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Mrac [35]
3 years ago
15

Uppose the reaction between naphthalene and 2-bromo-2-methylpropane did not run to completion, so that the limiting reagents wer

e not consumed. how would the two TLC plates have looked in your first and second run in this case?
A)there wouuld have been two spots in both cases

B)There would have been three spots in both cases

C)There would have been four pots in both cases

D)There would have been no distinct spots but only streaks.
Chemistry
1 answer:
ch4aika [34]3 years ago
3 0

Answer:

C)There would have been four pots in both cases

Explanation:

In an incomplete reaction of naphthalene and 2-bromo-2-methylpropane, we started with 2 spots (one each for the two starting materials) and we should end up with 3 spots If spots will appear clearly, one for naphthalene, one for the alkyl halide and one for the product. However, this is possible to have same number of spots initially and finally when we consider 2 spots each for naphthalene and 2-bromo-2-methylpropane then after reaction we have 2 spots of products as each 1 spot of reactants decreases. Finally making initial 4 and final 2+1+1 =4.

So, the answer is (C)

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A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If
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Answer:

<h3>25.0 grams is the mass of the steel bar.</h3>

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

Q_1=-Q_2

Mass of steel= m_1

Specific heat capacity of steel = c_1=0.452 J/g^oC

Initial temperature of the steel = T_1=2.00^oC

Final temperature of the steel = T_2=T=21.50^oC

Q_1=m_1c_1\times (T-T_1)

Mass of water= m_2= 105 g

Specific heat capacity of water=c_2=4.18 J/g^oC

Initial temperature of the water = T_3=22.00^oC

Final temperature of water = T_2=T=21.50^oC

Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))

On substituting all values:

(m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}

<h3>25.0 grams is the mass of the steel bar.</h3>
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3 years ago
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