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3 years ago

Butene boils at -6 degree. Use the kinetic particle theory to explain what happens when butene boils.

Chemistry

1 answer:

lina2011 [118]3 years ago

3 0

Answer:

The "kinetic particle theory" explains the properties of solids, liquids and gases. There are energy changes when changes in state occur.

When heat is added to a substance, the molecules and atoms vibrate faster.

As atoms vibrate faster, the space between atoms increases.

The motion and spacing of the particles determines the state of matter of the substance. The end result of increased molecular motion is that the object expands and takes up more space.

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g Using Newman projections, draw the most stable conformation for each of the following compounds. (a) 3-methyl pentane, viewed

Marysya12 [62]

Answer:

Using Newman projections, draw the most stable conformation for each of the following compounds.

(a) 3-methyl pentane, viewed along with the C2-C3 bond.

(b) 3,3-dimethyl hexane, viewed along with the C3-C4 bond.

Explanation:

(a) The structure of 3-methyl pentane is shown below:

In Newman projection, the most stable conformation is staggard conformation.

In staggard conformation, the torsional strain is very less compared to eclipsed conformation.

(b)3,3-dimethyl hexane, viewed along with the C3-C4 bond.

3 0

3 years ago

1.00 mL of a 3.55 × 10–4 M solution of oleic acid is diluted with 9.00 mL of petroleum ether, forming solution A. Then 2.00 mL o

Leto [7]

Answer:

Concentration of solution B is $7.10\times 10^{-6}M$

Explanation:

Solution A: Total volume of solution A = (9.00+1.00) mL = 10.00 mL

According to law of dilution, $C_{1}V_{1}=C_{2}V_{2}$

where $C_{1}$ and $C_{2}$ are initial and final concentration respectively. $V_{1}$ and $V_{2}$ are initial and final volume respectively.

Here $C_{1}$ = $3.55\times 10^{-4}M$ , $V_{1}$ = 1.00 mL, $V_{2}$ = 10.00 mL

So, $C_{2}=\frac{C_{1}V_{1}}{V_{2}}$ = $\frac{3.55\times 10^{-4}M\times 1.00mL}{10.00mL}=3.55\times 10^{-5}M$

So, concentration of solution A = $3.55\times 10^{-5}M$

Solution B: Total volume of solution B = (2.00+8.00) mL = 10.00 mL

Similarly as above, $C_{2}=\frac{C_{1}V_{1}}{V_{2}}$ = $\frac{3.55\times 10^{-5}M\times 2.00mL}{10.00mL}=7.10\times 10^{-6}M$

So, concentration of solution B = $7.10\times 10^{-6}M$

7 0

4 years ago

Rank the four gases (air, exhaled air, gas produced from the decomposition of H2O2, gas from decomposition of NaHCO3) in order o

Komok [63]

Astronomers use chemical signatures to determine the age and .... Gas mixtures that contain more than 4% hydrogen in air are potentially explosive. ... water vapor, carbon dioxide, and several other gases

6 0

3 years ago

Silver has 2 isotopes Ag-107 and Ag-109. The percent abundance of Ag-107 is 51.35 % and that of Ag-109 is 48.65%. Find the avera

7nadin3 [17]

Given:

Isotopes of silver with atomic masses:

Ag - 107 and Ag - 109

% Abundance of Ag-107 = 51.35 %

% Abundance of Ag-109 = 48.65 %

To determine:

The average atomic mass of Ag

Explanation:

The average atomic mass can be calculated using the formula-

Average Atomic Mass = f1M1 + f2M2 + .......+fnMn

where:

f = fraction representing the abundance of that isotope

M = atomic mass of that isotope

In the case of silver

Average atomic mass = (51.35/100)*107 + (48.65/100)*109

= 54.9445 + 53.0285 = 107.973 amu

Ans: the average atomic mass for silver is 107.973 amu

7 0

3 years ago

The decomposition of N2O5 in solution in carbon tetrachloride proceeds via the reaction 2 N2O5(soln) → 4 NO2(soln) + O2(soln) Th

Fantom [35]

Answer: The amount remained after 151 seconds are 0.041 moles

Explanation:

All the radioactive reactions follows first order kinetics.

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $4.82\times 10^{-3}s^{-1}$

t = time taken for decay process = 151 sec

$[A_o]$ = initial amount of the reactant = 0.085 moles

[A] = amount left after decay process = ?

Putting values in above equation, we get:

$4.82\times 10^{-3}=\frac{2.303}{151}\log\frac{0.085}{[A]}$

$[A]=0.041moles$

Hence, the amount remained after 151 seconds are 0.041 moles

7 0

4 years ago