Molarity of aqueous solution is 1.9 M
The molarity in given case can be calculated by the formula -
M = number of moles/Volume of solution in liters
Number of moles is further calculated by the formula -
Number of moles = mass/molar mass
As we know, molar mass of ethanol is 46 g/mol
Keep the values in formula to calculate number of moles.
Number of moles =
Number of moles = 0.02
As we know, 1000 ml = 1l
So, 10.5 ml = 0.0105 l
Keep the values in formula of molarity to calculate the molarity -
Molarity =
Molarity = 1.9 M
Hence, the molarity of given aqueous solution is 1.9 M.
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NH₃:
N = 8*10²²
NA = 6.02*10²³
n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol
O₂:
N=7*10²²
NA = 6.02*10²³
n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol
4NH₃ <span>+ 3O</span>₂ ⇒<span> 2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant
N₂:
n = 0.0665mol
M = 28g/mol
m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>
Answer:
NaOH → 0.015 M
Explanation:
We propose the neutralization formula to solve the problem:
N acid . Volume acid = N base . Volume base
N means normality. A sort of concentration that is defined as:
M / val where the val means the number of H⁺ and OH⁻ from the acid or the base.
In this case, we have NaOH and HCl, so the M = N
Let's replace data:
0.01 M . 22.4 mL = 15 mL . M NaOH
0.01 M . 22.4 mL / 15 mL = M NaOH → 0.015 M
A species whose oxidation number increases in a reaction is oxidized (reducing reagents or reactant).
For example, balanced chemical
reaction: 2KClO₃ → 2KCl + 3O₂<span>.
</span><span>Oxygen in the
reactant state has oxidation number -2 (+1 + 5 + 3 · x = 0) and it is
oxidized to 0 (in molecule of oxygen) in the product state.</span>