How many grams of solute are present in 635mL of 0.450 M KBr?

34.03 g

Explanation:

We'll begin by converting 635 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

635 mL = 635 mL × 1 L / 1000 mL

635 mL = 0.635 L

Thus, 635 mL is equivalent to 0.635 L

Next, we shall determine the number of mole of the solute (KBr) in the solution. This can be obtained as follow:

Volume = 0.635 L

Molarity = 0.450 M

Mole of KBr =?

Mole = Molarity × Volume

Mole of KBr = 0.450 × 0.635

Mole of KBr = 0.286 mole

Finally, we shall determine the mass of 0.286 mole of KBr. This can be obtained as follow:

Mole of KBr = 0.286 mole

Molar mass of KBr = 39 + 80

= 119 g/mol

Mass of KBr =?

Mass = mole × molar mass

Mass of KBr = 0.286 × 119

Mass of KBr = 34.03 g

Thus, the mass of the solute (KBr) in the solution is 34.03 g

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8