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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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Who discovered that certain stars produce electromagnetic radiation, as known as pulsars
PIT_PIT [208]

Answer:

Antony Hewish

Explanation:

4 0
3 years ago
Prestressed concrete is often used in modern construction because:
Basile [38]

Answer:

A.  Its unique strength allows predetermined engineering stresses to be placed in members to counteract stresses that occur when the unit is subjected to service loads

Explanation:

Prestressed concrete is a structural material that allows for predetermined, engineering stresses to be placed in members to counteract the stresses that occur when they are subject to loading.

It combines the high strength compressive properties of concrete with the high tensile strength of steel.

‘Building Construction Handbook’ (6th ed.), CHUDLEY, R., GREENO, R., Butterworth-Heinemann (2007)

Then the correct answer is:

A.  Its unique strength allows predetermined engineering stresses to be placed in members to counteract stresses that occur when the unit is subjected to service loads

5 0
3 years ago
If you push a crate across a factory floor at constant speed in a constant direction, what is the magnitude of the force of fric
poizon [28]

Answer:

The magnitude of the force of friction equals the magnitude of my push

Explanation:

Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.

Let F = push and f = frictional force and f' = net force

F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0

So, F - f = 0

Thus, F = f

So, the magnitude of the force of friction equals the magnitude of my push.

3 0
3 years ago
H20 (water) is an example of a __, as it is made up of two elements.
vova2212 [387]

Answer:

Compound.

Explanation:

A compound is a substance formed when two or more elements are chemically joined. Water, salt, and sugar are examples of compounds. When the elements are joined, the atoms lose their individual properties and have different properties from the elements they are composed of.

8 0
3 years ago
A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When
natita [175]

Answer:

a) \mu_{k} = 0.704, b) R = 0.312\,m

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s

\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}

\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}

\mu_{k} = 0.704

b) The speed of the block is determined by using the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}

v = x\cdot \sqrt{\frac{k}{m} }

v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }

v \approx 9.829\,\frac{m}{s}

The radius of the circular loop is:

\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}

\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}

R = 0.312\,m

5 0
4 years ago
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