Molecular weight of AgBr = 187.7
moles of Ag =
moles of Br = moles of Ag = 2.96 x 10⁻³ mol
concentration of HBr (Molarity) = conc. of Br (strong acid) =
Noble gases
Explanation:
Electronic configuration 1s² 2s² 2p⁶
The element belongs to the group of the noble gases.
- The noble gases have complete outer shell configuration of their atoms.
- we can infer that the configuration above is for an element in the p-block because the last sub-level filled is the p-orbital.
- The elements therefore belongs to the p-block
- The block is from group 111A to O
- Only the halogens and noble gases fits this picture from the option.
- The outer most p-subshell have three orbitals requiring 6 electrons to fill them up.
- This makes a complete and stable configuration.
- The highest energy level of 2 is also made up of 8 electrons, an octet.
- This is why we can conclude that they are noble gases.
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Many of the actual chemicals in flower petals that give them their different colors are called anthocyanins. These are water-soluble compounds that belong to a bigger class of chemicals known as flavonoids. Anthocyanins are responsible for creating the colors blue, red, pink, and purple in flowers.
Answer:
19.K, potassium
Explanation:
it has all properties of metals
Answer:
Theoretical yield of the reaction = 34 g
Excess reactant is hydrogen
Limiting reactant is nitrogen
Explanation:
Given there is 100 g of nitrogen and 100 g of hydrogen
Number of moles of nitrogen = 100 ÷ 28 = 3·57
Number of moles of hydrogen = 100 ÷ 2 = 50
Reaction between nitrogen and hydrogen yields ammonia according to the following chemical equation
N2 + 3H2 → 2NH3
From the above chemical equation for every mole of nitrogen that reacts, 3 moles of hydrogen will be required and 2 moles of ammonia will be formed
Now we have 3·57 moles of nitrogen and therefore we require 3 × 3·57 moles of hydrogen
⇒ We require 10·71 moles of hydrogen
But we have 50 moles of hydrogen
∴ Limiting reactant is nitrogen and excess reactant is hydrogen
From the balanced chemical equation the yield will be 2 × 3·57 moles of ammonia
Molecular weight of ammonia = 17 g
∴ Theoretical yield of the reaction = 2 × 3·57 × 17 = 121·38 g
