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3 years ago

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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Answer:

800 Watts

Explanation:

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Working in SI units, Power = Watts, Work = Joules, Time = seconds.

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3 years ago

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Explanation:

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4 years ago

Read 2 more answers

A damped harmonic oscillator consists of a mass on a spring, with a small damping force that is proportional to the speed of the

exis [7]

Answer:

2.19 N/m

Explanation:

A damped harmonic oscillator is formed by a mass in the spring, and it does a harmonic simple movement. The period of it is the time that it does one cycle, and it can be calculated by:

T = 2π√(m/K)

Where T is the period, m is the mass (in kg), and K is the damping constant. So:

2.4 = 2π√(0.320/K)

√(0.320/K) = 2.4/2π

√(0.320/K) = 0.38197

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3 years ago

A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside

neonofarm [45]

Answer:

$\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$ , level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

$\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}$

$\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt} = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}$

By replacing all known variables:

$\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$

The positive sign of the rate of change of the tank level indicates a rising behaviour.

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4 years ago