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Hitman42 [59]
2 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]2 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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Predict the enthalpy of reaction from the average bond enthalpies for the following reaction: 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O
Vitek1552 [10]

Answer:

B

Explanation:

8 0
2 years ago
If a pharmacist dissolves 1.2 grams of a medicinal agent in 60 ml of a cough syrup having a specific gravity of 1.20, what is th
guapka [62]

Mass of medicinal agent taken = 1.2 g

the volume is 60 mL

Specific gravity = 1.20

So the mass of solution = specific gravity X volume = 1.20 * 60 = 72g

Now if we have increased the volume by 0.2 so the new volume = 60.2

New mass = 72 + 1.2 = 73.2  

Specific gravity = mass /  volume = 73.2 / 60.2 = 1.22 g/mL

7 0
2 years ago
Missy Mae is given an unknown white ionic solid which contains one of the cations Na+, K+, Ca2+ and one of the anions Cl-, NO3-
topjm [15]

Answer:

NaNO₃

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Ca(NO₃)₂

Explanation:

The possible cations for the salt. = Na⁺ , K⁺ , Ca²⁺

The anion for which the formula has to written = NO₃⁻

The salt is of white color.

The possible salts may be:

<u>NaNO₃ </u>

<u>KNO₃</u>

<u>Ca(NO₃)₂</u>

Missy Mae can have one of these salts.

The color of NaNO₃ is from transparent to white.

The color of KNO₃ is white.

The color of NaNO₃ is from white to light grey.

7 0
2 years ago
For the following dehydrohalogenation (E2) reaction, draw the Zaitsev product(s) resulting from elimination involving C3–C4 (i.e
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Answer:

2-methyl-butene

Explanation:

For the E2 mechanism, we have an <u>anti-elimination</u>. The Br leaves the molecule and the base removes the hydrogen in the anti position to form the double that's why only one structure is produced. (See figure 1)

Since we have 2 hydrogens on the right carbon, we cannot indicate a <u>specific stereoisomer</u>, in other words, it is not possible to assign a <u>Z or E</u> configuration for this alkene.

8 0
2 years ago
Two bulbs are connected by a stopcock. The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 0.700 atm
ivolga24 [154]

Answer:

O_{2} and NO_{2}

Explanation:

For a given system at constant temperature, the number of moles of gas present in the system is proportional to the product of the system pressure and volume. Therefore, we have:

NO: 6 L * 0.7 atm = 4.2 L*atm

O:  1.5 L* 2.5 atm = 3.75 L*atm

For the given system based on a balanced chemical equation:

2.70 L*atm of nitric oxide reacts with (2.7/2) 1.35 L*atm of oxygen. This shows that there is more oxygen gas in the system than nitric oxide. Thus nitric oxide is the limiting reactant.

At the end of the experiment:

All the nitirc oxide has been used up, i.e. P_{NO} = 0

For the product: 2.70 L*atm NO produced  2.70 L*atm NO_{2}

The total volume of the system after the stopcock is opened = 6+1.5 = 7.5 L

The partial pressure of NO_{2}  = (2.70 L*atm NO_{2} ) / (7.5 L) = 0.36 atm NO_{2}  

Similarly for oxygen gas:

3.75 L*atm - 1.35 L*atm  = 2.40 L*atm oxygen gas remaining  

Partial pressure of oxygen is:

2.40 L*atm / 7.5 L = 0.32 atm  

Thus, the gases present at the end of the experiment are O_{2} and NO_{2}

3 0
2 years ago
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