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Would it be C???...idk<br> HELP

givi [52]

Answer:

C. Butanal , is the aldehyde

Explanation:

A . It is carboxylic acid : ---COOH group

B. It is Ester : ----COOR group , Here R = CH3

C. It is Aldehyde : -----CHO group

D. It is ketone : ----C=O group

See image :

8 0

3 years ago

How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56

neonofarm [45]

Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 0.0156

Thus,

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.0156=e^{-k\times t}$

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}$

$t = 6\times t_{1/2}$

<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>

6 0

3 years ago

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

3 years ago

A gas occupies 4.30 Liters at STP. What volume does it occupy at 2.56 atm and 302

laiz [17]

Answer:

V₂ = 1.86 L

Explanation:

Given data:

Initial volume = 4.30 L

Initial pressure = 1 atm

Initial temperature = 273.15 K

Final temperature = 302 K

Final volume = ?

Final pressure = 2.56 atm

Solution:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁T₂ /T₁ P₂

V₂ = 1 atm ×4.30 L × 302 K / 273.15 K × 2.56 atm

V₂ = 1298.6 atm.L.K / 699.26 K.atm

V₂ = 1.86 L

4 0

3 years ago

A student had a sample of pure water, and added an unknown substance to it. The student noticed that the hydroxide ion concentra

steposvetlana [31]

The answer is A it's a basic because once you add another substance to a neutral it either becomes acidic or basic. this one becomes basic because the hydroxide ion concentrate increased.

6 0

3 years ago

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