Answer:
1.Metals
These are very hard except sodium
These are malleable and ductile pieces
These are shiny
Electropositive in nature
Non-metals
These are soft except diamond
These are brittle and can break down into pieces
These are non-lustrous except iodine
Electronegative in nature
2. The electrochemical series helps to pick out substances that are good oxidizing agents and those which are good reducing agents.In an electrochemical series the species which are placed above hydrogen are more difficult to be reduced and their standard reduction potential values are negative.
3. Arrhenius theory, theory, introduced in 1887 by the Swedish scientist Svante Arrhenius, that acids are substances that dissociate in water to yield electrically charged atoms or molecules, called ions, one of which is a hydrogen ion (H+), and that bases ionize in water to yield hydroxide ions (OH−).
4. The common application of indicators is the detection of end points of titrations. The colour of an indicator alters when the acidity or the oxidizing strength of the solution, or the concentration of a certain chemical species, reaches a critical range of values.
If the results of the experiment on repeating are not same, it shows the results are not standard, there are some factors, which are not constant
Food code is like sort of a guide to ensure that the food is presented according decent standards, unadulterated and correctly presented to the customer.
It is published every four years by the " U.S Food and Drug Administration".
No. The number of a protons is not equal to it's atomic weight, instead it is equal to the 'atomic number'
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)
