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Diano4ka-milaya [45]
3 years ago
6

1. Put 0.000 034 into scientific notation.

Chemistry
2 answers:
xxMikexx [17]3 years ago
6 0

Answer:

3.4x10 to the power of 5

Explanation:

Marina CMI [18]3 years ago
5 0

Answer:

My answer to the question is 3.4×10^-5.

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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.

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An example can be perspiration or even rain.

Hope this helps.

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Answer:

<em>That's </em><em>because</em><em> </em><em>in </em><em>water</em><em> </em><em>NaF </em><em>will </em><em>dissolve</em><em> </em><em>to </em><em>produce </em><em>Na</em><em>+</em><em>,</em><em>the </em><em>conjugate </em><em>base </em><em>of </em><em>a </em><em>strong</em><em> </em><em>acid </em><em>which</em><em> </em><em>will </em><em>not </em><em>react </em><em>with </em><em>water.</em><em>h</em><em>o</em><em>w</em><em>e</em><em>v</em><em>e</em><em>r</em><em> </em><em>F- </em><em>will </em><em>behave </em><em>like </em><em>a </em><em>bronsted </em><em>base,</em><em> </em><em>and </em><em>accept</em><em> </em><em>a </em><em>proton </em><em>from </em><em>water.</em><em>t</em><em>h</em><em>i</em><em>s</em><em> </em><em>is </em><em>called </em><em>hydrolysis</em><em> </em><em>reaction,</em><em> because</em><em> </em><em>a </em><em>molecule</em><em> </em><em>of </em><em>water </em><em>is </em><em>broken </em><em>up.</em>

<em>a </em><em>conjugate</em><em> base</em><em> </em><em>is </em><em>what </em><em>I </em><em>leftover </em><em>after </em><em>an </em><em>acid </em><em>loses </em><em>a </em><em>hydrogen</em><em> </em><em>ion.</em>

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