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Ymorist [56]
3 years ago
12

Why is MnO2 added to the decomposition of H2O2 reaction?

Chemistry
2 answers:
timofeeve [1]3 years ago
8 0
Answer is: <span>to decrease activation energy.
Manganese(IV) oxide is catalyst in this reaction.
</span>Catalysis <span>is the increase in the </span>rate<span> of a </span>chemical reaction<span> due to the participation of an additional substance called a </span>catalyst. R<span>eactions occur faster with a catalyst because they require less </span>activation energy<span>.</span>
Elanso [62]3 years ago
7 0
<span>to decrease the activation energy

#Plato </span>
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Which answer below correctly identifies the type of change and the explanation for the boiling of water?
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Answer:

physical change because even though gas formation was observed the water was undergoing a state change which means that it's original properties are preserved

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I HAVE ANSWER
Agata [3.3K]
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2 years ago
A certain half-reaction has a standard reduction potential -1.33V . An engineer proposes using this half-reaction at the anode o
Paha777 [63]

Answer:

a. -0.63 V

b. No

Explanation:

Step 1: Given data

  • Standard reduction potential of the anode (E°red): -1.33 V
  • Minimum standard cell potential (E°cell): 0.70 V

Step 2: Calculate the required standard reduction potential of the cathode

The galvanic cell must provide at least 0.70V of electrical power, that is:

E°cell > 0.70 V    [1]

We can calculate the standard reduction potential of the cathode (E°cat) using the following expression.

E°cell = E°cat - E°an   [2]

If we combine [1] and [2], we get,

E°cat - E°an > 0.70 V

E°cat > 0.70 V + E°an

E°cat > 0.70 V + (-1.33 V)

E°cat > -0.63 V

The minimum E°cat is -0.63 V and there is no maximum E°cat.

3 0
3 years ago
3. In an experiment it was found that 40.0cm of 0.2M sodium hydroxide solution just neutralized 0.2g
hjlf

The relative molecular mass of acid A : 50 g/mol

<h3>Further explanation</h3>

Given

40.0 cm³(40 ml) of 0.2M sodium hydroxide

0.2g  of a dibasic acid

Required

the relative molecular mass of acid A

Solution

Titration formula

M₁V₁n₁=M₂V₂n₂

n=acid/base valence(number of H⁺/OH⁻)

NaOH ⇒ n = 1

Dibasic acid =  diprotic acid (such as H₂SO₄)⇒ n = 2

mol = M x V

Input the value in the formula :(1 = NaOH, 2=dibasic acid)

0.2 x 40 x 1 = M₂ x V₂ x 2

M₂ x V₂ = 4 mlmol = 4.10⁻³ mol ⇒ mol of Acid A

The relative molecular mass of acid A (M) :

\tt M_A=\dfrac{mass }{mol}=\dfrac{0.2~g}{4.10^{-3}}=50~g/mol

5 0
3 years ago
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