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3 years ago

Explain cross bedding

Chemistry

1 answer:

Varvara68 [4.7K]3 years ago

4 0

Answer:

In geology, cross-bedding, also known as cross-stratification, is layering within a stratum and at an angle to the main bedding plane. The sedimentary structures which result are roughly horizontal units composed of inclined layers.

Explanation:

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. <br> If the Moon orbited Earth twice as fast, how many high tides would occur in one day?

Ahat [919]

Because the Earth rotates through two tidal “bulges” every lunar day, coastal areas experience two high and two low tides every 24 hours and 50 minutes. ... This occurs because the moon revolves around the Earth in the same direction that the Earth is rotating on its axis.

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3 years ago

In 2.00 min, 29.7 mL of He effuse through a small hole. Under the same conditions of pressure and temperature, 9.28 mL of a mixt

sergiy2304 [10]

Answer : The percent composition by volume of mixture of $CO$ and $CO_2$ are, 18.94 % and 81.06 % respectively.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

And the relation between the rate of effusion and volume is :

$R=\frac{V}{t}$

or, from the above we conclude that,

$(\frac{V_1}{V_2})^2=\frac{M_2}{M_1}$ ..........(1)

where,

$V_1$ = volume of helium gas = 29.7 ml

$V_2$ = volume of mixture = 9.28 ml

$M_1$ = molar mass of helium gas = 4 g/mole

$M_2$ = molar mass of mixture = ?

Now put all the given values in the above formula 1, we get the molar mass of mixture.

$(\frac{29.8ml}{9.28ml})^2=\frac{M_2}{4g/mole}$

$M_2=40.97g/mole$

The average molar mass of mixture = 40.97 g/mole

Now we have to calculate the percent composition by volume of the mixture.

Let the mole fraction of $CO$ be, 'x' and the mole fraction of $CO_2$ will be, (1 - x).

As we know that,

$\text{Average molar mass of mixture}=\text{Mole fraction of }CO$

$\text{Average molar mass of mixture}=(\text{Mole fraction of }CO\times \text{Molar mass of } CO)+(\text{Mole fraction of }CO_2\times \text{Molar mass of } CO_2)$

Now put all the given values in this expression, we get:

$40.94g/mole=((x)\times 28g/mole)+((1-x)\times 44g/mole)$

$x=0.1894$

The mole fraction of $CO$ = x = 0.1894

The mole fraction of $CO_2$ = 1 - x = 1 - 0.1894 = 0.8106

The percent composition by volume of mixture of $CO$ = $0.1894\times 100=18.94\%$

The percent composition by volume of mixture of $CO_2$ = $0.8106\times 100=81.06\%$

Therefore, the percent composition by volume of mixture of $CO$ and $CO_2$ are, 18.94 % and 81.06 % respectively.

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3 years ago

Suppose two chemicals are reacted together. It takes 1000 kJ of energy to break the bonds of the reactants and 850 kJ of energy

Deffense [45]

The true statement is :
B.The reaction will need an external energy source to proceed. It requires 150kJ more for it to take place. The difference between the energy consumed and the energy produced is 150kJ.

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4 years ago

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Distinguish between the terms "insoluble" and "unsaturated." Give an example to illustrate each term.

Serjik [45]

Insoluble means that a substance does not dissociate in another, like mixing water and oil.

In an unsaturated solution, the solute does dissociate into solution but there is more "room" in the solution to hold more. Say if you have an amount of water that can dissolve 20g of salt but you only have 15g dissolved in it, that solution is unsaturated.

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3 years ago

What two elements are used to make a laser pen light?

irina1246 [14]

Answer:

Helium and Neon

Explanation:

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3 years ago