Answer:
[H⁺] = 1.58 x 10⁻⁷ M.
Explanation:
∵ pOH = - log[OH⁻]
7.20 = - log[OH⁻]
log[OH⁻] = - 7.20
∴ [OH⁻] = 6.31 x 10⁻⁸.
∵ [H⁺][OH⁻] = 10⁻¹⁴.
∴ [H⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/(6.31 x 10⁻⁸) = 1.585 x 10⁻⁷ M.
Answer:
Ethynylcyclopropane is the stable isomer for given alkyne.
Explanation:
In order to solve this problem we will first calculate the number of Hydrogen atoms. The general formula for alkynes is as,
CₙH₂ₙ₋₂
Putting value on n = 5,
C₅H₂.₅₋₂
C₅H₈
Also, the statement states that the compound contains one ring therefore, we will subtract 2 hydrogen atoms from the above formula i.e.
C₅H₈ ------------(-2 H) ----------> C₅H₆
Hence, the molecular formula for given compound is C₅H₆
Below, 4 different isomers with molecular formula C₅H₆ are attached.
The first compound i.e. ethynylcyclopropane is stable. As we know that alkynes are sp hybridized. The angle between C-C-H in alkynes is 180°. Hence, in this structure it can be seen that the alkyne part is linear and also the cyclopropane part is a well known moiety.
Compounds 3-ethylcycloprop-1-yne, <u>cyclopentyne </u>and 3-methylcyclobut-1-yne are highly unstable. The main reason for the instability is the presence of triple bond in three, five and four membered ring. As the alkynes are linear but the C-C-H bond in these compound is less than 180° which will make them highly unstable.
w/w percentage <span>
= mass of the pure compound /
total mass of the sample x 100%
70% HNO₃
contains by mass means every 100 g of sample has 70 g of HNO₃.</span><span>
The mass of solution = 103.8 g
Hence the mass of HNO₃ = 103.8 g x 70%</span><span>
= 103.8 g x (70 / 100)
<span>
= 72.66 g = 72.7 g.</span></span>