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2 years ago

What is the role of a catalyst in a chemical reaction?

1 answer:

4 0

A catalyst is a substance that increases the rate of a chemical reaction by lowering the activation energy without itself being consumed by the reaction.

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Which of these answers best describe a fossil ?

Mekhanik [1.2K]

You would have to show me the answers

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3 years ago

Hybridization of NH3

nadya68 [22]

Answer:

Sp3 hybridization

Explanation:

The NH3 molecule, which consists of one lone pairs and three bond pair of electron on its valance shell due to lone pair bond pair repulsion makes bond angle of 107.5°resulting distorted tetrahedral geometry.

Hybridization =no. of bond pair +lone pair=3+1=4=sp3 hybridization

6 0

3 years ago

Many chemical processes involve substances in either the liquid or gaseous state. Which of the following statements is correct?

Mrac [35]

A specific mass of a gas has a fixed volume at room temperature

7 0

3 years ago

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.650 M, [B] = 1.35 M, and [C] = 0.300 M. The fo

dimaraw [331]

Answer: The value of $K_c$ for given reaction is 0.465

Explanation:

We are given:

Initial concentration of A = 0.650 M

Initial concentration of B = 1.35 M

Initial concentration of C = 0.300 M

Equilibrium concentration of A = 0.550 M

Equilibrium concentration of B = 0.400 M

For the given chemical equation:

$A+2B\rightarrow C$

Initial: 0.65 1.35 0.30

At eqllm: 0.65-x 1.35-2x 0.30+x

Evaluating the value of 'x'

$0.650-x=0.550\\\\x=0.100$

So, equilibrium concentration of B = 1.35 - 2x = [1.35 - 2(0.100)] = 1.15 M

Equilibrium concentration of C = (0.30 + x) = (0.300 + 0.100) = 0.400 M

The expression of $K_c$ for above equation follows:

$K_c=\frac{[C]}{[A][B]^2}$

Putting values in above equation, we get:

$K_c=\frac{0.400}{0.650\times (1.15)^2}\\\\K_c=0.465$

Hence, the value of $K_c$ for given reaction is 0.465

7 0

3 years ago

Hydrogen chloride gas and oxygen gas react to form water and chlorine gas. A reaction mixture initially contains 53.2 g of hydro

Sergio [31]

Answer: The mass of the excess reactant (oxygen gas) is 3.136 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For HCl:

Given mass of HCl = 53.2 g

Molar mass of HCl = 36.5 g/mol

Putting values in equation 1, we get:

$\text{Moles of HCl}=\frac{53.2g}{36.5g/mol}=1.46mol$

For oxygen gas:

Given mass of oxygen gas = 26.5 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{26.5g}{32g/mol}=0.828mol$

The chemical equation for the reaction of HCl and oxygen gas follows:

$2HCl+O_2\rightarrow H_2O+Cl_2$

By Stoichiometry of the reaction:

2 moles of HCl reacts with 1 mole of oxygen gas

So, 1.46 moles of HCl will react with = $\frac{1}{2}\times 1.46=0.73mol$ of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, HCl is considered as a limiting reagent because it limits the formation of product.

Excess moles of oxygen gas = (0.828 - 0.73) = 0.098 moles

Now, calculating the mass of oxygen gas from equation 1, we get:

Molar mass of oxygen gas = 32 g/mol

Excess moles of oxygen gas = 0.098 moles

Putting values in equation 1, we get:

$0.098mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.098mol\times 32g/mol)=3.136g$

Hence, the mass of the excess reactant (oxygen gas) is 3.136 grams

4 0

3 years ago