Determine the type of each chemical equation describing a precipitation reaction. C a 2 + ( a q ) + S O 4 2 − ( a q ) ⟶ C a S O

Answer: The change in enthalpy for each mole of zinc involved in the reaction is 152.4 kJ/mol.

Explanation:

First, we need to determine the moles of Zn and HCl.

The molar mass of Zn is 65 g/mole

and,

Next, we must identify the limiting reagent and the excess reagent.

The chemical reaction given is:

According to the balanced reaction we find that

1 mole of reacts with 2 moles of

Thus, 0.0206 moles of react with moles of

This leads us to determine that is the excess reagent because the moles provided exceed the required moles, while is limiting and restricts product formation.

Now to find the enthalpy change for each mole of zinc reacting in this reaction.

From the reaction we gather that,[ [TAG_59]]

0.0206 moles of Zn yield heat = 3.14 kJ

This implies that 1 mole of Zn generates heat =

Hence, the enthalpy change per mole of zinc involved in this reaction amounts to 152.4 kJ/mol.

Answer:

Explanation:

Assuming all calculations occur at standard pressure and a temperature of -1.72°C :

Where

is the number of moles of hydrogen

is the mass of hydrogen

is the density of hydrogen

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The result is:

The new volume is:

Rationale:

Because the temperature remains constant, we can apply Boyle's Law to solve this issue.

Boyle's Law stipulates that:

Where,

P is the gas's pressure.

V is the gas's volume.

According to the information provided:

Let's put the values into the equation:

Consequently, the new volume is:

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Answer :

The percentage ionic character (%IC) equals 10%, indicating the bond is mostly covalent with slight polarity.

Percent Ionic Character:

This reflects the fraction of ionic nature within a polar covalent bond. The formula for %IC (% ionic character) is:

Here, Xa is the electronegativity of atom A and Xb is that of atom B.

Given: The compound is TiAl₃.

Electronegativity of Ti = 2.0

Electronegativity of Al = 1.6 (as shown in the provided image)

Substitute these values into the formula:

The value of e⁻¹ equals 0.90.

Therefore, percent ionic character = (1 - 0.90) × 100

Percent Ionic Character = 10%

Because the % IC is only 10%, which is relatively low, the bond is classified as covalent with minimal polarity.

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An experiment that directly tests the hypothesis

Variables to be tested by an experiment