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damaskus [11]
2 years ago
9

PLEASE HELP I WILL MARK YOU BRAINLIEST !!If you react 4.25 moles of FeCl2, how many moles of Cl2 did you also react?

Chemistry
1 answer:
Ksju [112]2 years ago
7 0
Lol yeah that’s what I’m doing I just want you know I got a lot of you guys to go and you don’t get
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An ionic bond is a bond between
OleMash [197]

between two oppositely charged ions

7 0
3 years ago
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Determine the acid dissociation constant for a 0.0250 M weak acid solution that has a pH of 2.37 . The equilibrium equation of i
Oksanka [162]

Answer:

For part (a): pHsol=2.22

Explanation:

I will show you how to solve part (a), so that you can use this example to solve part (b) on your own.

So, you're dealing with formic acid, HCOOH, a weak acid that does not dissociate completely in aqueous solution. This means that an equilibrium will be established between the unionized and ionized forms of the acid.

You can use an ICE table and the initial concentration ofthe acid to determine the concentrations of the conjugate base and of the hydronium ions tha are produced when the acid ionizes

HCOOH(aq]+H2O(l]⇌ HCOO−(aq] + H3O+(aq]

I 0.20 0 0

C (−x) (+x) (+x)

E (0.20−x) x x

You need to use the acid's pKa to determine its acid dissociation constant, Ka, which is equal to

3 0
2 years ago
A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg
Brut [27]

Answer:  1.0\times 10^2g

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like urea)

K_f = freezing point constant = 3.96^0C/m

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}

Weight of solvent (X)= 950 g = 0.95 kg  

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}

x=1.0\times 10^2g

Thus 1.0\times 10^2g urea was dissolved.

8 0
3 years ago
A solution contains 15 grams of salt per litter of solution. How many grams of salt are in 3.3 litters of solution?​
Verizon [17]

49.5g

Explanation:

Mass of salt = 15g

Volume of solution = 1L

Volume of given solution = 3.3L

Unknown:

Mass of salt in the solution = ?

Solution:

Since we have been given the concentration of the salt in the solution, we can use it to solve the problem.

Concentration; 15g/L

Given;

    In 1L of the solution we have 15g of salt,

In    3.3L of the solution we will have 3.3 x 15 = 49.5g

The salt is the solute and it represents the dissolved substances.

learn more:

Concentration brainly.com/question/4641902

#learnwithBrainly

7 0
3 years ago
A brick has it dimensions of 25 cm x 5 cm x 15 cm so what is the volume of the brick in cubic meters
Sveta_85 [38]
Well the formula for the volume of a cube is

V=length×width×height

so input it

V=25×5×15

V=1875 cubic meters
8 0
3 years ago
Read 2 more answers
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