Answer: E=∆H*n= -40.6kj
Explanation:
V(CO) =15L=0.015M³
P=11200Pa
T=85C=358.15K
PV=nRT
n=(112000×0.015)/(8.314×358.15)
n(Co)= 0.564mol
V(Co)= 18.5L = 0.0185m³
P=744torr=98191.84Pa
T= 75C = 388.15k
PV=nRT
n= (99191.84×0.0185)/(8.314×348.15)
n(H2) = 0.634mol
n(CH30H) =1/2n(H2)=1/2×0.634mol
=0.317mol
∆H =∆Hf{CH3OH}-∆Hf(Co)
∆H=-238.6-(-110.5)
∆H = 128.1kj
E=∆H×n=-40.6kj.
<u>Answer: </u>The reaction between iodine and strontium is less reactive than the reaction of bromine and strontium.
<u>Explanation:</u>
Iodine and bromine belong to the same group which is Group 17 of the periodic table. Iodine lies in Period 5 and bromine lies in Period 4.
Reactivity of non-metals is defined as the tendency of the elements to gain electrons easily. This depends on the electronegativity values. More the electronegativity, more will be the reactivity and vice-versa.
Reactivity of non-metals decreases from moving down the group and increases across the period.
The electronegativity of iodine is less than bromine. Hence, it is less reactive than the bromine.
The reaction between strontium and bromine is more favorable because it is the reaction between two reactive elements.
Hence, the reaction between iodine and strontium is less reactive than the reaction of bromine and strontium.
The biggest example is Pressure.
- If area will increase pressure decrease.
This compound contains these two nonmetal elements: carbon (C) and fluorine (F). Since this compound only contains two types of nonmetal elements, we will be using Type 3 naming (for binary compounds).
In type 3 naming we use prefixes for the second element named (in this case, fluorine) and the first element (carbon) is just left alone.
So, since we have four fluorine atoms, the prefix will be "tetra." Also, we turn the "ine" in fluorine into "ide"
So here is the name: carbon tetrafluoride.
(Note that carbon tetrafluoride is also known as: tetrafluoromethane)
