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3 years ago

Apart from boiling point and flammability, what other quality does the size of the molecules in hydrocarbons affect?

Chemistry

2 answers:

Sonbull [250]3 years ago

6 0

Answer:

Viscosity

Explanation:

Sijui kuexpalain lakini

ratelena [41]3 years ago

5 0

Answer:

Explanation:

apart from boiling point and flammability viscosity density and solubility are those qualities of hydrocarbon which are effected by their size

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kumpel [21]

Answer:

3.14

Explanation:

3.14 is known as pie

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3 years ago

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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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3 years ago

Which of the following is not a correct chemical equation for a double displacement reaction? (D?)

emmasim [6.3K]

The correct answer would be the last option. A double displacement type of reaction involves the switching of places the cations and anions accordingly. The given reaction is erroneous since in the product side the anions and cations are being paired which would not make sense. The correct reaction should be

4NaBr + Co(SO3)2 yields <span>CoBr4 + 2Na2SO3</span>

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4 years ago

BALANCING CHEMICAL EQUATIONS WORKSHEET

kherson [118]

Answer:

Put that in y butt ps

Explanation:

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3 years ago

Consider the reaction of 19.0 g of zinc metal with excess silver nitrate to produce silver metal and zinc nitrate. The reaction

Alexeev081 [22]

Answer:

14.5 g silver

Explanation:

This is a problem using the stoichiometry of the reaction. First thing we need is the balanced equation:

Zn + 2 AgNO3 ----------------------- 2 Ag + Zn(NO3)2

We know that 14.6 g of Zn did not reacted, then we can calculate the amount of Zn reacted and do the calculation given the above reaction.

amount Zn reacted: 19.0 -14.6 g Zn = 4.4 g Zn

atomic weight of Zn: 65.37 g/mol

mol Zn reacted: 4.4 g Zn x ( 1 mol Zn/ 65.37 g Zn) = 0.067 mol Zn

We know from the balanced equation that moles of Ag are produced from 1 mol Zn therefore the mol of Ag produced are:

0.067 mol Zn x 2 mol Ag/ 1mol Zn = 0.135 mol Ag

and the mass of silver then will be given by multiplying by the atomic weight of silver:

0.135 mol Ag x 107.9 g/mol = 14.5 g Ag

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3 years ago

Apart from boiling point and flammability, what other quality does the size of the molecules in hydrocarbons affect?​

Apart from boiling point and flammability, what other quality does the size of the molecules in hydrocarbons affect?