+1
Explanation:
To solve this problem, we need to set up an algebraic equation. Let us first understand the meaning of oxidation number.
The oxidation number is the formal charge assigned to an atom present in a molecule or formula unit
The algebraic sum of all oxidation numbers of atoms in an ion containing more than one kind of atom is the charge on the ion.
The algebraic sum of all oxidation number of atoms in a neutral compound is zero;
The radical NO₃ has a formal charge of -1;
let the oxidation number of Li = x
x + (-1) = 0
x = + 1
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Answer:
M HCl sln = 12.0785 M
Explanation:
- molarity (M) [=] mol/L
- %mm = ((mass compound)/(mass sln))*100
∴ mass sln = 100.0 g
∴ δ sln = 1.19 g/mL
∴ % m/m = 37 %
⇒ 37 % =((mass HCl/mass sln))*100
⇒ 0.37 = mass HCl / 100.0 g
⇒ 37 g = mass HCl
∴ molar mass HCl = 36.46 g/mol
⇒ mol HCl = (37 g)*(mol/36.46 g) = 1.015 mol
⇒ volume sln = (100 g sln)*(mL/1.19 g) = 84.034 mL = 0.084034 L
⇒ M HClsln = 1.015 mol/0.084034 L
⇒ M HCl sln = 12.0785 M
Answer:
The chemical equation needs to be balanced so that it follows the law of conservation of mass.
Explanation:
Answer:
We can do the nitration of benzene by treating the benzene with a mixture of nitric acid and sulphuric acid by not extending the temperature of 50°C
Explanation:
Nitration of benzene takes place by treating the benzene with a mixture of nitric acid and sulphuric acid at low temperatures such as the temperatures below 50°C
The nitration of benzene takes place through electrophilic substitution reaction
In this reaction the electrophile is nitronium ion (NO2+) which performs an electrophilic substitution reaction on the benzene ring and during the reaction an intermediate will also be formed in which there will be positive charge distributed in the benzene
These electrophile is generated when nitric acid is treated with sulphuric acid
As nitric acid is a strong oxidising agent, here in this case the oxidation state of nitrogen will change from +5 to +3
The reactions regarding the nitration of benzene is present in the file attached
Answer:
i assume that it would be a gametophyte.
Explanation:
