All categories

2 years ago

Calculate the amount of solute excess out when the 60 gm of saturated solution of

Chemistry

1 answer:

kogti [31]2 years ago

8 0

Answer:

weight of salt = 90 gram,

solubility = 50

sing formula

solubility of solute =

weightofsolventinsaturatedsolution

weightofsoluteinsaturatedsolution

×100

or 50=

weightofsolventinsaturatedsolutio

90gm

×100

so weight of solvent in saturated solution =

90gm

×100=180gm

Hence 180 gm of water is required to prepare a saturated solution.

You might be interested in

Juanita dissolves 46 g of MgBr2 (molar mass: 184.11 g/mol) in 0.5 kg of distilled water. What is the molality of the solution?

irina [24]

Formula: molality, m = n solute / kg solvent

n solute = # of moles of solute = mass(g) / molar mass

Molar mass of Mg Br2 = 184.11 g/mol

m = [46g / 184.11 g/mol] / 0.5 kg = 0.50 mol/kg

3 0

3 years ago

1. A dining hall had a total of 25 tables-some long rectangular tables and some round

TEA [102]

x = 20 long tables

y = 5 round table

Explanation:

We have the following system of equations:

x + y = 25

8x + 6y = 190

From the first equation we have:

x = 25 - y

And we replace x in the second equation:

8(25 - y) + 6y = 190

200 - 8y + 6y = 190

200 - 2y = 190

200 - 190 = 2y

10 = 2y

y = 5

Now we insert the value of y in the next equation:

x = 25 - y

x = 25 - 5

x = 20

Learn more about:

system of equations

brainly.com/question/1748197

brainly.com/question/1658819

#learnwithBrainly

7 0

3 years ago

A chunk of tin weighing 18.5 grams and originally at 97.38 °C is dropped into an insulated cup containing 75.7 grams of water at

weqwewe [10]

Answer:

22.44°C will be the final temperature of the water.

Explanation:

Heat lost by tin will be equal to heat gained by the water

$-Q_1=Q_2$

Mass of tin = $m_1=18.5 g$

Specific heat capacity of tin = $c_1=0.21 J/g^oC$

Initial temperature of the tin = $T_1=97.38^oC$

Final temperature = $T_2$ =T

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2=75.7 g$

Specific heat capacity of water= $c_2=4.184 J/g^oC$

Initial temperature of the water = $T_3=21.52^oC$

Final temperature of water = $T_2$ =T

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

On substituting all values:

$-(18.5 g\times 0.21 J/g^oC\times (T-97.38^oC))=75.7 g\times 4.184 J/g^oC\times (T-21.52 ^oC)$

we get, T = 22.44°C

22.44°C will be the final temperature of the water.

5 0

3 years ago

Identify the name of the smallest particle of an element that still retains all the chemical properties of that element?

Nezavi [6.7K]

The smallest particle of an element that still retains the chemical properties of it is an atom.

4 0

2 years ago

A chemist dissolves 239 mg of pure potassium hydroxide in enough water to make up of solution. Calculate the pH of the solution.

PSYCHO15rus [73]

Answer: the pH is 11.78

Explanation:Please see attachment for explanation

7 0

2 years ago