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sweet [91]
4 years ago
6

The thick soil accumulation of fine-grained particles is called: O Sediment Loess O Gravel O Silt​

Chemistry
1 answer:
DaniilM [7]4 years ago
5 0

The thick soil accumulation of fine-grained particles is called as silt

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18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c
Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

   Percentage by mass of hydrogen: 0.77%

b)  Percentage by mass of chlorine: 80.37%

c) Molecular formula: C_{2} H Cl_{3}

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}

The same process is used to calculate the amount of hydrogen:

0.089g H_{2}O*\frac{2g H}{18g H_{2}O }  = 0.010g H

The percentage by mass of carbon and hydrogen are calculated as follows:

%C\frac{0.238g}{1.3g} *100%= 18.3%

%H\frac{0.010g}{1.3g} *100%=0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}= 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl=\frac{0.43g}{0.535g} * 100%= 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

18.3g C*\frac{1mol C}{12g C} = 1.52mol C

0.77g H*\frac{1mol H}{1g H} = 0.77mol H

80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl

Dividing each of the quantities above by the smallest (0.77mol), the  subscripts in a tentative formula would be

C=\frac{1.52}{0.77} = 1.97 ≈ 2

H = \frac{0.77}{0.77} = 1

Cl =\frac{2.27}{0.77}=2.94≈3

The empirical formula for the compound is:

C_{2} H Cl_{3}

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

5 0
3 years ago
Nitrogen dioxide is produced by combustion in an automobile engine. For the following reaction, 0.377 moles of nitrogen monoxide
Contact [7]

Answer:

The amount of NO₂ that can be produced 8.533 g

Explanation:

       According to question

                                2 NO(g) + O₂(g) → 2 NO₂(g)

Given

Moles of nitrogen monoxide = 0.377

Moles of oxygen = 0.278

'For NO'=\frac{Mole}{Stoichiometry}=\frac{0.377}{2} =0.1855\\'For O_{2} '=\frac{0.278}{1}= 0.278\\

Since 'NO' is the limiting reagent according to this ratio.

According to equation

         2 moles NO reacts to form 2 moles NO₂

So,  0.1855 moles NO give  = 0.1855 moles of NO₂

            Mass of 1 mole NO₂ = 46 g/mole

            Mass of 0.1855 moles = 46 x 0.1855 = 8.533 g

5 0
4 years ago
A theory is:
SpyIntel [72]
A theory is a proposed explanation for an observation
3 0
2 years ago
How many grams of solute are needed to prepare a 3.50% mass/mass solution that has a solution mass of 2.50x102 grams.
Elis [28]
One way of expressing concentration is by percent. It may be on the basis of mass, mole or volume. Percent is expressed as the amount of solute per amount of the solution. For this case, we are given the percent by mass. In order to solve the amount of solute, we multiply the percent with the amount of the solution.

Mass of solute = percent by mass x mass solution
Mass of solute = 0.0350 x 2.50 x10^2 = 8.75 grams of solute
5 0
4 years ago
2. This graph shows the energy involved in a reaction involving two molecules. Describe what is happening or what the state of t
blagie [28]
If you think of it endothermic is when there is energy needed for the reaction to occur and exothermic is when the reaction releases energy
6 0
3 years ago
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