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STALIN [3.7K]
2 years ago
12

Write the chemical equation: hydrogen peroxide decomposes to yield oxygen gas and water.

Chemistry
1 answer:
Doss [256]2 years ago
6 0

Explanation:

2 H2O 2 → 2H 2 O + O2.........

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Using the mmoles of (35)-2,2,-dibromo-3,4-dimethylpentane calculated earlier and the molecular weight of the product (962 g/mol)
Sedbober [7]

Answer:

The yield of the product in gram is \mathsf{{w_P}=0.26 \ gram}

Explanation:

Given that:

the molecular mass weight of the product = 96.2 g/mol

the molecular mass of the reagent (3S)-2,2,-dibromo-3,4-dimethylpentane is 257.997 g

given that the millimoles of the reagent = 2,7 millimoles = 2.7 \times 10^{-3} \ moles

We know that:

Number of moles = mass/molar mass

Then:

2.7 \times 10^{-3} = \dfrac{ mass}{257.997}

mass = 2.7 \times 10^{-3} \times 257.997

mass = 0.697

Theoretical yield = (number of moles of the product/ number of moles of reactant) × 100

i.e

Theoretical yield = \dfrac{n_P}{n_R}\times 100\%

where;

n_P = \dfrac{w_P}{m_P}    and n_R = \dfrac{w_R}{m_R}

Theoretical yield = \dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%

Given that the theoretical yield = 100%

Then:

100\% =\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%

\dfrac{w_P}{m_P}=\dfrac{w_R}{m_R}

{w_P}=\dfrac{w_R \times m_P}{m_R}

where,

w_P = derived weight of the product

m_P =the molecular mass of the derived product

m_R = the molecular mass of the reagent

w_R = weight in a gram of the reagent

{w_P}=\dfrac{w_R \times m_P}{m_R}

{w_P}=\dfrac{0.697 \times 96.2}{257.997}

\mathsf{{w_P}=0.26 \ gram}

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2 years ago
Describe the quantitative characteristics of all chemical reactions. Explain how we can use these features to make quantitative
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Places with _______ altitude will have ______ climates. low; cold high; cold low; freezing high; warm
vlada-n [284]
Places with high altitudes will have cold climates.
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2 years ago
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the scientific law of conservation of mass states that the total amount of mass in an isolated system remains constant. This mea
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It describes how, when particles/mc/elements react, despite forming different substances the mass is neither created nor destroyed but only converted.
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Why can some metals be extracted from compounds by heating with carbon and why can some not
pentagon [3]

Answer:

Some metals can be extracted from compounds by heating with carbon atom because they are less reactive than carbon and some metals cannot be extracted because they are more reactive than carbon atom.

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