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Hitman42 [59]
3 years ago
6

Standard Thermodynamic Quantities for Selected Substances at 25 ∘C Reactant or product ΔHf∘(kJ/mol) Al(s) 0.0 MnO2(s) −520.0 Al2

O3(s) −1675.7 Mn(s) 0.0 Part A The thermite reaction, in which powdered aluminum reacts with manganese oxide, is highly exothermic. 4Al(s)+3MnO2(s)→2Al2O3(s)+3Mn(s) Use standard enthalpies of formation to find ΔH∘rxn for the thermite reaction.
Chemistry
1 answer:
Svet_ta [14]3 years ago
4 0

Answer:

-1815.4 kJ/mol

Explanation:

Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:

∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)

This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.

In this case:

ΔH rxn = [(2 * -1675.7) - (3 * -520.0)] kJ/mol = -1815.4 kJ/mol

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I am Lyosha [343]

Answer:

The substance with a weaker molecular attraction

4 0
2 years ago
1) A common experiment to determine the relative reactivity of metallic elements is to place a pure sample of one metal into an
Annette [7]

Answer: Zn(s)+CuSO_4(aq)\rightarrow ZnSO_4+Cu

Explanation:-

Single replacement reaction is a chemical reaction in which more reactive element displaces the less reactive element from its salt solution.

As zinc is more reactive than copper, it could easily displace copper from its aqueous solution and thus leads to formation of zinc (II) sulfate and pure copper.

The chemical reaction can be represented as :

Zn(s)+CuSO_4(aq)\rightarrow ZnSO_4+Cu

The phases are represented as (s) for solid sate, (l) for liquid state, (g) for gaseous state and (aq) for aqueous state.

7 0
3 years ago
Read 2 more answers
5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time
11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen =d_{O_2}=\frac{1 m}{t}

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane=d_{CH_4}=\frac{y }{t}

\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

8 0
3 years ago
A candle slowly burns until the last bit of wax and wick are gone. The two reactants in this chemical reaction are the wick and
Aleonysh [2.5K]

The component of the candle burning in the surrounding has been the oxygen in the air.

The burning of candle wax and wick has been the chemical reaction. It has been based on the reaction of wick with the atmospheric oxygen, resulting in the formulation of the wax burning.

<h3>Chemical reaction of burning of wax</h3>

The wax has been vaporizes by the heat of the flame, that has been resulted by the burning. The wick has been able to react with the oxygen and form the byproducts that helps in flame burning.

The end products have been wick and oxygen as the wax has been consumed in the reaction. The air in the surrounding has oxygen as the part of the system, as it has been involved in the reaction.

Learn more about candle burning, here:

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6 0
2 years ago
How many particles are in a 151 g sample of Li2O?
neonofarm [45]

Answer:

3.052 × 10^24 particles

Explanation:

To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)

The mass of Li2O given in this question is as follows: 151grams.

To convert this mass value to moles, we use;

moles = mass/molar mass

Molar mass of Li2O = 6.9(2) + 16

= 13.8 + 16

= 29.8g/mol

Mole = 151/29.8g

mole = 5.07moles

number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23

= 30.52 × 10^23

= 3.052 × 10^24 particles.

4 0
2 years ago
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