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Masja [62]
2 years ago
8

When 1.14 g of octane (molar mass = 114 g/mol) reacts with excess oxygen in a constant volume calorimeter, the temperature of th

e calorimeter increases by 10.0°C. The heat capacity of the calorimeter is 6.97 kJ/°C. Determine the energy flow, q (reaction).
Physics
1 answer:
balandron [24]2 years ago
7 0

Answer:

Q (reaction) = -69.7 kJ

Explanation:

Octane reacts with oxygen to give carbon dioxide and water.

C₈H₁₈ + 25 O₂ ---> 16 CO₂ +18 H₂O

This reaction is exothermic in nature. Therefore, the energy is released into the atmosphere. This reaction took place in a calorimeter, there the temperature (T) increases by 10 C. The heat capacity of the calorimeter is 6.97 kJ/C

The heat (q) of the reaction is calculated as follows:

Q= -cT, where c is the heat capacity of the calorimeter and T is the increase in temperature

q = -(6.97) x (10) = -69.7kJ

<em>Since the heat capacity is given in kilo -joule per degree Celsius, therefore, the mass of octane is not required </em>

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An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 6
lara [203]

To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.

The intensity of the wave at the receiver is

I = \frac{P_{avg}}{A}

I = \frac{P_{avg}}{4\pi r^2}

I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m

I = 6.529*10^{-6}W/m^2

The amplitude of electric field at the receiver is

I = \frac{E_{max}^2}{2\mu_0 c}

E_{max}= \sqrt{2I\mu_0 c}

The amplitude of induced emf by this signal between the ends of the receiving antenna is

\epsilon_{max} = E_{max} d

\epsilon_{max} = \sqrt{2I \mu_0 cd}

Here,

I = Current

\mu_0 = Permeability at free space

c = Light speed

d = Distance

Replacing,

\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}

\epsilon_{max} = 0.05434V

Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V

6 0
3 years ago
Perform the calculation and report your answer using sig figs. 657.70 - 26.543
Anton [14]

Answer:

The answer is 631.157

Explanation:

The question requested that the answer to the subtraction of 26.543 from 657.70 must be written using significant figures.

Here are a few tips about how to Identify significant figures.

1) It should be noted that <u>the number "0" is what is usually (but not always) affected</u> while trying to identify significant figures. Hence, <u>all other numbers/digits are always significant</u>. For example, 26.543 has five significant figures.

2) The zeros found between these "other numbers/digits" are also significant. For example, 2202 has four significant figures.

3) In the case of a decimal, the tailing zeros or the final zero is also significant. 657.70 and 657.07 have five significant figures.

Now, back to the question

657.70  - 26.543  = 631.157.

Our final answer does not have a zero, hence all the digits (six) are significant.

8 0
2 years ago
Explain how soilds non metals are different from solid metals
marysya [2.9K]
Solid metal is all different types of metals or some thing that a magnet can pick up that's a full hard solid

but a non metal is everything not metal that's a solid like plastic a hot glue gun can burn throught plastic and not metal

so in conclusion metal is stronger and thicker than non modal things
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2 years ago
A car is accelerating at 30 m/s2, if the car is 400 kg how much force
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It would be 12,000 because newton’s third 2nd law states F=ma (force=matter x acceleration) so 30x400 would be your force .

please mark brainliest and i hope this helps!
6 0
3 years ago
The speed of sound in aluminum is 5200 m/s. Can you hear a sound with a
romanna [79]

Answer:

v = wavelength * frequency

frequency = 5200 m/s / .2 m = 26000 / sec

20,000 / sec is optimistic for the upper frequency of human hearing

So 26,000 is above the hearing range for human ears

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2 years ago
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