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77julia77 [94]
3 years ago
6

270kJ are needed to keep a 75 watt light bulb burning for 1 hour. How many kilocalories are needed to burn this light bulb for 3

hours?
Chemistry
1 answer:
elena-s [515]3 years ago
8 0

Answer:

For 1 hour 75w light bulb requires 270 kj  for burning

for 3 hours 75 w light bulb requires 270*3 = 810kj for burning

Explanation:

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Calculate the amount of heat required to raise the temperature of a 65-g sample of water from 32 ∘C to 65 ∘C. (The specific heat
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The amount of heat required is 9.0 kJ.

<em>q = mC</em>Δ<em>T </em>

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<em>q</em> = 65 g × 4.184 J·°C⁻¹g⁻¹ × 33 °C = 9000 J = 9.0 kJ

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An object was measured by a worker as
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<h2>Answer - 0.73%</h2>

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Certain hydrocarbon is 92.3 % carbon and 7.7 % hydrogen by mass.If the molar mass of the hydrocarbon is approximately 40 g/mol,
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Answer:

The hydrocarbon has a molecular formula of C3H3

Explanation:

Step 1: Data given

Suppose the mass of the compound is 100 grams

A hydrocarbon contains:

 ⇒ 92.3 % carbon = 92.3 grams

 ⇒ 7.7 % hydrogen = 7.7 grams

Molar mass of the compound is 40 g/mol

Molar mass of carbon = 12.01 g/mol

Molar mass of hydrogen = 1.01 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles carbon = 92.3 grams / 12.01 g/mol

Moles carbon = 7.69 moles

Moles hydrogen = 7.70 grams / 1.01 g/mol

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Step 3: Calculate mol ratio

We divide by the smallest amount of moles

Carbon: 7.69/7.62 = 1

Hydrogen: 7.62/7.62 = 1

The empirical formula is CH

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