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2 years ago

Predict the products of the combustion of methanol, CH3OH(l).

Physics

2 answers:

gregori [183]2 years ago

8 0

Answer:

Carbon dioxide and water

Explanation:

The products of complete combustion are always carbon dioxide and water.

The balanced reaction is:

4 CH₃OH + 3 O₂ → 4 CO₂ + 2 H₂O

morpeh [17]2 years ago

8 0

Answer : The products of the combustion of methanol are, carbon dioxide $(CO_2)$ and water $(H_2O)$

Explanation :

Combustion reaction : It is a type of reaction in which the hydrocarbon react with the oxygen gas to give carbon dioxide and water as products.

The balanced chemical reaction of combustion of methanol $(CH_3OH)$ is :

$2CH_3OH(l)+3O_2(g)\rightarrow 2CO_2(g)+4H_2O(g)$

By the stoichiometry we can say that, 2 mole of methanol react with 3 moles of oxygen gas to give 2 mole of carbon dioxide and 4 moles of water as products.

In this reaction, methanol and oxygen gas are the reactants and carbon dioxide and water are the products.

Therefore, the products of the combustion of methanol are, carbon dioxide $(CO_2)$ and water $(H_2O)$

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A plane wall of thickness 0.1 mm and thermal conductivity 25 W/m K having uniform volumetric heat generation of 0.3 MW/m3 is ins

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Answer:

The maximum temperature is 90.06° C

Explanation:

Given that

t= 0.1 mm

Heat generation

$q_g=0.3\ MW/m^3$

Heat transfer coefficient

$h=500\ W/m^2K$

Here one side(left side) of the wall is insulated so the all heat will goes in to right side .

The maximum temperature will at the left side.

Lets take maximum temperature is T

Total heat flux ,q

$q=q_g\times t$

$q=0.3\times 1000000\times 0.1 \times 10^{-3}\ W/m^2$

$q=30\ W/m^2$

So the total thermal resistance per unit area

$R=\dfrac{t}{K}+\dfrac{1}{h}$

$R=\dfrac{0.1\times 10^{-3}}{25}+\dfrac{1}{500}$

R=0.002 K/W

We know that

q=ΔT/R

30=(T-90)/0.002

T=90.06° C

The maximum temperature is 90.06° C

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2 years ago

What is the wavelength of an earthquake wave if it has a speed of 12 km/s and a frequency of 15 Hz

professor190 [17]

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A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal

dsp73

Answer:

$v\approx 8.570\,\frac{m}{s}$

Explanation:

The equation of equlibrium for the box is:

$\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a$

The formula for the acceleration, given in $\frac{m}{s^{2}}$ , is:

$a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}$

Velocity can be derived from the following definition of acceleration:

$a = v\cdot \frac{dv}{dx}$

$v\, dv = a\, dx$

$\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx$

$\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg} \int\limits^{17\,m}_{0\,m}\, dx - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx$

$\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}$

$v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }$

The speed after the box has travelled 17 meters is:

$v\approx 8.570\,\frac{m}{s}$

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2 years ago

A skier is pulled by a towrope up a frictionless ski slope that makes an angle of 12 degrees with the horizontal. The rope moves

MArishka [77]

Answer:

Explanation:

Given,

Work done by the rope 900 m/s.
Angle of inclination of the slope = $\theta\ =\ 12^o$
Initial speed of the skier = v = 1.0 m/s
Length of the inclined surface = d = 8.0 m

part (a)

The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity

$\therefore W_r\ =\ W_g\ =\ 900\ J$

In both cases the height attained by the skier is equal. and the work done by gravity does not depend upon the speed of the skier.

part (b)

Initial speed of the skier = v = 1.0 m/s.

Rate of the work done by the rope is power of the rope.

$Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 1.0}{8.0}\\\Rightarrow P\ =\ 112.5\ Watt$

Part (c)

Initial speed of the skier = v = 2.0 m/s.

Rate of the work done by the rope is power of the rope.

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2 years ago