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Gennadij [26K]
3 years ago
12

Do anyone know how to do question B

Chemistry
1 answer:
Over [174]3 years ago
7 0

Answer:

a) IUPAC Names:

                   1) (<em>trans</em>)-but-2-ene

                   2) (<em>cis</em>)-but-2-ene

                   3) but-1-ene

b) Balance Equation:

                       C₄H₁₀O + H₃PO₄   →   C₄H₈ + H₂O + H₃PO₄

As H₃PO₄ is catalyst and remains unchanged so we can also write as,

                                    C₄H₁₀O   →   C₄H₈ + H₂O

c) Rule:

           When more than one alkene products are possible then the one thermodynamically stable is favored. Thermodynamically more substituted alkenes are stable. Furthermore, trans alkenes are more stable than cis alkenes. Hence, in our case the major product is trans alkene followed by cis. The minor alkene is the 1-butene as it is less substituted.

d) C is not Geometrical Isomer:

        For any alkene to demonstrate geometrical isomerism it is important that there must be two different geminal substituents attached to both carbon atoms. In 1-butene one carbon has same geminal substituents (i.e H atoms). Hence, it can not give geometrical isomers.

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Calculate the specific heat capacity for a 22.7-g sample of lead that absorbs 237 J when its temperature increases from 29.8 °C
soldier1979 [14.2K]

Answer:

\boxed {\boxed {\sf c\approx 0.159 \ J/ g \textdegree C}}

Explanation:

We are asked to find the specific heat capacity of a sample of lead. The formula for calculating the specific heat capacity is:

c= \frac{Q}{m \times \Delta T}

The heat absorbed (Q) is 237 Joules. The mass of the lead sample (m) is 22.7 grams. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature. The temperature increases <em>from</em> 29.8 °C <em>to </em>95.6 °C.

  • ΔT = final temperature -inital temperature
  • ΔT= 95.6 °C - 29.8 °C = 65.8 °C

Now we know all three variables and can substitute them into the formula.

  • Q= 237 J
  • m= 22.7 g
  • ΔT = 65.8 °C

c= \frac {237 \ J}{22.7 \ g  \ \times  \ 65.8 \textdegree C}

Solve the denominator.

  • 22.7 g * 65.8 °C = 1493.66 g °C

c= \frac {237 \  J}{1493.66 \ g \textdegree C}

Divide.

c= 0.1586706479 J /g \textdegree C

The original values of heat, temperature, and mass all have 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place tells us to round the 8 up to a 9.

c \approx 0.159 \ J/g \textdegree C

The specific heat capacity of lead is approximately <u>0.159 Joules per gram degree Celsius.</u>

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3 years ago
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