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Elodia [21]
3 years ago
13

Two samples of sodium chloride with different masses were decomposed into their constituent elements. One sample produced 2.55 g

of sodium and 3.93 g of chlorine. Being consistent with the law of constant composition, also called the law of definite proportions or law of definite composition, which set of masses could be the result of the decomposition of the other sample
Chemistry
1 answer:
Charra [1.4K]3 years ago
3 0

Answer:

4.71 g of sodium and 7.25 g of chlorine

Explanation:

<em>Note: The question is incomplete. the complete question is given below:</em>

<em>Two samples of sodium chloride with different masses were decomposed into their constituent elements. One sample produced 2.55 g of sodium and 3.93 g of chlorine. Being consistent with the law of constant composition, also called the law of definite proportions or law of definite composition, which set of masses could be the result of the decomposition of the other sample?</em>

<em>4.71 g of sodium and 3.30 g of chlorine</em>

<em>4.71 g of sodium and 7.25 g of chlorine</em>

<em>4.71 g of sodium and 1.31 g of chlorine</em>

<em>4.71 g of sodium and 13.7 g of chlorine</em>

The law of definite proportions states that all pure samples of a particular chemical compound contain the same elements combined in the same proportion by mass.

This means that irrespective of the source of any sample of a pure chemical compound, the constituents elements are always combined in the same mass ratio.

In the first sample, the mass ratio of Sodium to chlorine is given below:

mass of sodium = 2.55 g

mass of chlorine = 3.93 g

mass ratio; sodium to chlorine = 2.55 / 3.93 = 0.65

From the set of masses give above, we can determine the result of the decomposition of the second sample.

4.71 g of sodium and 3.30 g of chlorine

mass ratio; sodium to chlorine = 4.71 / 3.30 = 1.43

4.71 g of sodium and 7.25 g of chlorine

mass ratio; sodium to chlorine = 4.71 / 7.25 = 0.65

4.71 g of sodium and 1.31 g of chlorine

mass ratio; sodium to chlorine = 4.71 / 1.31 = 3.59

4.71 g of sodium and 13.7 g of chlorine

mass ratio; sodium to chlorine = 4.71 / 13.7 = 0.34

From the results above, the correct set of masses for the second sample is 4.71 g of sodium and 7.25 g of chlorine

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artcher [175]

Answer:

Keq=0.866

Explanation:

Hello,

In this case, the undergone chemical reaction is:

N_2O_42NO_2

In such a way, since 0.0055 mol of N₂O₄ remains in the flask, one infers that the reacted amount (x) was:

x=0.04mol-0.0055mol=0.0345mol

In addition, the produced amount of NO₂ is:

2*0.0345mol=0.069mol

Finally, considering the flask's volume, the equilibrium constant is then computed as follows:

Keq=\frac{(2*0.0345M)^2}{0.0055M}=0.866

Best regards.

6 0
3 years ago
If the [H+] = 0.01 M, what is the pH of the solution, and is the solution a strong acid, weak acid, strong base, or weak base?
lana66690 [7]

Answer:

2, strong acid

Explanation:

Data obtained from the question. This includes:

[H+] = 0.01 M

pH =?

pH of a solution can be obtained by using the following formula:

pH = –Log [H+]

pH = –Log 0.01

pH = 2

The pH of a solution ranging between 0 and 6 is declared to be an acid solution. The smaller the pH value, the stronger the acid.

Since the pH of the above solution is 2, it means the solution is a strong acid.

5 0
3 years ago
The reaction N O space plus thin space O subscript 3 space rightwards arrow space N O subscript 2 space plus thin space O subscr
nikklg [1K]

Answer:

4.3 × 10⁻⁵ M s⁻¹

Explanation:

Step 1: Given data

  • Rate constant (k):  2.20 × 10⁷ M⁻¹s⁻¹
  • Concentration of NO ([NO]): 3.3 × 10⁻⁶ M
  • Concentration of O₃ ([O₃]): 5.9 × 10⁻⁷ M
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Step 2: Write the balanced reaction

NO + O₃ ⇒ NO₂ + O₂

Step 3: Calculate the reaction rate

The rate law is:

rate = k × [NO] × [O₃]

rate = 2.20 × 10⁷ M⁻¹s⁻¹ × 3.3 × 10⁻⁶ M × 5.9 × 10⁻⁷ M

rate = 4.3 × 10⁻⁵ M s⁻¹

6 0
4 years ago
Consider the following equilibrium system involving SO2, Cl2, and SO2Cl2 (sulfuryl dichloride): SO2(g) + Cl2(g) ⇌ SO2Cl2(g) Pred
Andrei [34K]

Answer:

(a) forward direction

(b) forward direction

(c) backward direction.

Explanation:

Given , the chemical reaction in equilibrium is,

SO₂(g)  + Cl₂(g)  ⇄  SO₂Cl₂ (g)

The direction of the reaction by changing the concentration can be determined by Le Chatelier's principle,

It states that ,

When a reaction is at equlibrium , Changing the concentration , pressure,  temperature disturbs the equilibrium , and the reaction again tries to attain equilibrium by counteracting the changes.

(a)

For the reaction , Cl₂ is added to the system , i.e. , increasing the concentration of Cl₂ ,Now, according to Le Chatelier , The reaction will move in forward direction , to reduce the increased amount of Cl₂.

Hence, reaction will go in forward direction.

(b)

Removing SO₂Cl₂ from the system ,i.e. , decreasing the concentration of SO₂Cl₂  , according to Le Chatelier , the reaction will move in forward direction , to increase the amount of reduced SO₂Cl₂.

Hence, reaction will go in forward direction.

(c)

Removing SO₂ from the system , i.e. decreasing the concentration of SO₂ ,  according to Le Chatelier , the reaction will move in backward direction , to increase the amount of reduced SO₂.

Hence, reaction will go in backward direction.

3 0
3 years ago
The normal boiling point of liquid ethyl acetate is 350 K. Assuming that its molar heat of vaporization is constant at 34.4 kJ/m
Roman55 [17]

Answer:

337.22 K

Explanation:

Given that:

P₁ = 1 atm

T₁ = 350 K

P₂ = 0.639 atm

T₂ = ??? (unknown)

R(rate constant) =  8.34 J k⁻¹ mol⁻¹

Using Clausius-Clapeyron equation, we can determine the final boiling point of the process.

Clausius-Clapeyron equation can be written as:

In\frac{P_2}{P_1}=\frac{\delta H_{vap}}{R}[\frac{T_2-T_1}{T_2T_1}]

Substituting our values given; we have:

In\frac{0.639}{1}=(\frac{34.4*10^3J/mol}{8.314 J K^{-1}mol^{-1}})[\frac{T_2-350}{350T_2}]

In({0.639})=(\frac{34.4*10^3}{8.314K^{-1}})[\frac{T_2-350}{350T_2}]

- 0.4479 = 41317.599 [\frac{T_2-350}{350T_2} ]K

-\frac{0.4479}{4137.599} = [\frac{T_2-350}{350T_2} ]

- 1.0825118*10^{-4} = [\frac{T_2-350}{350T_2} ]

- 1.0825118*10^{-4} *350T_2 =T_2-350

- 0.037889T_2=T_2-350

350 = 0.03789T_2+T_2

350 =1.03789T_2

T_2= \frac{350}{1.03789}

T_2 = 337.22K

∴ the boiling point of CH3COOC2H5 when the external pressure is 0.639 atm is <u>337.22</u> K.

5 0
3 years ago
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