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rewona [7]
3 years ago
15

Explain why one molecule of nabh4 will reduce only two molecules

Chemistry
1 answer:
12345 [234]3 years ago
4 0
<span>m- acetylbenzaldehyde which has both an aldehyde and a ketone. Thus each molecule takes two hydrogens and borohydride. Borohydride has four hydrides to give, so it will reduce two molecule.</span>
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How many H2O molecules are contained in 0.0016 moles of the hydrate Na2B4O7.10H2O?
mamaluj [8]
Maybe this example could help you to understand this problem.

https://image.slidesharecdn.com/121howmanyatoms-091201144624-phpapp02/95/12-1-how-many-atoms-17-728....
8 0
3 years ago
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Will sodium iodide react with bromine to produce sodium bromide and iodine? Why or why not?
Gre4nikov [31]

Answer:

C

Explanation:

The higher the period the higher the activity of an element, therefore, since iodine is in period 6 and bromine is in period 5, the described reaction is not possible due to the fact that bromine is less active

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2 years ago
1 point
inysia [295]

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8 0
3 years ago
.<br> How many moles are there in 8.5 x 1024 molecules of sodium sulfate, Na2SO4?
NeTakaya

Answer: To solve this question, we need to use the Avogadro's Number, which is a constant first discovered by Amadeo Avogadro, an Italian scientist. He discovered that in a mole of a substance, there are 6,02*10²³ molecules. Using this relationship, we apply the following conversion factor:

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5 0
2 years ago
Calculate ΔH for the reaction: C(graphite) + 2H 2(g) + 1/2 O 2(g) =&gt; CH 3OH(l) Using the following information: C(graphite) +
Alika [10]

Answer:

\Delta H for the given reaction is -238.7 kJ

Explanation:

The given reaction can be written as summation of three elementary steps such as:

C(graphite)+O_{2}(g)\rightarrow CO_{2}(g) \Delta H_{1}= -393.5 kJ

2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l) \Delta H_{2}= (2\times -285.8)kJ

CO_{2}(g)+2H_{2}O(l)\rightarrow CH_{3}OH(l)+\frac{3}{2}O_{2}(g)  \Delta H_{3}= 726.4 kJ

---------------------------------------------------------------------------------------------------

C(graphite)+2H_{2}(g)+\frac{1}{2}O_{2}(g)\rightarrow CH_{3}OH(l)

\Delta H=\Delta H_{1}+\Delta H_{2}+\Delta H_{3}=-238.7 kJ

4 0
3 years ago
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