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3 years ago

How do sediments form?

2 answers:

6 0

Sedimentary<span> rocks are formed when </span>sediment<span> is deposited out of air, ice, wind, gravity, or water flows carrying the particles in suspension</span>

Ksju [112]3 years ago

4 0

Sediments is a term used for small peices of rocks, dirt, sand, etc. sediments can be formed from larger things.

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A solution is 40.00% by volume benzene (C6H6) in carbon tetrachloride at 20°C. The vapor pressure of pure benzene at this temper

finlep [7]

Answer:

The total vapor pressure is 84.29 mmHg

Explanation:

Step 1: Data given

Solution = 40.00 (v/v) % benzene in CCl4

Temperature = 20.00 °C

The vapor pressure of pure benzene at 20.00 °C = 74.61 mmHg

Density of benzene is 0.87865 g/cm3

The vapor pressure of pure carbon tetrachloride is 91.32 mmHg

We suppose the total volume = 100 mL

Step 2: Calculate volume benzene and CCl4

40 % benzene = 40 mL

60 % mL CCl4 = 60 mL

Step 3: Calculate mass benzene

Mass = density * volume

Mass of benzene = 40.00 mL * 0.87865 g/mL = 35.146 g

Step 4: Calculate moles of benzene

Moles = mass / molar mass

Number of moles of benzene = 35.146 grams / 78 g/mol = 0.45059 mol

Step 5: Calculate mass of CCl4

Mass of CCl4 = 60 mL * 1.5940 g/mL = 95.64 g

Step 6: Calculate moles CCl4

Number of moles of CCl4 = 95.64 grams / 154g/mol = 0.62104 mol

Step 7: Calculate total number of moles

Total number of moles = moles benzene + moles CCl4

0.45059 moles + 0.62104 moles = 1.07163 mol

Step 8: Calculate mole fraction benzene and CCl4

Mole fraction = moles benzene / total moles

Mole fraction of benzene = 0.45059 / 1.07163 = 0.4205

Mole fraction of CCl4 = 0.62104 / 1.07163 = 0.5795

Step 9: Calculate partial pressure

Partial pressure of benzene = 0.4205 * 74.61 = 31.37 mmHg

Partial pressure of CCl4 = 0.5795 * 91.32 = 52.92 mmHg

Total vapor pressure = 31.37 + 52.92 = 84.29 mmHg

The total vapor pressure is 84.29 mmHg

7 0

3 years ago

What is the molarity of a solution that contains 125 g nacl in 4.00 l solution?

bogdanovich [222]

Answer:

[NaCl[ = 0.535M

Explanation:

We determine the moles of solute:

125 g . 1 mol/ 58.45 g = 2.14 moles

Molarity (mol/L) → 2.14 mol / 4L → 0.535M

Molarity is a sort of concentration that indicates the moles of solute in 1L of solution

4 0

4 years ago

A sample of 1.000 g of a compound containing carbon and hydrogen reacts with oxygen at elevated temperature to yield 0.692 g H₂O

ollegr [7]

Answer :

(a) 1.000 g of compound containing carbon and hydrogen is, 0.922 g and 0.0769 g respectively.

(b) There is no other element present in the compound.

Explanation :

(a) Now we have to determine the masses of C and H in the sample.

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of Carbon and hydrogen respectively.

We are given:

Mass of $CO_2=3.381g$

Mass of $H_2O=0.692g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.381 g of carbon dioxide, $\frac{12}{44}\times 3.381=0.922g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.692 g of water, $\frac{2}{18}\times 0.692=0.0769g$ of hydrogen will be contained.

Thus, 1.000 g of compound containing carbon and hydrogen is, 0.922 g and 0.0769 g respectively.

(b) Now we have to determine the compound contain any other elements or not.

Mass carbon + Mass of hydrogen = 0.922 g + 0.0769 g = 0.999 g ≈ 1 g

This means that there is no other element present in the compound.

3 0

3 years ago

A particular reaction, A- products, has a rate that slows down as the reaction proceeds. The half-life of the reaction is found

Thepotemich [5.8K]

Explanation:

Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.

a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement a is wrong.

b. Expression for second order reaction is as follows:

$\frac{1}{[A]} =\frac{1}{[A]_0} +kt$

the above equation can be written in the form of Y = mx + C

so, the plot between 1/[A] and t is linear. So the statement b is true.

Expression for half life is as follows:

$t_{1/2}=\frac{1}{k[A]_0}$

As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.

Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong

8 0

3 years ago

What is the total number of joules required to freeze a 10 g sample of water at 0°C?

Lunna [17]

Explanation:

the answer will be 98.4 kJ

8 0

3 years ago