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Deffense [45]
2 years ago
5

Chalk markings cannot be easily rubbed off from a black board if kept for a long time.why?

Chemistry
1 answer:
snow_tiger [21]2 years ago
3 0

The chalk particles embed themselves into the small pores on the surface.

Although a chalkboard seems smooth to the touch, it is quite rough at the microscopic level, with <em>pores</em> that reach below the surface.

When you drag chalk across the board, friction causes small particles of chalk to rub off onto the surface.

If you leave the markings for a long time, some of the chalk particles will work their way into the pores.

A brush will remove the surface particles, but <em>it will not be able to get at the particles in the pores</em>.

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Assume that all of this fossil fuel is in the form of octane (C8H18) and calculate how much CO2 in kilograms is produced by worl
AnnZ [28]

The given question is incomplete. the complete question is:

The world burns the fossil fuel equivalent of approximately 9.50\times 10^{12} kg of petroleum per year. Assume that all of this petroleum is in the form of octane. Calculate how much CO2 in kilograms is produced by world fossil fuel combustion per year.( Hint: Begin by writing a balanced equation for the combustion of octane.)

Answer: 29\times 10^{12}kg

Explanation:

Combustion is a chemical reaction in which hydrocarbons are burnt in the presence of oxygen to give carbon dioxide and water.

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

2_C8H_{18}+17O_2\rightarrow 16CO_2+18H_2O

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of octane}=\frac{9.50\times 10^{12}\times 10^3g}{114g/mol}=0.083\times 10^{15}moles

According to stoichiometry :

As 2 moles of octane give = 16 moles of CO_2

Thus 0.083\times 10^{15}moles of octane give =\frac{16}{2}\times 0.083\times 10^{15}=0.664\times 10^{15}moles  of CO_2

Mass of CO_2=moles\times {\text {Molar mass}}=0.664\times 10^{15}moles\times 44g/mol=29.2\times 10^{15}g=29.2\times 10^{12}kg

Thus 29\times 10^{12}kg of CO_2 is produced by world fossil fuel combustion per year.

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Answer: B) metals, non-metals, metalloids

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A metalloid is a bit of a mix between a metal and non-metal element. It's sorta like an element that has both properties of metals and non-metals, or it's in a murky gray area. An example of a metalloid would be silicon.

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