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2 years ago

Classify the following as an element, compound, homogeneous mixture, or heterogeneous mixture: iron (Fe)

Chemistry

1 answer:

Sunny_sXe [5.5K]2 years ago

7 0

Answer:

A. Element

Explanation:

Iron (Fe), chemical element and one of the transition elements, the most-used and cheapest metal.

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Part iv. Is the neutralization reaction enthalpy favored?

Burka [1]

Yes, it is a special case of enthalpy of neutralization.

The enthalpy of neutralization (ΔHn) is the change in enthalpy that occurs when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to form water and a salt.

The standard enthalpy change of neutralization is the enthalpy change when solutions of an acid and an alkali react together under standard conditions to produce 1 mole of water.

5 0

2 years ago

The percentage of keys you type correctly is A. accuracy B. correctness C. exactness D. precision WHO EVER GETS THIS EARNS BRAIN

e-lub [12.9K]

The answer is A: accuracy

5 0

1 year ago

)

xenn [34]

Answer:

Its C hope it helped

Explanation:

3 0

2 years ago

A bomb calorimetric experiment was run to determine the enthalpy of combustion of ethanol. The reaction is The bomb had a heat c

gulaghasi [49]

Answer:

The enthalpy of combustion of ethanol in kJ/mol is -1419.58 kJ/mol.

Explanation:

The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change. Working with this equation, and assuming no heat is lost to the surroundings, we write :

qcal= Ccal × ΔT= 490 J/K × 276.7 K= 135,583 J = 135.58 kJ

Note we expressed the temperature change in K, because the heat capacity is written in K.

Now that we have the heat of combustion, we need to calculate the molar heat.

Because qsystem = qrxn + qcal and qrxn = -qcal, the heat change of the reaction is -135.58 kJ.

This is the heat released by the combustion of 4.40 g of ethanol ; therefore, we can write the conversion factor as 135.58 kJ/ 4.40 g.

The molar mass of ethanol is 46.07 g, so the heat of combustion of 1 mole of ethanol is :

molar heat of combustion= -135.58 kJ/4.40 g x 46.07 g/ 1 mol= -1419.58 kJ/mol

Therefore, the enthalpy of combustion of ethanol in kJ/mol is -1419.58 kJ/mol.

3 0

2 years ago

I have 50.00 mL of 0.100 M ethyl amine (C2H5NH2). I gradually add a solution of 0.025 M nitric acid (HNO3) to the ethyl amine so

sergeinik [125]

Answer:

4.00 is the pH of the mixture

Explanation:

The ethyl amine reacts with HNO3 as follows:

C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻

To solve this question we need to find the moles of ethyl amine and the moles of HNO3:

Moles C2H5NH2:

0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine

Moles HNO3:

0.201L * (0.025mol/L) = 0.005025 moles HNO3

That means HNO3 is in excess. The moles in excess are:

0.005025 moles HNO3 - 0.00500 moles ethyl amine =

2.5x10⁻⁵ moles HNO₃

In 50 + 201mL = 251mL = 0.251L:

2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]

As pH = -log [H+]

pH = -log 9.96x10⁻⁵M

pH = 4.00 is the pH of the mixture

6 0

2 years ago