The vapor pressure of 25 milliliters of water at 25 degrees celcius will be the same as

(*Anohter one* For my friends!!! ^^)

Sliva [168]

Answer:

Thx Have a Fantastic day :)

Explanation:

7 0

3 years ago

Read 2 more answers

Calculate the area of a 3.0 inch by 5.0 inch index card in square millimeters (mm). (You can look up the formula for the area of

meriva

Answer:

The area of the given rectangular index card = 9677.4 mm²

Explanation:

Area is defined as the space occupied by a two dimensional shape or object. The SI unit of area is square metre (m²).

The area of a rectangle (A) = length (l) × width (w)

Given dimensions of the rectangle: Length (l) = 5.0 inch, Width (w) = 3.0 inch

Since, 1 inch = 25.4 millimetres (mm)

Therefore, l = 5 × 25.4 = 127 mm, and w = 3 × 25.4 = 76.2 mm

Therefore, the area of the given rectangular index card = A= l × w = 127 mm × 76.2 mm = 9677.4 mm²

5 0

2 years ago

Select the compound that is most likely to increase the solubility of ZnSe when added to water.NaCnMgBr2NaClKClO4

vlada-n [284]

Answer:

NaCl.

Explanation:

In the solution, ZnSe ionizes to $Zn^2^+$ and $Se^2^-$ . Following reaction represents the ionization of ZnSe in solution -

$ZnSe$ ⇄ $Zn^2^+ + Se^2^-$

As we want to increase the solubility of ZnSe, we must decrease the concentration of dissociated ions so that the reaction continues to forward direction.

If we add NaCl to this solution, then we have $Na^+$ and $Cl^-$ in the solution which will be formed by the ionization of NaCl.

Now, $Zn^2^+$ in the solution will react with two $Cl^-$ ions to form $ZnCl_2$ as follows -

$Zn^2^++2Cl^-$ ⇄ $ZnCl_2$

Due to this reaction the concentration of $Zn^2^+$ will decrease in the solution and more ZnSe can be soluble in the solution.

6 0

2 years ago

Please help me with this question

vampirchik [111]

Remember that

For being a bond covalent ∆E<1.8
For being a bond ionic ∆E>1.8

#1

∆EN=0.5

Carbon is present so it's covalent

#2

OH bond is also covalent

#3

P-H will hardly form a bond

#4

Ionic bond

#5

Ionic bond

7 0

1 year ago

The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be

Mrac [35]

Answer: The freezing point of solution is 5.35°C

Explanation:

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 = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 4.90°C/m

$m_{solute}$ = Given mass of solute (naphthalene) = 2.60 g

$M_{solute}$ = Molar mass of solute (naphthalene) = 128.2 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC$

Hence, the freezing point of solution is 5.35°C

3 0

3 years ago