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Vlad1618 [11]
2 years ago
10

64. Which of the followings can dissociate in water?

Chemistry
2 answers:
Arisa [49]2 years ago
8 0

Answer:

<h2>Hi there !</h2>

<h2>C. HCl</h2>

Explanation:

<h2>Reason :-</h2>

<h2>Salts are strong electrolytes, so they undergo complete dissociation.</h2><h3>Hope it helps u.....</h3><h3>Stay safe, stay healthy and blessed</h3><h3>Have a good day</h3><h3>Thank you ~</h3>
MAXImum [283]2 years ago
8 0

Answer:

HCl

Explanation:

Dissociation denotes to increase in mols

Vant Hoff factor for dissociation is always Greater than 1

Here the reaction is given by

  • HCl—»H+ +Cl-
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ololo11 [35]

The molality of the solution = 17.93 m

<h3>Further explanation</h3>

Given

6.00 L water with 6.00 L of  ethylene glycol(ρ=1.1132 g/cm³= 1.1132 kg/L)

Required

The molality

Solution

molality = mol of solute/ 1 kg solvent

mol of solute = mol of ethylene glycol

  • mass of ethylene glycol :

= volume x density

= 6 L x 1.1132 kg/L

= 6.6792 kg

= 6679.2 g

  • mol of ethylene glycol (MW=62.07 g/mol)

=mass : MW

=6679.2 : 62.07

=107.608

  • mass of water

6 L water = 6 kg water(ρ= 1 kg/L)

  • molality

\tt =\dfrac{107.608}{6}=17.93~m

5 0
3 years ago
What is the concentration of a salt water solution with 15 grams of salt in 100 mL of water?
DanielleElmas [232]
It’s 5% because without the 5% you wouldn’t make it to 100 equally
8 0
3 years ago
Read 2 more answers
How many moles are in 2g of H2O<br>​
Studentka2010 [4]

Answer:

1/9 moles

Explanation:

No of moles = mass/molar mass

No of moles = 2g/18gmol-1

No of moles = 1/9 moles

7 0
3 years ago
Be sure to answer all parts.
MrRissso [65]

Answer: The molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

Explanation:

Molarity is the number of moles of a substance present in liter of a solution.

And, moles is the mass of a substance divided by its molar mass.

(a) Moles of ethanol (molar mass = 46 g/mol) is as follows.

Moles = \frac{mass}{molar mass}\\= \frac{28.5 g}{46 g/mol}\\= 0.619 mol

Now, molarity of ethanol solution is as follows.

Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.619 mol}{4.50 \times 10^{2} \times 10^{-3}L}\\= 1.38 M

(b) Moles of sucrose (molar mass = 342.3 g/mol) is as follows.

Moles = \frac{mass}{molar mass}\\= \frac{21.6 g}{342.3 g/mol}\\= 0.063 mol

Now, molarity of sucrose solution is as follows.

Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.063 mol}{0.067 L}  (1 mL = 0.001 L)\\= 0.94 M

(c) Moles of sodium chloride (molar mass = 58.44 g/mol) are as follows.

Moles = \frac{mass}{molar mass}\\= \frac{6.65 g}{58.44 g/mol}\\= 0.114 mol

Now, molarity of sodium chloride solution is as follows.

Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.114 mol}{0.0962 L}\\= 1.182 M

Thus, we can conclude that the molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

4 0
3 years ago
Need help please it must be right thank u
vova2212 [387]

Answer:

13. c) Net

14. a) Gravity

15. d) Watt

16. b) the speed will suddenly decrease.

Explanation:

13. The <u>net </u> force is combination of all individual forces.

14. When we sit on the chair , then <u>gravitational force</u> acts on our body due to our mass.

F = mg

here, F = force due to gravity , g = acceleration due to gravity  and m= mass of the body

15.Power : It is the force acting per unit area .It is discovered by James Watt

Power =\frac{force}{area}

Power =\frac{N}{m^{2}}

Its unit is Nm^{-2} .In honor of the Scientist , its unit is Watt.

Watt = Nm^{-2}

16. When the parachute first opens , the <u>speed of the skydiver suddenly decreases</u> because of the resistance offered by the air particles.(For more details look at the figure attached)

17. Transfer of momentum from one object to other is t<u>he law of conservation of momentum</u><u> </u>(NOT ENERGY).So the answer is<u> </u><u>False</u>

<u> Law of Conservation of Momentum : </u> Total momentum of a system is conserved if no external force attack on it. Momentum can neither be created nor destroyed but transferred from one form to other.

7 0
3 years ago
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