C) powdered sugar in hot water because the sugar is already broken so it would be more faster than sugar cube in hot water
Salt might be a pure substance but sugar is a compound
Answer:
concentration of ![[O_2]](https://tex.z-dn.net/?f=%5BO_2%5D)
= 0.0124 = 12.4 ×10⁻³ M
concentration of ![[CO]](https://tex.z-dn.net/?f=%5BCO%5D)
= 0.0248 = 2.48 ×10⁻² M
concentration of ![[CO_2]](https://tex.z-dn.net/?f=%5BCO_2%5D)
= 0.4442 M
Explanation:
Equation for the reaction:
⇄ 
+ 
Concentration of 
= 
= 0.469
For our ICE Table; we have:
⇄ +
Initial 0.469 0 0
Change - 2x +2x +x
Equilibrium (0.469-2x) 2x x
K = ![\frac{[CO]^2[O]}{[CO_2]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO%5D%5E2%5BO%5D%7D%7B%5BCO_2%5D%5E2%7D)
K = ![\frac{[2x]^2[x]}{[0.469-2x]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2x%5D%5E2%5Bx%5D%7D%7B%5B0.469-2x%5D%5E2%7D)
Since the value pf K is very small, only little small of reactant goes into product; so (0.469-2x)² = (0.469)²
![x=\sqrt[3]{1.9929*10^{-6}}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B1.9929%2A10%5E%7B-6%7D%7D)
x = 0.0124
∴ at equilibrium; concentration of = 0.0124 = 12.4 ×10⁻³ M
concentration of = 2x = 2 ( 0.0124)
= 0.0248
= 2.48 ×10⁻² M
concentration of = 0.469-2x
= 0.469-2(0.0124)
= 0.469 - 0.0248
= 0.4442 M
<span>You are given O2 and C3H8, this is a combustion
reaction. The chemical reaction is C3H8 + 10O2 à 3CO2 + 4H2O. You are also given the molar mass
of O2 which is 32.00 g/mol and C3H8 which is 44.1 g/mol. You are required to
find the mass of O2 in grams. Since you have the reaction, oe mole of C3H8 is
required to completely react 10 moles of O2. So,</span>
0.025g C3H8(1 mol C3H8/44.1 g C3H8)(10 mol O2/1
mol C3H8)(32 g O2/1 mol O2) = <u>0.1802 g O2
</u>
<span> </span>
