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12345 [234]
3 years ago
7

When 0.1375 g of solid magnesium is burned in a constant-volume bomb calorimeter, the temperature increases by 1.126°C. The heat

capacity of the bomb calorimeter, determined in a separate experiment, is 3024 J/°C. Calculate the heat given off by the burning magnesium in kJ/mol.
Physics
1 answer:
Bumek [7]3 years ago
6 0

Answer:

-24.76 kJ/mol

Explanation:

given,

mass of solid magnesium burned = 0.1375 g

the temperature increases by(ΔT) 1.126°C

heat capacity of of bomb calorimeter (C_{cal})= 3024 J/°C

heat absorbed by the calorimeter

    q_{cal} = C_{cal}\DeltaT

    q_{cal} = 3024 \times 1.126

    q_{cal} =3405.24\ J

    q_{cal} =3.405\ kJ

heat released by the reaction

    q_{rxn} = -q_{cal}

    q_{rxn} = -3.405\ kJ

energy density will be equal to heat released by the reaction divided by the mass of magnesium

Energy density = \dfrac{-3.405}{0.1375}

Energy density = -24.76 kJ/mol

heat given off by burning magnesium is equal to -24.76 kJ/mol

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Consider two copper wires of the same length. One has twice the cross-sectional area of the other. How do the resistances of the
JulsSmile [24]

Answer: a) when the cross section is doubled the resultant resistence  is a half. This means the thicker wire have half resistence than the thinner wire.

Explanation: In order to explain this behaviur we have to consider the expresion for the resistence which is given by:

R=\frac{\epsilon o  L}{A} where L and A are the length and the cross section for the wire, respectively.

From this expresssion we can conclude the above, this means

R=εo*L/A  if A is now 2A we have

R' = εo*L/2*A= R/2

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3 years ago
(BRAINLIEST)<br><br> Question: Name 2 waves that you use in your everyday life.
Archy [21]

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4 years ago
2 Points
goldfiish [28.3K]

Answer:

I believe it's called rolling friction

8 0
3 years ago
Read 2 more answers
The spreading of waves behind an aperture ismore for long wavelengths and less for short wavelengths.less for long wavelengths a
kompoz [17]
<h2>Answer: The spreading of waves behind an aperture ismore for long wavelengths and less for short wavelengths</h2>

Here we are talking about Diffraction and, in fact, waves diffract the most when their wavelength is about the same size of the gap or aperture.

Diffraction happens when a wave (mechanical or electromagnetic wave) meets an obstacle or a slit .When this occurs, <u>the wave bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming multiple patterns with the shape of the aperture of the slit. </u>

<u />

Note that the principal condition for the occurrence of this phenomena is that the obstacle must be comparable in size (similar size) to the size of the wavelength.

In other words, when the gap (or slit) size is larger than the wavelength, the wave passes through the gap and does not spread out much on the other side, but when the gap size is equal to the wavelength, maximum diffraction occurs and the waves spread out greatly.

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6 0
3 years ago
The air in a room with volume 180 m3 contains 0.25% carbon dioxide initially. fresher air with only 0.05% carbon dioxide flows i
Butoxors [25]

Answer:

p(t)=\frac{1}{20}(1+2e^{-t/90})

Explanation:

Let p(t) be % of CO_2 at time t(time in minutes).

Amount of CO_2 in room=Volume of room \times p

Rate of change of CO_2 in room=Volume of room x \frac{dp}{dt}

=180\times \frac{dp}{dt}

Rate of inflow of CO_2=(\% CO_2 \ in\  Fresh\  Air \times \frac{1}{100})\times(Rate \ of  Inflow\ of \ air)

=(0.05\times \frac{1}{100})\times 2=\frac{1}{100}

Rate of CO_2 outflow

(\% CO_2 \ in\  Fresh\  Air \times \frac{1}{100})\times(Rate \ of  Outflow\ of \ air)\\=(p\times\frac{1}{100})\times2=\frac{p}{50}

Therefore rate of change of CO_2 is \frac{1}{100}-\frac{p}{50}

% rate of change is100\times \ Rate of \ Change=\frac{1}{10}-2p

Therefore we have:

180\frac{dp}{dt}=\frac{1}{10}-2p\\\frac{dp}{dt}=\frac{1-20p}{1800}\\\\\frac{dp}{20p-1}=-\frac{dt}{1800}

Integrate both:

\int\limits \frac{dp}{20p-1}=\int\limits-\frac{dt}{1800}

\frac{1}{20}In|20p-1|=-\frac{t}{1800}+In \ C

In|20p-1|=-\frac{t}{90}+InK

+20In \ C=+In \ K

Raise both sides to base e,

20p-1=e^{-\frac{t}{90}+In \ K}\\\\p=\frac{1}{20}(1+Ke^{-t/90})

Initially, there's only 0.25% of CO_2,

Now substitute p=0.25 and t=0

0.25=\frac{1}{20}(1+K)\\K=4

p=\frac{1}{20}(1+2e^{-t/90})

8 0
3 years ago
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