Possibly wet and unstable
Answer: Fluorine
Explanation: It belongs in the same group as Bromine
Answer:
The standard enthalpy of formation of methanol is, -238.7 kJ/mole
Explanation:
The formation reaction of CH_3OH will be,
![C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?](https://tex.z-dn.net/?f=C%28s%29%2B2H_2%28g%29%2B%5Cfrac%7B1%7D%7B2%7DO_2%5Crightarrow%20CH_3OH%28g%29%2C%5CDelta%20H_%7Bformation%7D%3D%3F)
The intermediate balanced chemical reaction will be,
..[1]
..[2]
..[3]
Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:
We get :
..[1]
..[2]
[3]
The expression for enthalpy of formation of
will be,
![\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3](https://tex.z-dn.net/?f=%5CDelta%20H_%7Bformation%7D%3D%5CDelta%20H_1%2B2%5Ctimes%20%5CDelta%20H_2%2B%5CDelta%20H_3)
![\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)](https://tex.z-dn.net/?f=%5CDelta%20H%3D%28-393.5kJ%2Fmole%29%2B%28-571.6kJ%2Fmole%29%2B%28726.4kJ%2Fmole%29)
![\Delta H=-238.7kJ/mole](https://tex.z-dn.net/?f=%5CDelta%20H%3D-238.7kJ%2Fmole)
The standard enthalpy of formation of methanol is, -238.7 kJ/mole
Answer:
a metal and a nonmetal element
Explanation:
Answer:
There is 54.29 % sample left after 12.6 days
Explanation:
Step 1: Data given
Half life time = 14.3 days
Time left = 12.6 days
Suppose the original amount is 100.00 grams
Step 2: Calculate the percentage left
X = 100 / 2^n
⇒ with X = The amount of sample after 12.6 days
⇒ with n = (time passed / half-life time) = (12.6/14.3)
X = 100 / 2^(12.6/14.3)
X = 54.29
There is 54.29 % sample left after 12.6 days