) =?
Next, we shall convert 75 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
75 mL = 75 mL × 1 L / 1000 mL
75 mL = 0.075 L
Next, we shall convert 22.5 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Temperature (T) = 22.5 °C
Temperature (T) = 22.5 °C + 273
Temperature (T) = 295.5 K
Finally, we shall determine the amount of oxygen molecules present in steel calorimeter. This can be obtained as follow:
Volume (V) = 0.075 L
Pressure (P) = 255 kPa
Temperature (T) = 295.5 K
Gas constant (R) = 8.314 KPa.L/Kmol
Number of mole (n) =?
PV = nRT
255 × 0.075 = n × 8.314 × 295.5
19.125 = n × 2456.787
Divide both side by 2456.787
n = 19.125 / 2456.787
n = 7.78×10¯³ mole
Thus, amount of oxygen molecules present in steel calorimeter is 7.78×10¯³ mole
Answer:
KOH molar mass = 39 + 16 + 1 = 56g
To make 1 L of 1M soln needs 56g KOH
To make 500mL 1M needs 56/2 = 28g
To make 500mL 0.2M needs 28 x 0.2gn:
Paradoxes and Russell's Type Theories.
Simple Type Theory and the λ -Calculus.
Ramified Hierarchy and Impredicative Principles.
Type Theory/Set Theory.
Type Theory/Category Theory.
Extensions of Type System, Polymorphism, Paradoxes.
Univalent Foundations.
Its Na20
the formula for sodium oxide
<span>KCl<span>O3</span><span>(s)</span>+Δ→KCl<span>(s)</span>+<span>32</span><span>O2</span><span>(g)</span></span>
Approx. <span>3L</span> of dioxygen gas will be evolved.
Explanation:
We assume that the reaction as written proceeds quantitatively.
Moles of <span>KCl<span>O3</span><span>(s)</span></span> = <span><span>10.0⋅g</span><span>122.55⋅g⋅mo<span>l<span>−1</span></span></span></span> = <span>0.0816⋅mol</span>
And thus <span><span>32</span>×0.0816⋅mol</span> dioxygen are produced, i.e. <span>0.122⋅mol</span>.
At STP, an Ideal Gas occupies a volume of <span>22.4⋅L⋅mo<span>l<span>−1</span></span></span>.
And thus, volume of gas produced = <span>22.4⋅L⋅mo<span>l<span>−1</span></span>×0.0816⋅mol≅3L</span>
Note that this reaction would not work well without catalysis, typically <span>Mn<span>O2</span></span>.
