Answer:
88.46%
Explanation:
Percentage yield is actual/theoretical * 100
138/156 * 100 = 88.4615385
Explanation:
Groups are numbered 1–18 from left to right. The elements in group 1 are known as the alkali metals; those in group 2 are the alkaline earth metals; those in 15 are the pnictogens; those in 16 are the chalcogens; those in 17 are the halogens; and those in 18 are the noble gases.
Cell theory states that all the living organisms are made up of basic structural and functional units, cells. All cells divide and give rise to new cells. All the cells have similar chemical composition. The nuclei acid DNA present in the chromosomes helps in carrying hereditary information from generation to generation. So, in order to sustain life the cells must contain DNA either enclosed in the nucleus or freely floating in the cytoplasm.
Answer:
D. ![K_{a} = \frac{[\text{H}^{+}][\text{NO}_{2}^{-}]}{[\text{HNO}_{2}]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%5B%5Ctext%7BNO%7D_%7B2%7D%5E%7B-%7D%5D%7D%7B%5B%5Ctext%7BHNO%7D_%7B2%7D%5D%7D)
Explanation:
The general form of an equilibrium constant expression is
![K = \frac{[\text{Products}]}{[\text{Reactants}]}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5B%5Ctext%7BProducts%7D%5D%7D%7B%5B%5Ctext%7BReactants%7D%5D%7D)
In the equilibrium
HNO₂ ⇌ H⁺ + NO₂⁻
The products are H⁺ and NO₂⁻, and the reactant is HNO₂.
∴
Answer:
Aluminum iodide (AlI₃)
Explanation:
The synthesis reaction of aluminum (Al) and iodine (I) can be illustrated as shown below:
Aluminium exhibit trivalent positive ion (Al³⁺)
Iodine exhibit univalent negative ion (I¯)
During reaction, there will be an exchange of ion as shown below:
Al³⁺ + I¯ —> AlI₃
Thus, we can write the balanced equation for the reaction as follow:
Al + I₂ —› AlI₃
There are 2 atoms of I on the left side and 3 atoms on the right side. It can be balance by putting 2 in front of AlI₃ and 3 in front of I₂ as shown below:
Al + 3I₂ —› 2AlI₃
There are 2 atoms of Al on the right side and 1 atom on the left side. It can be balance by putting 2 in front of Al as shown below:
2Al + 3I₂ —› 2AlI₃
Thus the equation is balanced.
The product on the reaction is aluminum iodide (AlI₃)
