Which of the following compounds would you expect to be an electrolyte?

75 mL of water is added to a 360 mL solution of acetic acid with a concentration of 0.87 M. Determine the molarity of the new so

kramer

Answer:

The molarity of the new solution is 0.72 M

Explanation:

Step 1: Data given

Volume of the original solution = 360 mL =.360 L

Molarity = 0.87 M

We add 75 mL = 0.075 L

Step 2: Calculate moles

Moles = molarity * volume

Moles = 0.87 M * 0.360 L

Moles = 0.3132 moles

Step 3: Calculate new molarity

The number of moles stays constant

Molarity = moles / volume

Molarity = 0.3132 moles / (0.36+0.075)

Molarity = 0.3132 moles / 0.435 L

Molarity = 0.72 M

The molarity of the new solution is 0.72 M

3 0

4 years ago

If a star is 33.11 trillion kilometers away how much light years would that be

abruzzese [7]

33.11 trillion kilometers is equivalent to 3.5 light years

hope this helps :)

8 0

3 years ago

Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres

yulyashka [42]

Answer: The freezing point of solution is -0.454°C

Explanation:

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (KCl) = 5.0 g

$M_{solute}$ = Molar mass of solute (KCl) = 74.55 g/mol

$W_{solvent}$ = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC$

Hence, the freezing point of solution is -0.454°C

3 0

3 years ago

The density of Diamond is 3.51 g/cm3, and the density of platinum is 21.43 g/cm3. If equal masses of diamond and platinum are tr

Reptile [31]

Answer:

Explanation:

Diamond has lesser density than platinum . So , if we take equal mass of both , the volume of mass of platinum will be far less .

The density of both diamond and platinum are more than water so both of them will be drowned in water completely . They will not float . On being drowned , platinum will displace lesser volume of water because of its less volume . So volume change in case of platinum mass will be far less . The volume change for diamond will be more because of its bigger size.

4 0

3 years ago

What is the oxidation number of chlorine in HClO4?

Elza [17]

CIO4-=-1

CI=4O=-1

O has a 2- oxidation change so

CI+4(-2)=-1

CI+(-8)=-1

CI=-1+8=7

So the oxidation number of chlorine is 7 in this case

8 0

3 years ago