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3 years ago

What is the molality of an aqueous solution that contains 29.5 g of glucose (C6H12O6) dissolved in 950 g of water (H2O)?

1 answer:

7 0

Molality is one way of expressing concentration of a solute in a solution. It is expressed as the mole of solute per kilogram of the solvent. To calculate for the molality of the given solution, we need to convert the mass of solute into moles and divide it to the mass of the solvent.

Molality = 29.5 g glucose (1 mol / 180.16 g ) / .950 kg water
Molality = 0.1724 mol / kg

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Draw the products of the complete hydrolysis of an acetal. Draw all products of the reaction.

jolli1 [7]

Answer: the product is ketone or aldehyde

Explanation:

The first step is the conversion of acetal to hemiacetal in the presence of H3O+/ ROH, and then the final conversion of hemiacetal to ketone/aldehyde using

H3O+/ ROH...

Attached is the structural conversion

8 0

3 years ago

Draw this image below. It shows one ton of feathers and one ton of bricks a. Compare the mass of the two substances:

bogdanovich [222]

I don't understand the question

Can you elaborate further

8 0

3 years ago

Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

4 0

3 years ago

Help!! < PHYSICAL SCIENCE>

Vlad1618 [11]

The answer is B or the second answer

3 0

3 years ago

If 45.8 grams of potassium chlorate decomposes, how many grams of oxygen gas can be produced? 2KClO3 → 2KCl + 3O2

natta225 [31]

To do this problem, we must first look at the balanced chemical equation for the decomposition of potassium chlorate:

2KClO3 --> 2KCl + 3O2 

We can take the given amount of grams, and use the molar mass of KClO3 to convert to moles. Then, we can use the stoichiometric ratios to relate moles of KClO3 to moles of O2. 

(39.09)+(35.45)+(3*15.99)= 122.51 g/ mol = molar mass of KClO3 
45.8 g KClO3/ 122.51 g/ mol KClO3 = .374 moles KClO3 
.374 mol KClO3 *(3 moles O2/2 mol KClO3)= .560 moles O2 

Once we have moles of O2, we can convert to grams of O2. 

(2*15.99)= 31.98 g/mol = molar mass of O2 
(.560 moles O2) (31.98 g/mol)= 17.91 g O2 

Hope this helps :)

3 0

3 years ago