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3 years ago

The second reaction in the formation of sulfuric acid occurs slowly.

Chemistry

1 answer:

rjkz [21]3 years ago

5 0

Answer: The correct option is $NO_2(g)$

Explanation:

A catalyst is defined as the chemical species that increases the reaction rate but does not participate in it and is left behind after the completion.

A homogeneous catalyst is one that is present in the same phase as the reactants and products.

A heterogeneous catalyst is one that is present in different phase as that of reactants and products.

For the given chemical reaction:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

As all the reactants and products are in gaseous state so, the homogeneous catalyst must also be in the gaseous state only.

Hence, the correct option is $NO_2(g)$

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3 years ago

Write a balanced chemical equation for the complete combustion of C3H7BO3, a gasoline additive. The products of combustion are C

nexus9112 [7]

Answer:

2C₃H₇BO₃ + 8O₂ → 6CO₂ + 7H₂O + B₂O₃.

Explanation:

For balancing a chemical equation, we should apply the law of conversation of mass. It states that the no. of atoms in the reactants side is equal to that of the products side.

So, the balanced equation:

2C₃H₇BO₃ + 8O₂ → 6CO₂ + 7H₂O + B₂O₃.

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3 years ago

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omeli [17]

Answer:

Answer -b

Explanation:

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3 years ago

Read 2 more answers

The powder mixture (Cu, Al and Fe) = 10g was oxidized from sufficient chloride acid. 1) What are the possible reactions to the p

Mandarinka [93]

Answer:

1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

Explanation:

1) Possible reactions

2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Mass of each metal

a) Mass of Cu

The waste was the unreacted copper.

Mass of Cu = 2.5 g

b) Masses of Al and Fe

We have two relations :

Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g

H₂ from Al + H₂ from Fe = 6.38 L at NTP

i) Calculate the moles of H₂

NTP is 20 °C and 1 atm.

$\begin{array}{rcl}pV & = & n RT\\\text{1 atm} \times \text{6.38 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{293.15 K}\\6.38 & = & 24.06n \text{ mol}^{-1} \\n & = & \dfrac{6.38}{24.06 \text{ mol}^{-1} }\\\\ & = & \text{0.2652 mol}\\\end{array}$