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miv72 [106K]
2 years ago
8

If 2. 5 moles of h2o are produced how many moles of hydrogen gas must be used.

SAT
1 answer:
iVinArrow [24]2 years ago
3 0

Answer:

2.5 moles of H2

Explanation:

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Break down of rocks and minerals into smaller particles without a change in composition is called_
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Weathering breaks down the rocks, erosion moves the particles, and deposition drops the sediments in another location. 1. Physical Weathering – the breakdown of rocks and minerals into smaller pieces without a change in chemical composition. You're welcome.

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2. The organization that certifies laboratory animal technicians (LATs) is the
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Which option best describes your approach to taking notes as you read.
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The best option that describes one's approach to taking notes as one reads is: <em>you take good notes that helps you to recall </em><em>important information.</em>

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<h3>Taking Notes When Reading</h3>
  • When reading a book, textbook, or any other book, a technique for easily recalling to mind important details after reading is taking notes.
  • Taking of notes helps you to remember important information.

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Newton’s law of cooling states that for a cooling substance with initial temperature t0, the temperature t(t) after t minutes ca
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The cooling rate of the substance is approximately 0.0732.

According to the statement, the Newton's law of cooling is defined by the following formula:

T(t) = T_{s} + (T_{o}-T_{s})\cdot e^{-k\cdot t} (1)

Where:

  • T_{s} - Final temperature, in degrees Celsius.
  • T_{o} - Initial temperature, in degrees Celsius.
  • t - Time, in minutes.
  • k - Cooling rate, in \frac{1}{min}.
  • T(t) - Current temperature, in degrees Celsius.

Please notice that substance reaches thermal equilibrium when T(t) = T_{s}, that is when temperature of the substance is equal to the temperature of surrounding air.

If we know that T_{o} = 80\,^{\circ}C, t = 15\,min, T_{s} = 50\,^{\circ}C and T(15) = 60\,^{\circ}C, then the cooling rate of the substance is:

60 = 50 + (80 - 50)\cdot e^{-15\cdot k}

\frac{60-50}{80-50}= e^{-15\cdot k}

\frac{1}{3} = e^{-15\cdot k}

k = -\frac{1}{15}\cdot \ln \frac{1}{3}

k \approx 0.0732

The cooling rate of the substance is approximately 0.0732.

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