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Mars2501 [29]
3 years ago
13

The reducing agent in this catalytic hydrogenation reaction was molecular hydrogen (H2), which was produced in situ (in the reac

tion mixture) during the course of the reaction. a) During the course of the experiment, how did you know that H2 had been formed? b) Write a balanced chemical equation for the reaction that produced H2. c) Was H2 the limiting reactant? Why or why not? Show the necessary calculations to support your answer.
Chemistry
1 answer:
creativ13 [48]3 years ago
4 0

Answer:

a)  After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) 4NaBH_{4} + 2HCl + 7H_{2}O ⇒ 16H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7} c) No H_{2} was not the limiting reactant.

Explanation:

Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.

a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.

b)  4NaBH_{4} + 2HCl + 7H_{2}O ⇒ 16H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}

c) H_{2} was not the limiting reactant based on the mol to mol ratio of H_{2} and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of H_{2} would also be produced.

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tresset_1 [31]

Answer:

.125 M

Explanation:

.15 M/L   * .125 L = .01875 moles

now dilute to 150 cc  (by adding 25 cc)

.01875M / (150/1000) = .125M

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A solution of aminoheptanoic acid, H2N-(CH2).COOH, in m-cresol at a concentration of 3.3 mol amino acid per kg of solution is po
garik1379 [7]

Explanation:

The given data is as follows.

       Concentration = 3.3 mol

      Rate constant = 2.74 \times 10^{3} kg/mol min

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      \tau = C_{A_{0}} \times \int_{0}^{X_{A}} \frac{dX_{A}}{-r_{A}}

Initial concentration (C_{A_{0}}) of aminoheptanoic is 3.3 kg/mol.

       -r_{A} = 2.74 \times 10^{3} kg/mol min

Hence, putting the given values into the above formula as follows.

    \tau = C_{A_{0}}\times \int_{0}^{X_{A}} \frac{dX_{A}}{-r_{A}}

              = \frac{3.3}{2.74 \times 10^{3} kg/mol min} [X_{A}]^{0.1}_{0}

              = 120.43 min

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3 0
3 years ago
Which statement best describes electrons?(1) They are positive subatomic particles and are found in the nucleus.(2) They are pos
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The answer should be (4) they are negative subatomic particles and are found surrounding the nucleus. The nucleus is formed by neutrons and protons. Neutrons are neutral and protons are positive.
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8 0
3 years ago
2. A 10.0 g ice cube, initially at 0.0ºC, is melted in 100.0 g of water that was initially 20.0ºC. After the ice has melted, the
arlik [135]

Answer:

The heat lost  by the water  Q_{water}  = 3.8 KJ

The heat gain by ice  Q_{ice} = 228.76 J

The heat required to melt the ice Q_{melt} = 3340 J

Explanation:

Mass of ice cube m_{ice} = 10 gm

Initial temperature of ice cube T_{ice} = 0 °c

Mass of water m_{water} = 100 gm

Initial temperature of water T_{w} = 20 °c

Final temperature of mixture T_{f} = 10.93 °c

(a). Total heat lost by the water Q_{water}  =  m_{water} C_{w} ( T_{f} - T_{w} )

⇒ Q_{water}  = 100 × 4.184 (20 - 10.93)

⇒ Q_{water}  = 3.8 KJ

This is the heat lost  by the water.

(b). Heat gained by the ice cube Q_{ice} = m_{ice} C_{ice} (T_{f} - T_{ice}  )

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⇒ Q_{melt} = 3340 J

This is the heat required to melt the ice.

7 0
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