We find the weight of the empirical formula:
12.0107 + 2 x 1.00794 + 15.9994
= 30.03

Now, we divide the molecular weight by the weight of the empirical formula to find the number of times the empirical formula repeats:
90.09 / 30.03
= 3

The formula is 3(CH₂O)
C₃H₆O₃

6 0

4 years ago

Write a balanced net ionic equation for the reaction of aqueous solutions of baking soda (NaHCO3) and acetic acid.(A) HCO3–(aq)

IgorC [24]

Answer: The correct answer is Option C.

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of sodium bicarbonate and acetic acid is given as:

$NaHCO_3(aq.)+CH_3COOH(aq.)\rightarrow CH_3COONa(aq.)+H_2O(l)+CO_2(g)$

Ionic form of the above equation follows:

$Na^+(aq.)+HCO_3^-(aq.)+CH_3COO^-(aq.)+H^+(aq.)\rightarrow CH_3COO^-(aq.)+Na^+(aq.)+H_2O(l)+CO_2(g)$

As, sodium and acetate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$HCO_3^-(aq.)+H^+(aq.)\rightarrow CO_2(g)+H_2O(l)$

Hence, the correct answer is Option C.

8 0

3 years ago

Which metal is more active than H2? (1) Ag (3) Cu (2) Au (4) Pb

astraxan [27]

Answer is (4) - Pb.

According to the reactivity series of elements

- the elements which are above the hydrogen are more reactive than hydrogen.

- the elements which are below the hydrogen are less reactive than hydrogen.

Among the given choices, only Pb is placed above the hydrogen in the reactivity series and rest are below the hydrogen.

Hence, Pb is more active than hydrogen.

5 0

3 years ago

The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=k[H2][I2] Some proposed mechanism

MatroZZZ [7]

Answer:

Mechanism A and B are consistent with observed rate law

Mechanism A is consistent with the observation of J. H. Sullivan

Explanation:

In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.

In the proposed mechanisms:

Mechanism A

(1) H2(g)+I2(g)→2HI(g)(one-step reaction)

Mechanism B

(1) I2(g)⇄2I(g)(fast, equilibrium)

(2) H2(g)+2I(g)→2HI(g) (slow)

Mechanism C

(1) I2(g) ⇄ 2I(g)(fast, equilibrium)

(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)

(3) H(g)+I(g)→HI(g) (fast)

The rate laws are:

A: rate = k₁ [H2] [I2]

B: rate = k₂ [H2] [I]²

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]

Where K' = K1 * K2

C: rate = k₁ [H2] [I]

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]^1/2

Thus, just mechanism A and B are consistent with observed rate law

In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just mechanism A is consistent with the observation of J. H. Sullivan

5 0

3 years ago