All categories

2 years ago

How to prepare super saturated solution of copper sulphate ? Explain the experiment

Chemistry

1 answer:

expeople1 [14]2 years ago

6 0

Explanation:

Monitor the temperature of the water with the thermometer. Stop heating the water once it nears the boiling point of 100 degrees Celsius. Add copper(II) sulfate and stir until the heated solution is saturated. When the solution is saturated, copper(II) sulfate will not dissolve anymore

You might be interested in

Under which conditions are gases most soluble in water?

Volgvan

Hi,

Gases are most soluble in water under high pressure and low pressure.

6 0

3 years ago

A marble rolls off horizontally from the edge of table top 1.50 m above the floor. it strikes the floor 2.0 m from the base of t

GrogVix [38]

a. t=0.553 s

b. vox(horizontal speed) = 3.62 m/s

<h3>Further explanation</h3>

Given

h = 1.5 m

x = 2 m

Required

a. time

b. vo=initial speed

Solution

Free fall motion

a. h = 1/2 gt²(vertical motion=h=voyt+1/2gt²⇒voy = 0)

$\tt t=\sqrt{\dfrac{2h}{g} }$

t = √2h/g

t = √2.1.5/9.8

t=0.553 s

b. x=vox.t(horizontal motion)

$\tt x=v_{ox}\times t$

vox=x/t

vox=2/0.553

vox=3.62 m/s

3 0

2 years ago

How many protons and neutrons in the atom K-42

katovenus [111]

There are 20n nutrons in the atom k-42.

3 0

3 years ago

At 2°C, the vapor pressure of pure water is 23.76 mmHg and that of a certain seawater sample is 23.09 mmHg. Assuming that seawat

wariber [46]

Answer:

0.808 M

Explanation:

Using Raoult's Law

$\frac{P_s}{Pi}= x_i$

where:

$P_s$ = vapor pressure of sea water( solution) = 23.09 mmHg

$P_i$ = vapor pressure of pure water (solute) = 23.76 mmHg

$x_i$ = mole fraction of water

∴

$\frac{23.09}{23.76}= x_i$

$x_i = 0.9718$

$x_i+x_2=1$

$x_2 = 1- x_i$

$x_2 = 1- 0.9718$

$x_2 = 0.0282$

$x_i = \frac{n_i}{n_i+n_2}$ ------ equation (1)

$x_2 = \frac{n_2}{n_i+n_2}$ ------ equation (2)

where; $(n_2) =$ number of moles of sea water

$(n_i) =$ number of moles of pure water

equating above equation 1 and 2; we have :

$\frac{n_2}{n_i}$ $= \frac{0.0282}{0.9178}$

$= 0.02901$

NOW, Molarity = $\frac{moles of sea water}{mass of pure water }*1000$

$= \frac{0.02901}{18}*1000$

$= 0.001616*1000$

$= 1.616 M$

As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have $\frac{1.616}{2} =0.808 M$

3 0

3 years ago

Read 2 more answers

What factors affect freezing-point depression of a solution? (select all that apply)

ANTONII [103]

Answer:

A and F i think

Explanation:

Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all lower than that of the pure solvent and is directly proportional to the molality of the solute

8 0

3 years ago