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2 years ago

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. For the following example, identify the following. H2O(l) → H2O(g) A) a positive ΔH and a positive ΔS B) a negative ΔH and a n

egative ΔS C) a positive ΔH and a negative ΔS D) a negative ΔH and a positive ΔS E) It is not possible to determine without more in

1 answer:

IrinaK [193]2 years ago

4 0

Answer:

A) a positive ΔH and a positive ΔS

Explanation:

Entropy -

In a system, the randomness is measured by the term entropy .

Randomness basically refers as a form of energy that can not be used for any work.

The change in entropy is given by amount heat per change in temperature.

When solid is converted to gas entropy increases,

As the molecules in solid state are tightly packed and has more force of attraction between the molecules, but as it is converted to gas, the force of attraction between the molecule decreases and hence entropy increases.

So,

The particles of the substance , if are tightly held by strong force of attraction will decrease the entropy ,

And

If the particles are loosely held , the entropy will increase .

And , hence the value for ΔS will be positive .

The process of conversion of solid to gas , require some energy , and hence , is a endothermic process ,

Therefore ,

ΔH is positive .

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How are exothermic and endothermic reactions linked in the process of refining metal ore?

Alex777 [14]

The intended sense is that of a reaction that depends on absorbing heat if it is to proceed. The opposite of an endothermic process is an exothermic process, one that releases "gives out" energy in the form of heat

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2 years ago

Determine the empirical formula for compounds that have the following analyses: a. 66.0% barium and 34.0% chlorine b. 80.38% bis

almond37 [142]

Answer: a) $BaCl_2$

b) $BiO_3H_3$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

a) Mass of Ba= 66.06 g

Mass of Cl = 34.0 g

Step 1 : convert given masses into moles.

Moles of Ba = $\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{66.06g}{137g/mole}=0.48moles$

Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Ba = $\frac{0.48}{0.48}=1$

For O = $\frac{0.96}{0.48}=2$

The ratio of Ba: Cl= 1:2

Hence the empirical formula is $BaCl_2$

b) Mass of Bi= 80.38 g

Mass of O= 18.46 g

Mass of H = 1.16 g

Step 1 : convert given masses into moles.

Moles of Bi = $\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{80.38g}{209g/mole}=0.38moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{18.46g}{16g/mole}=1.15moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.16g}{1g/mole}=1.16moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Bi= $\frac{0.38}{0.38}=1$

For O = $\frac{1.15}{0.38}=3$

For H= $\frac{1.16}{0.38}=3$

The ratio of Bi: O: H= 1:3: 3

Hence the empirical formula is $BiO_3H_3$

6 0

2 years ago

1.05 Quiz: Measure Angles

Lelechka [254]

There’s no question for me to answer ?

6 0

2 years ago

Read 2 more answers

ethylene glycol used in automobile antifreeze and in the production of polyester. The name glycol stems from the sweet taste of

Luden [163]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=9.06g$

Mass of $H_2O=5.58g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 9.06 g of carbon dioxide, $\frac{12}{44}\times 9.06=2.47g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 5.58 g of water, $\frac{2}{18}\times 5.58=0.62g$ of hydrogen will be contained.

Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.206 moles.

For Carbon = $\frac{0.206}{0.206}=1$

For Hydrogen = $\frac{0.62}{0.206}=3$

For Oxygen = $\frac{0.206}{0.206}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

Hence, the empirical formula for the given compound is $C_1H_{3}O_1=CH_3O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 124 amu = 124 g/mol

Mass of empirical formula = 31 g/mol

Putting values in above equation, we get:

$n=\frac{124g/mol}{31g/mol}=4$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 4)}H_{(3\times 4)}O_{(1\times 4)}=C_4H_{12}O_4$

Thus, the empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

3 0

2 years ago

The compound AX2 decomposes according to the equation, 2 AX2(g) => 2 AX(g) + X2(g). In one experiment, AX2 was measured at va

Karolina [17]

Answer:

0.0011 mol/L.s

Explanation:

The average rate of disappearing of the reagent is the variation of the concentration of it divided by the time that this variation is being measured. The reaction rate, is proportional to the coefficient of the substance, so, for a generic reaction:

aA + bB --> cC + dD

rate = -(1/a)Δ[A]/Δt = -(1/b)Δ[B]/Δt = (1/c)Δ[C]/Δdt = (1/d)Δ[D]/dt

The minus sign is because of the reagent is desapering, so:

rate = -(1/2)*(0.0209 - 0.0300)/(10 - 6)

rate = 0.0011 mol/L.s

8 0

2 years ago