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Ivenika [448]
3 years ago
12

A 7.73 mass % aqueous solution of ethylene glycol (HOCH2CH2OH) has a density of 1.19 g/mL. Calculate the molarity of the solutio

n. Give your answer to 2 decimal places.
Chemistry
1 answer:
butalik [34]3 years ago
8 0

Answer:

Molarity for the solution is 1.48 M

Explanation:

Molarity involves the moles of solute that are contained in 1L of solution. It is a sort of concentration. The most usual.

7.73 by mass involves the mass of solute that are contained in 100 g of solution, so this one of ethylene glycol contains 7.73 g of it.

Let's determine the moles of solute

7.73 g / molar mass of ethylene glycol = moles

7.73 g / 62 g/mol = 0.124 mol

If the mass of solution is 100 g, we can determine the volume with density.

Density = Mass / Volume

Volume = Mass / Density

Volume = 100 g / 1.19 g/mL = 84.03 mL

In conclusion, 0.124 moles are contained in 84.03 mL

Molarity is mol/L, so let's convert the volume in L

84.03 mL . 1L / 1000 mL = 0.08403 L

0.124 mol / 0.08403 L = 1.48 M

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A compound with chemical formula na2cx3 has formula mass 106 amu .. what is the atomic mass of element x
maria [59]

The element "X" is "O" (oxygen).

<h3>Calculation:</h3>

Given,

Chemical formula = Na₂CX₃

Formula mass = 106 amu

Molar mass of Na = 23 amu

Molar mass of C = 12 amu

To find,

Element X =?

We will equate the equation as follows,

2(23) + 12 + 3(y) = 106

46 + 12 + 3y =106

58 + 3y = 106

3y = 106 - 58

3y = 48

y = 48/3

y = 16

We know that Oxygen has molecular mass of 16. Therefore the element "X" is "O".

Learn more about molar mass here:

brainly.com/question/22997914

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5 0
1 year ago
A chemist prepares a solution of silver(I) nitrate by measuring out of silver(I) nitrate into a volumetric flask and filling the
leva [86]

Answer:

3.33 M

Explanation:

It seems your question is incomplete, however, that same fragment has been found somewhere else in the web:

" <em>A chemist prepares a solution of silver nitrate (AgNO3) by measuring out 85.g of silver nitrate into a 150.mL volumetric flask and filling the flask to the mark with water.</em>

<em>Calculate the concentration in mol/L of the chemist's silver nitrate solution. Be sure your answer has the correct number of significant digits.</em> "

In this case, first we <u>calculate the moles of AgNO₃</u>, using its molecular weight:

  • 85.0 g AgNO₃ ÷ 169.87 g/mol = 0.500 mol AgNO₃

Then we<u> convert the 150 mL of the volumetric flask into L</u>:

  • 150 / 1000 = 0.150 L

Finally we <u>divide the moles by the volume</u>:

  • 0.500 mol AgNO₃ / 0.150 L = 3.33 M
4 0
3 years ago
In the reaction fecl2 + 2naoh fe(oh)2(s) + 2nacl, if 6 moles of fecl2 are added to 6 moles of naoh, how many moles of fecl2 woul
pychu [463]
The balanced chemical reaction would be 

<span>fecl2 + 2naoh =  fe(oh)2(s) + 2nacl

Initial amounts of the reactants are given, so, we need to determine which of the reactants is the limiting reactant and use this amount to determine what is asked. However, what is being asked is how many of the FeCl2 is used in the reaction, showing that it is NaOH that is the limiting reactants. Thus, we just use the initial amount of NaOH and relate the substances by the chemical reaction as follows:

6 mol NaOH ( 1 mol FeCl2 / 2 mol NaOH ) = 3 mol FeCl2

Therefore, 3 moles of FeCl2 is used up and 3 moles of FeCl2 is also left after the reaction.</span>
5 0
3 years ago
Read 2 more answers
21 Which process is used to determine the concentration of an acid?
Inessa05 [86]

Answer:

4) Titration

Explanation:

Titration is a common process used to determine the concentration of acids. It does this by adding a solution of base with a known concentration to the acid until it reaches neutralization.

8 0
3 years ago
Calculate the dissociation constant of nh4oh(aq) if the degree of dissociation of 0.006 mol/kg solution is 0.053 and the activit
Anastasy [175]

The dissociation equation will be

                         NH4OH   --->        NH4+                   + OH-

Initial                 0.006                        0                          0

Change         -0.006 X 0.053        +0.006 X 0.053      -0.006 X 0.053

Equlibrium     0.006 -0.006 X 0.053      0.006 X 0.053    0.006 X 0.053

Ka = [NH4+] [ OH-] / [NH4OH] = (0.006 X 0.053)^2 / 0.006 -0.006 X 0.053

Ka = 1.78 X 10^-5

7 0
3 years ago
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