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3 years ago

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The balanced equation for the reaction of ammonia and oxygen is the following. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) The stand

ard molar entropies for the products and reactants are listed below. Calculate the change in standard molar entropy for this reaction at 298.0 K and standard pressure, in J/mol·K. (Enter your answer to the tenths place. Include the sign of the value in your answer.)

1 answer:

ale4655 [162]3 years ago

7 0

Answer:

ΔS° = 180.5 J/mol.K

Explanation:

Let's consider the following reaction.

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)

The standard molar entropy of the reaction (ΔS°) can be calculated using the following expression.

ΔS° = ∑np × S°p - ∑nr × S°r

where,

ni are the moles of reactants and products

S°i are the standard molar entropies of reactants and products

ΔS° = 4 mol × S°(NO(g)) + 6 × S°(H₂O(g)) - 4 mol × S°(NH₃(g)) - 5 mol × S°(O₂(g))

ΔS° = 4 mol × 210.8 J/K.mol + 6 × 188.8 j/K.mol - 4 mol × 192.5 J/K.mol - 5 mol × 205.1 J/K.mol

ΔS° = 180.5 J/K

This is the change in the entropy per mole of reaction.

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A metal sample weighing 129.00 grams and at a temperature of 97.8 degrees Celsius was placed in 45.00 grams of water in a calori

Nitella [24]

Answer : The specific heat of metal is $0.481J/g^oC$ .

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of metal = 129.00 g

$m_2$ = mass of water = 45.00 g

$T_f$ = final temperature = $39.6^oC$

$T_1$ = initial temperature of metal = $97.8^oC$

$T_2$ = initial temperature of water = $20.4^oC$

Now put all the given values in the above formula, we get

$129.00g\times c_1\times (39.6-97.8)^oC=-45.00g\times 4.184J/g^oC\times (39.6-20.4)^oC$

$c_1=0.481J/g^oC$

Therefore, the specific heat of metal is $0.481J/g^oC$ .

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3 years ago

If you are using 3.00% (mass/mass) hydrogen peroxide solution and you determine that the mass of solution required to reach the

Maurinko [17]

Answer:

0.004522 moles of hydrogen peroxide molecules are present.

Explanation:

Mass by mass percentage of hydrogen peroxide solution = w/w% = 3%

Mass of the solution , m= 5.125 g

Mass of the hydrogen peroxide = x

$w/w\% = \frac{x}{m}\times 100$

$3\%=\frac{x}{5.125 g}\times 100$

$x=\frac{3\times 5.125 g}{100}=0.15375 g$

Mass of hydregn pervade in the solution = 0.15375 g

Moles of hydregn pervade in the solution :

$=\fraC{ 0.15375 g}{34 g/mol}=0.004522 mol$

0.004522 moles of hydrogen peroxide molecules are present.

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2 years ago

A monoprotic weak acid when dissolved in water is 0.66% dissociated and produces a solution with a pH of 3.04. Calculate the Ka

raketka [301]

Answer:

Ka = 6.02x10⁻⁶

Explanation:

The equilibrium that takes place is:

HA ⇄ H⁺ + A⁻

Ka = [H⁺][A⁻]/[HA]

We <u>calculate [H⁺] from the pH</u>:

pH = -log[H⁺]

[H⁺] = $10^{-pH}$

[H⁺] = 9.12x10⁻⁴ M

Keep in mind that [H⁺]=[A⁻].

As for [HA], we know the acid is 0.66% dissociated, in other words:

[HA] * 0.66/100 = [H⁺]

We <u>calculate [HA]</u>:

[HA] = 0.138 M

Finally we <u>calculate the Ka</u>:

Ka = $\frac{[9.12x10^{-4}]*[9.12x10^{-4}]}{[0.138]}$ = 6.02x10⁻⁶

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2 years ago

The weather report says that if the temperature drops 10°, it will reach 25° below zero. The current temperature T can be found

Neporo4naja [7]

Answer:

The current temperature is -15° (15° below zero)

Explanation:

The temperature drops 10°:

T-10°

It will reach 25° below zero:

T - 10° = -25°

We add 10° in both members of the equation:

T - 10° +10° = -25° +10°

The equation is simplified as follows:

T = -25° +10°

T = -15°

The integer -15° can be expressed as 15° below zero.

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Answer:

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Explanation:

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