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AfilCa [17]
3 years ago
10

A student collected a 47.5 mL sample of gas in the lab at 0.8 atm pressure and 29.00C. What volume would this gas sample occupy

at standard conditions?
Chemistry
1 answer:
Sav [38]3 years ago
8 0

Answer:

the volume would be 69.034mL

Explanation:

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What change would you expect on the rate of theSN2 reaction of 1-iodo-2-methylbutane withcyanide ion if the nucleophile concentr
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Answer:

The reaction rate of the both questions remain unchanged.

Explanation:

      For question 1:  The reaction 1-iodo -2- methylbutane with cyanide ion is an SN2 reaction because the Alkyl halide is a primary alkyl halide. The rate of reaction is dependent on concentration of the nucleophile and the alkyl halide at the same. For the rate of reaction to be affected (increased or decreased), the concentration of nucleophile and the alkyl halide have to be altered.

      For question 2: The reaction of 2-iodo -2- methylbutane with ethanol is an SN1 reaction because the Alkyl halide is a tertiary alkyl halide. There are two-step reaction mechanism in this reaction. The first step is the rate determining step which determines the extent of the reaction and hence the rate of reaction. For the rate of reaction to be affected (increased or decreased), the concentration of the Alkyl halide alone will be altered. The rate of reaction is independent of the concentration of the nucleophile.

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The density of pure diamond is 3.5 g/cm°. What is the volume of a diamond with a mass of 0.25 g?
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18. When a chemical reaction takes place in an open system, matter_____.
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A) can enter from the surroundings, but cannot escape to the surroundings
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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
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