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ohaa [14]
3 years ago
11

How much so you can be dissolved and 50 mL of water at 30 Celsius see chart:)

Chemistry
1 answer:
RoseWind [281]3 years ago
8 0

Answer:

The correct answer is - 100 grams.

Explanation:

The solubility graph is a very useful tool because it tells you the amount of solute that can be dissolved per  100 mL  of water in order to have a saturated solution of potassium nitrate at a given temperature.

In order to find the solubility of the salt at  30 ∘ C , start from the  30 ∘ C

mark on the graph and move up until you intersect the curve. At the point of intersection, move left until you intersect the  y  axis and read off the value that you land on which is 200. However, it is for 100 ml so half the value you read:

200/2 = 100 gram .

So, solubility  ≈  100 g / 50 mL water

So, a saturated solution of solute will hold about  100 g  of dissolved solute, that is of dissociated solute, per 50 mL  of water at  30 ∘ C .

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At 20 ?C the vapor pressure of benzene (C6H6)is 75 torr, and that of toluene (C7H8) is 22 torr
tia_tia [17]

Answer:

(a) Benzene = 0.26; toluene = 0.74

(b) Benzene = 0.55

Explanation:

1. Calculate the composition of the solution

For convenience, let’s call benzene Component 1 and toluene Component 2.

According to Raoult’s Law,  

p_{1} = \chi_{1}p_{1}^{\circ}\\p_{2} = \chi_{2}p_{2}^{\circ}

where

p₁ and p₂ are the vapour pressures of the components above the solution

χ₁ and χ₂ are the mole fractions of the components

p₁° and p₂° are the vapour pressures of the pure components.

Note that

χ₁ + χ₂ = 1

So,  

\begin{array}{rcl}p_{\text{tot}} &=& p_{1} + p_{2}\\p_{\text{tot}} & = & \chi_{1}p_{1}^{\circ}  + (1 - \chi_{1})p_{2}^{\circ}\\36 & = & \chi_{1}\times 75  + (1 - \chi_{1}) \times 22 \\36 & = & 75\chi_{1} + 22 -22\chi_{1}\\14 & = & 53\chi_{1}\\\chi_{1} & = & \mathbf{0.26}\\\end{array}

χ₁ = 0.26 and χ₂ = 0.74

2. Calculate the mole fraction of benzene in the vapour

In the liquid,  

p₁ = χ₁p₁° = 0.26 × 75 mm = 20 mm

∴ In the vapour

\chi_{1} = \dfrac{p_{1}}{p_{\text{tot}} } = \dfrac{\text{20 mm} }{\text{36 mm}}  = 0.55

Note that the vapour composition diagram below has toluene along the horizontal axis. The purple line is the vapour pressure curve for the vapour. Since χ₂ has dropped to 0.45, χ₁ has increased to 0.55.

7 0
3 years ago
The hydroboration of an alkene occurs in ___________ which places the boron of the borane on the ___________ carbon of the doubl
Nana76 [90]

Answer: The hydroboration of an alkene occurs in TWO CONCERTED STEP which places the boron of the borane on the LESS SUBSTITUTED carbon of the double bond. The oxidizing agent then acts as a nucleophile, attacking the electrophilic BORON and resulting in the placement of a hydroxyl group on the attached carbon. Thus, the major product of the hydroboration oxidation reaction DOES NOT follow Markovnikov's rule.

Explanation:

Hydroboration is defined as the process which allows boron to attain the octet structure. This involves a two steps pathway which leads to the production of alcohol.

--> The first step: this involves the initiation of the addittion of borane to the alkene and this proceeds as a concerted reaction because bond breaking and bond formation occurs at the same time.

--> The second step: this involves the addition of boron which DOES NOT follow Markovnikov's rule( that is, Anti Markovnikov addition of Boron). This is so because the boron adds to the less substituted carbon of the alkene, which then places the hydrogen on the more substituted carbon.

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