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3 years ago

Question 3 (1 point)

Chemistry

2 answers:

bazaltina [42]3 years ago

5 0

Explanation:

Question 3:

Mass of anhydrous compound = 7.26g

Mass of water = 8.50 - 7.26 = 1.24 g

Percentage of water of crystallization in the compound is = (1.24 / 8.50) * 100 = 14.59%

The mass ratio of Na2CO3 : H2O = 7.26 : 1.24

Using mole = mass / molar mass;

The mole ratio of Na2CO3 : H2O = 7.26 /105.99 : 1.24 / 18 = 0.069 : 0.069 or 1 : 1

The formular is; Na2CO3.H2O

Question 4:

Mass of anhydrous compound = 48.23 g

Mass of water = 60.98 - 48.23 = 12.75 g

Percentage of water of crystallization in the compound is = (12.75 / 60.98) * 100 = 20.91%

The mass ratio of CaSO4 : H2O = 48.23 : 12.75

Using mole = mass / molar mass;

The mole ratio of CaSO4 : H2O = 48.23 / 136.14 : 12.75 / 18 = 0.3543 : 0.708 which is 1 : 2

The formular is; CaSO4.2H2O

larisa86 [58]3 years ago

4 0

Answer:

1. Formulae of hydrate is Na₂CO₃.10H₂O

2. Formulae of hydrate is CaSO₄.2H₂O

Explanation:

Q1. Mass of hydrated salt = 8.50 g mass

mass of anhydrous salt = 7.26 grams

mass of water lost = 1.24

Formula of hydrated salt = Na₂CO₃.xH₂O

Formula of anhydrous salt = Na₂CO₃

Molar mass of anhydrous salt is obtained as below where Na = 23, C = 12, O = 16, H = 1

Molar mass of Na₂CO₃ = (23 *2 + 12 + 16 * 3) = 106 g

molar mass of water = (1 *2 + 16) = 18 g

Mole ratio of Na₂CO₃ to H₂O = 7.26/106 : 1.24/18

Mole ratio of Na₂CO₃ to H₂O = 0.0068 : 0.068

Mole ratio of Na₂CO₃ to H₂O = 1 :10

Therefore, formulae of hydrate is Na₂CO₃.10H₂O

Q2. Mass of hydrated salt = 60.98 g mass

mass of anhydrous salt = 48.23 grams

mass of water lost = 12.75

Formula of hydrated salt = CaSO₄.xH₂O

Formula of anhydrous salt = CaSO₄

Molar mass of anhydrous salt is obtained as below where Ca = 40, S = 32, O = 16, H = 1

Molar mass of CaSO₄= (40 + 32 + 16 * 4) = 136 g

molar mass of water = (1 *2 + 16) = 18 g

Mole ratio of CaSO₄ to H₂O = 48.23/136 : 12.75/18

Mole ratio of CaSO₄ to H₂O = 0.35 : 0.70

Mole ratio of CaSO₄ to H₂O = 1 : 2

Therefore, formulae of hydrate is CaSO₄.2H₂O

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1 year 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

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3 years ago

At a certain temperature the rate of this reaction is first order in HI with a rate constant of :0.0632s

Shalnov [3]

Answer : The time taken for the reaction is, 28 s.

Explanation :

Expression for rate law for first order kinetics is given by :

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = 0.0632

t = time taken for the process = ?

$[A_o]$ = initial amount or concentration of the reactant = 1.28 M

$[A]$ = amount or concentration left time 't' = $1.28\times \frac{17}{100}=0.2176M$

Now put all the given values in above equation, we get:

$0.0632=\frac{2.303}{t}\log\frac{1.28}{0.2176}$

$t=28s$

Therefore, the time taken for the reaction is, 28 s.

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