The moles of butane gas and oxygen gas reacted if 2.50 moles of H2O is produced is calculated as below
the equation for reaction
2C4H10 +13 O2 = 8CO2 +10 H2O
the moles of butane (C4H10) reacted calculation
by use of mole ratio between C4H10: H2O which is 2 : 10 the moles of C4H10= 2.50 x2/10 = 0.5 moles of C4H10 reacted
The moles of O2 reacted calculation
by use of mole ratio between O2 : H2O which is 13:10 the moles of O2
= 2.50 x 13/10= 3.25 moles of O2 reacted
I think so, <span> how much of all chemistry is to do with numbers? A lot! So being good at mathematics makes a huge aspect of chemistry so much easier but if you aren't mathematical then you can work at it and become reasonably proficient at it.</span>
Answer:
i think the answer is D-Transpirtation
Answer:
Hydration (of an alkene)
Mechanism : Electrophilic addition.
Answer:
The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.
Explanation:
Consider the ICE take for the solubility of the solid, CuF₂ as:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - -
At t =equilibrium (x-s) s 2s
The expression for Solubility product for CuF₂ is:
![K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5Cleft%20%5B%20Cu%5E%7B2%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20F%5E-%20%5Cright%20%5D%5E2)


Given s = 7.4×10⁻³ M
So, Ksp is:


Ksp = 1.6209×10⁻⁶
Now, we have to calculate the solubility of CuF₂ in NaF.
Thus, NaF already contain 0.20 M F⁻ ions
Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - 0.20
At t =equilibrium (x-s') s' 0.20+2s'
The expression for Solubility product for CuF₂ is:
![K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5Cleft%20%5B%20Cu%5E%7B2%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20F%5E-%20%5Cright%20%5D%5E2)

Solving for s', we get
<u>s' = 4.0×10⁻⁵ M</u>
<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.</u>