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Llana [10]
3 years ago
13

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

Physics
2 answers:
gregori [183]3 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]3 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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FromTheMoon [43]

Answer:

Part a)

g = 1 \times 10^{-5} m/s^2

Part b)

v = 0.011 m/s

Explanation:

Part a)

As we know that the acceleration due to gravity is given as

g = \frac{GM}{R^2}

g = \frac{G(\frac{4}{3}\pi R^3 \rho)}{R^2}

g = \frac{4}{3}\pi \rho G R

now we know that

\rho = \frac{M}{\frac{4}{3}\pi R^3}

\rho = \frac{5.98 \times 10^{24}}{\frac{4}{3}\pi (6.37 \times 10^6)^3}

\rho = 5523.2 kg/m^3

now we have

g = \frac{4}{3}\pi (5523.2) (6.67 \times 10^{-11})(6.5)

g = 1 \times 10^{-5} m/s^2

Part b)

As we know that the escape speed is given as

v = \sqrt{\frac{2GM}{R}}

v = \sqrt{2gR}

v = \sqrt{2(1\times 10^{-5})(6.5)}

v = 0.011 m/s

8 0
3 years ago
A train moves with a uniform velocity of 10 m/s for 8s find the distance traveled by it
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The distance traveled is: 80 m
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A consistent tendency for forecasts to be greater or less than the actual values is called​ ________ error.
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The Answer is: Bias Error
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the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity
NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

where

a is acceleration

s is distance

v is velocity

Given;

a = 6 + 0.02s

so,

\int\limits^s_0 {a} \, ds  = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds  = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\

Remember that

v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t=  (5\sqrt{2} ) ln  \frac{| [s + 300 + \sqrt{(s^{2}  + 600s)} ] |}{300} .......2

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s

8 0
3 years ago
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