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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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How many moles of FeCI3 are present in 345.0 g FeCI3

Sphinxa [80]

Answer:

$\boxed {\boxed {\sf About \ 2.127 \ moles \ of \ FeCl_3}}$

Explanation:

To convert from moles to grams, the molar mass must be used.

1. Find Molar Mass

The compound is iron (III) chloride: FeCl₃

First, find the molar masses of the individual elements in the compound: iron (Fe) and chlorine (Cl).

Fe: 55.84 g/mol

Cl: 35.45 g/mol

There are 3 atoms of chlorine, denoted by the subscript after Cl. Multiply the molar mass of chlorine by 3 and add iron's molar mass.

FeCl₃: 3(35.45 g/mol)+(55.84 g/mol)=162.19 g/mol

This number tells us the grams of FeCl₃ in 1 mole.

2. Calculate Moles

Use the number as a ratio.

$\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}$

Multiply by the given number of grams, 345.0.

$345.0 \ g \ FeCl_3 *\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}$

Flip the fraction so the grams of FeCl₃ will cancel.

$345.0 \ g \ FeCl_3 *\frac{1 \ mol \ FeCl_3}{162.19 \ g \ FeCl_3}$

$345.0 *\frac{1 \ mol \ FeCl_3}{162.19 }$

$\frac{345.0 \ mol \ FeCl_3}{162.19 }$

Divide.

$2.12713484 \ mol \ FeCl_3$

3. Round

The original measurement of grams, 345.0, has 4 significant figures. We must round our answer to 4 sig figs.

For the answer we calculated, that is the thousandth place.

The 1 in the ten thousandth place tells us to leave the 7 in the thousandth place.

$\approx 2.127 \ mol \ FeCl_3$

There are about <u>2.127 mole</u>s of iron (III) chloride in 345.0 grams.

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2 years ago

A 44.0 g sample of an unknown metal at 99.0 oC was placed in a constant-pressure calorimeter of negligible heat capacity contain

tatiyna

Answer:

$C_m=0.474\frac{J}{g\°C}$

Explanation:

Hello.

In this case, since this is a system in which the water is heated up and the metal is cooled down in a calorimeter which is not affected by the heat lose-gain process, we can infer that the heat lost by the metal is gained be water, it means that we can write:

$Q_m=-Q_w$

Thus, in terms of masses, specific heats and temperatures we can write:

$m_mC_m(T_{eq}-T_m)=-m_wC_w(T_{eq}-T_w)$

Whereas the equilibrium temperature is the given final temperature of 28.4 °C and we can compute the specific heat of the metal as shown below:

$C_m=\frac{-m_wC_w(T_{eq}-T_w)}{m_m(T_{eq}-T_m)}$

Plugging the values in and since the density of water is 1.00 g/mL so the mass is 80.0g, we obtain:

$C_m=\frac{-80.0g*4.184\frac{J}{g\°C} (28.4\°C-24.0\°C)}{44.0g(28.4\°C-99.0\°C)}\\\\C_m=0.474\frac{J}{g\°C}$

Best regards!

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2 years ago