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3 years ago

Question 2

Chemistry

1 answer:

Mazyrski [523]3 years ago

7 0

There is an increase in the frequency of particle collisions which leads to the increased rate of the reaction.

<u>Explanation:</u>

Hydrochloric acid reacts faster with the powdered zinc than with the equal mass of zinc strips.

Since the surface area of the Zinc particle increases when it is powdered and has more active spots when compared to less number of active sites in Zinc strips and so the number of collision between the Zinc particles and the particles of Hydrochloric acid also increases and consequently the rate of the reaction also enhances.

So there is an increase in the frequency of particle collisions.

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Write the expression for the equilibrium constant Kp for the following reaction. (Enclose pressures in parentheses and do NOT wr

Liono4ka [1.6K]

Answer and Explanation:

For the following balanced reaction:

PCl₅(g) ↔ PCl₃(g) + Cl₂(g)

We can see that all reactants and products are gases, so it is an homogeneous equilibrium. The expression for the equilibrium constant Kp can be written from the partial pressures (P) of reactants and products as follows:

$Kp=\frac{(P PCl_{3})(P Cl_{2})}{(P PCl_{5})}$

Where PPCl₃ is the partial pressure of PCl₃ (reactant), PCl₂ is the partial pressure of Cl₂ (reactant) and PPCl₅ is the partial pressure of PCl₅ (product).

6 0

2 years ago

If a liquid has a Ph of 3 what is it

kvasek [131]

Lower than 7 is acid greater than 7 is a base

6 0

3 years ago

Match each term to its description. Match Term Definition Excess reactant A) Reactant that can produce more of the product Limit

faust18 [17]

Answer:

Explanation:

A) Reactant that can produce more of the product

Excess reactant:

In a given reaction, the reactant that is in excess supply is the excess reactant. If the amount of the excess reactant is match, more of the product will be produced.

B) Reactant that can produce a lesser amount of the product

Limiting reactant

The limiting reactant restricts the progress of the reaction. It determines the amount of product that can be formed.

C) Amount of product predicted to be produced by the given reactants

Theoretical yield

For a given amount of reactants, the theoretical yield determines the amount of products that can be produced.

5 0

3 years ago

4. Magnesium and oxygen undergo a chemical reaction to

erastovalidia [21]

Answer:

About 16.1 grams of oxygen gas.

Explanation:

The reaction between magnesium and oxygen can be described by the equation:
$\displaystyle 2\text{Mg} + \text{O$_2$} \longrightarrow 2\text{MgO}$

24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.

Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:

Convert grams of MgO to moles of MgO.
Moles of MgO to moles of O₂
And moles of O₂ to grams of O₂.

The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.

Dimensional analysis:

$\displaystyle 40.5\text{ g MgO} \cdot \frac{1\text{ mol MgO}}{40.31\text{ g MgO}} \cdot \frac{1\text{ mol O$_2$}}{2\text{ mol MgO}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}$

In conclusion, about 16.1 grams of oxygen gas was reacted.

You will obtain the same result if you compute with the 24.4 grams of Mg instead:

$\displaystyle 24.4\text{ g Mg}\cdot \frac{1\text{ mol Mg}}{24.31\text{ g Mg}} \cdot \frac{1\text{ mol O$_2$}}{1\text{ mol Mg}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}$

3 0

2 years ago

I NEED HELP PLEASE, THANKS! :)

krok68 [10]

Answer:

$\large \boxed{\text{2.20 g Pb}}$

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 239.27 32.00 207.2

2PbS + 3O₂ ⟶ 2Pb + 2SO₃

m/g: 2.54 1.88

2. Calculate the moles of each reactant

$\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}$

3. Calculate the moles of Pb from each reactant

$\textbf{From PbS:}\\\text{Moles of Pb} = \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}$

4. Calculate the mass of Pb

$\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}$

5 0

2 years ago