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bearhunter [10]
2 years ago
6

A burning match will burn more vigorously in pure oxygen than in air because _________ . Select one: a. oxygen is a catalyst for

combustion b. nitrogen is a reactant in combustion and its low concentration in pure oxygen catalyzes the combustion c. oxygen is a product of combustion d. nitrogen is a product of combustion and the system reaches equilibrium at a lower temperature e. oxygen is a reactant in combustion and pure oxygen increases the reactant concentration
Chemistry
1 answer:
Westkost [7]2 years ago
8 0

Answer:

e. oxygen is a reactant in combustion and pure oxygen increases the reactant concentration

Explanation:

The reaction of a burning match is combustion. In this combustion, the organic components of the match (such as cellulose, C₆H₁₀O₅) react with oxygen, producing water and carbon dioxide:

  • C₆H₁₀O₅(s) + 6O₂(g) → 5H₂O(g) + 6CO₂(g)

Seeing as oxygen is a reactant and not a catalyst nor product, and that nitrogen plays no part in the reaction, <em>the only correct answer is option e</em>.

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Answer:

If you want to separate black grapes from the mixture of black and green grapes, then you will simply pick black grapes using your hands from the mixture. In this way you are actually using handpicking separation method.

Explanation:

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7 0
3 years ago
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3. A student measured 15.0 grams of ice in a beaker. The beaker was then
gregori [183]

Answer:

The heat that was used to melt the 15.0 grams of ice at 0°C is 4,950 Joules

Explanation:

The mass of ice in the beaker = 15.0 grams

The initial temperature of the ice = 0°C

The final temperature of the ice = 0°C

The latent heat of fusion of ice = 330 J/g

The heat required to melt a given mass of ice = The mass of the ice to be melted × The latent heat of fusion of ice

Therefore, the heat, Q, required to melt 15.0 g of ice = 15.0 g × 330 J/g = 4,950 J

The heat that was used to melt the 15.0 grams of ice = 4,950 Joules.

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3 years ago
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A 3000 MWt fast reactor has a 42% efficiency. This reactor operates for 23 months followed by a 1 month down time for refueling
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Answer:

capacity factor = 0.952

Availability factor = 0.958

Explanation:

1) capacity factor

capacity factor = actual power output /  maximum power output

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                       = \frac{1200}{\frac{42}{100}*3000}

= 0.952

2) Availability factor

Availability factor  = Actual operation time period/ total time period

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8 0
2 years ago
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