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3 years ago

Determine the molality of an aqueous solution that is 16.5 percent urea by mass

Chemistry

1 answer:

yarga [219]3 years ago

7 0

Take a hypothetical sample of exactly 100 grams of the solution.

(16g urea) / (60.06 g urea/mol) = 0.2664 mol urea

((100 g total) - (16g urea)) = 84.0 g H2O = 0.0840 kg H2O

(0.2664 mol) /0.0840 (kg) = 3.17143mol/kg = 3.18m urea

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Does anyone know the anserw the symbol of the element

Komok [63]

Cesium.

Groups are the vertical columns that run up and down while periods are the horizontal rows. So to find the answer to this, go to the first column (Group 1) and find the sixth period (row 6) which will land you on Cesium, element 55.

Hope this helps!

3 0

3 years ago

Which substance is most likely metallic

Montano1993 [528]

I don’t know “which substance” are you talking about??

You can put the picture there so we can see.

8 0

3 years ago

Read 2 more answers

A reaction was performed in which 3.6 g 3.6 g of benzoic acid was reacted with excess methanol to make 1.3 g 1.3 g of methyl ben

Anit [1.1K]

Answer: The percent yield of the reaction is 32.34 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For benzoic acid:

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 122.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol$

The chemical equation for the reaction of benzoic acid and methanol is:

$\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}$

By Stoichiometry of the reaction

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0295 moles of benzoic acid will produce = $\frac{1}{1}\times 0.0295=0.108$ moles of methyl benzoate

Now, calculating the mass of methyl benzoate from equation 1, we get:

Molar mass of methyl benzoate = 136.15 g/mol

Moles of methyl benzoate = 0.0295 moles

Putting values in equation 1, we get:

$0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g$

To calculate the percentage yield of methyl benzoate, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of methyl benzoate = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

$\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%$

Hence, the percent yield of the reaction is 32.34 %

6 0

3 years ago

CONCLUSIONS: Suppose a student titrated a sample of monoprotic acid of unknown concentration using a previously standardized sol

Zolol [24]

Concentration of unknown acid is 0.061 M

Given:

Concentration of NaOH = 0.125 M

Volume of NaOH = 24.68 mL

Volume of acid solution = 50.00 mL

To Find:

Concentration of the unknown acid

Solution: Concentration is the abundance of a constituent divided by the total volume of a mixture. The concentration of the solution tells you how much solute has been dissolved in the solvent

Here we will use the formula for concentration:

M1V1 = M2V2

0.125 x 24.68 = 50 x M2

M2 = 0.125 x 24.68 / 50

M2 = 0.061 M

Hence, the concentration of unknown acid is 0.061 M

Learn more about Concentration here:

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3 0

2 years ago

Help in my exam plzz now

swat32

Answer:

yes your answer is correct

I hope it helps mate

I will always help you understanding your assingments have a great day

#Captainpower:)

8 0

3 years ago