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Nikitich [7]
3 years ago
10

8. Estimate the maximum volume percent of Methanol vapor that can exist at standard conditions. Vapor pressure = 88.5 mm Hg in a

ir.
Chemistry
1 answer:
sashaice [31]3 years ago
6 0

Explanation:

Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.

It is known that at standard condition, vapor pressure is 760 mm Hg.

And, it is given that methanol vapor pressure in air is 88.5 mm Hg.

Hence, calculate the volume percentage as follows.

                  Volume percentage = \frac{\text{given vapor pressure}}{\text{standard vapor pressure}} \times 100

                                                    = \frac{88.5}{760} \times 100

                                                    = 11.65%

Thus, we can conclude that the maximum volume percent of Methanol vapor that can exist at standard conditions is 11.65%.

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Answer:

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Explanation:

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The elements in Groups 1A(1) and 7A(17) are all quite reactive. What is a major difference between them?
matrenka [14]

The elements in Groups 1A(1) and 7A(17) are all quite reactive.

<h3>Major difference between Groups 1A(1) and 7A(17) : </h3>

Group 7's halogens, which are non-metal elements, become less reactive as you move down the group. In contrast to the alkali metals in Group 1 of the periodic table, this trend is the opposite. The most reactive element in Group 7 is fluorine.

Alkali metals are soft and reactive metals. They react vigorously with water and become more reactive. And other hand halogens are reactive non metals.

  • Elements of group 1A are known as alkali metals. Elements of this group are lithium, sodium, potassium, rubidium, cesium.
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To know more about Groups 1A(1) and 7A(17) please click here :

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6 0
1 year ago
Determine the missing species:
Galina-37 [17]

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Explanation:

a) The given reaction is ^{41}_{20}\textrm{Ca}+^{x}_{y}\textrm{X}\rightarrow ^{41}_{19}\textrm {K}

As the mass on both reactant and product side must be equal:

41+x=41

x=0

As the atomic number on both reactant and product side must be equal:

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b) ^{15}_{8}\textrm{O}\rightarrow ^{15}_{7}\textrm{N}+^{x}_{y}\textrm{X}

Total mass on reactant side = total mass on product side

15 =15 + x

x = 0

Total atomic number on reactant side = total atomic number on product side

8 = 7 + y

y = 1

^{15}_{8}\textrm{O}\rightarrow ^{15}_{7}\textrm{N}+^{0}_{1}\textrm{e}

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Answer:

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Explanation:

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