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USPshnik [31]
2 years ago
13

Sulfur reacts with fluorine to produce three different compounds. The mass ratio of fluorine to sulfur for each compound is give

n in the following table:
Compound mass F : mass S
S2F10 2.962
SF4 2.370
SF6 3.555
Show that these data are consistent with the law of multiple proportions.
Chemistry
1 answer:
NARA [144]2 years ago
8 0

Answer:

Explanation:

Mass of F / Mass of S = 2.962/1 =2.962 X 32 / 32 = 94.78/32

Mass of F / Mass of S = 2.370 /1 = 2.370 X 32 / 32 = 75.84 /32

Mass of F /Mass of S = 3.555/1 = 3.555 x 32 / 32 = 113.76 / 32 .

Now constant mass of S that is 32 g reacts with different mass of F. They are as follows :

94.78 g , 75.84 g , and 113.76 g

Their ratio = 94.78 : 75.84 : 113.76

divide them by 19

their ratio = 5 : 4 : 6

So these data are consistent with law of multiple proportion.

You might be interested in
Which response includes all of the following processes that are accompanied by an increase in entropy?
dlinn [17]

Answer:

1 and 3.

Explanation:

The entropy measures the randomness of the system, as higher is it, as higher is the entropy. The randomness is associated with the movement and the arrangement of the molecules. Thus, if the molecules are moving faster and are more disorganized, the randomness is greater.

So, the entropy (S) of the phases increases by:

S solid < S liquid < S gases.

1. The substance is going from solid to gas, thus the entropy is increasing.

2. The substance is going from a disorganized way (the molecules of I are disorganized) to an organized way (the molecules join together to form I2), thus the entropy is decreasing.

3. The molecules go from an organized way (the atom are joined together) to a disorganized way, thus the entropy increases.

4. The ions are disorganized and react to form a more organized molecule, thus the entropy decreases.

7 0
3 years ago
A volume of 90.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What
Gemiola [76]

Answer:

The answer to your question:  0.7 M

Explanation:

Data

V of KOH = 90 ml

[KOH] = ?

V H2SO4 = 21.2 ml

[H2SO4] = 1.5 M

                       2KOH(aq)  +  H₂SO₄(aq)   →   K₂SO₄(aq)  +  2H₂O(l)

Molarity = moles / volume

moles of H₂SO₄ = (1.5) (21.2)

                           = 31.8

                    2 moles of KOH --------------  1 mol of H₂SO₄

                   x                           --------------  31.8 mol of H₂SO₄

                    x = (31.8)(2) / 1

                    x = 63.8 moles of KOH

Molarity = 63.8 / 90

             = 0.7 M

7 0
2 years ago
If an equal number of moles of the weak acid HF and the strong base KOH are added to water, is the resulting solution acidic, ba
s344n2d4d5 [400]

Answer:

Being a weak acid and a strong base, where it is diluted in a neutral medium such as water, the basic medium predominates, almost alkaline pH.

Explanation:

The acidity of the solution, being weak, means that its pH is not so low, therefore it will be easier to reach the values of 7 or 7 where alkalinity or basity is indicated.

6 0
2 years ago
Is this an example of decomposition<br> 2CH4 + 402 → 2C02 + 4H2O
Molodets [167]

Answer:

Identify each equation as a composition reaction, a decomposition reaction, or neither.

Fe2O3 + 3 SO3 → Fe2(SO4)3

NaCl + AgNO3 → AgCl + NaNO3

(NH4)2Cr2O7 → Cr2O3 + 4 H2O + N2

Solution

In this equation, two substances combine to make a single substance. This is a composition reaction.

Two different substances react to make two new substances. This does not fit the definition of either a composition reaction or a decomposition reaction, so it is neither. In fact, you may recognize this as a double-replacement reaction.

A single substance reacts to make multiple substances. This is a decomposition reaction.

Test Yourself

Identify the equation as a composition reaction, a decomposition reaction, or neither.

C3H8 → C3H4 + 2 H2

Explanation:

I hope I help :)))

5 0
3 years ago
Calculate the ph at of a solution of sodium hypochlorite . note that hypochlorous acid is a weak acid with a of . round your ans
loris [4]

The question is incomplete, here is the complete question:

Calculate the pH at 25°C of a 0.39 M solution of sodium hypochlorite NaClO. Note that hypochlorous acid HClO is a weak acid with a pKa of 7.50. Round your answer to 1 decimal place.

<u>Answer:</u> The pH of the solution is 10.4

<u>Explanation:</u>

We are given:

Molarity of sodium hypochlorite = 0.39 M

pK_a of HClO = 7.50

We know that:

pK_a=-\log K_a

K_a  of HClO = 10^{-7.50}=3.16\times 10^{-8}

To calculate the base dissociation constant for the given acid dissociation constant, we use the equation:

K_w=K_b\times K_a

where,

K_w = Ionic product of water = 10^{-14}

K_a = Acid dissociation constant  = 3.16\times 10^{-8}

K_b = Base dissociation constant

Putting values in above equation, we get:

10^{-14}=3.16\times 10^{-8}\times K_b\\\\K_b=\frac{10^{-14}}{3.16\times 10^{-8}}=3.16\times 10^{-7}

The chemical equation for the reaction of hypochlorite ion with water follows:

                    ClO^-+H_2O\rightarrow HClO+OH^-

<u>Initial:</u>           0.39

<u>At eqllm:</u>      0.39-x                   x           x

The expression of K_b for above equation follows:

K_b=\frac{[HClO][OH^-]}{[ClO^-]}

Putting values in above equation, we get:

3.16\times 10^{-7}=\frac{x\times x}{(0.39-x)}\\\\x=-0.00035,0.00035

Neglecting the negative value of 'x' because concentration cannot be negative

To calculate the pOH of the solution, we use the equation:

pOH=-\log[OH^-]

We are given:

[OH^-]=0.00035M

Putting values in above equation, we get:

pOH=-\log (0.00035)=3.6

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH=14-3.6=10.4

Hence, the pH of the solution is 10.4

3 0
2 years ago
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