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Nat2105 [25]
2 years ago
5

Calculate the molarity of the sodium acetate solution as described below.

Chemistry
1 answer:
Airida [17]2 years ago
4 0

Answer:

This question is incomplete.

Explanation:

This question is incomplete because of the absence of given mass and volume, however, the steps below will help solve the completed question. The molarity (M) of a solution is the number of moles of solute per liter of solvent. The formula is illustrated below;

Molarity = number of moles (n) / volume (in liter or dm³)

To calculate the number of moles of NaC₂H₃O₂, we say

number of moles (n) =

given or measured mass of NaC₂H₃O₂ ÷ molar mass of NaC₂H₃O₂

The volume of the solvent must be in liter (same as dm³). Thus, to convert mL to liter, we divide by 1000

The unit for Molarity is M (Molar concentration), mol/L or mol/dm³

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Explanation:

8 0
3 years ago
Can this reaction take place/does it exist - . CaCl2+CO2+H20----> CaCO3 + 2HCl
mash [69]
Yes the reaction given above does exist
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so i conclude it does exist
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6 0
3 years ago
Transition metals tend to
bonufazy [111]
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7 0
2 years ago
In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide Al2O3 dissolved in molten c
Darya [45]

Answer:

0.382g

Explanation:

Step 1: Write the reduction half-reaction

Al³⁺(aq) + 3 e⁻ ⇒ Al(s)

Step 2: Calculate the mass of Al produced when a current of 100. A passes through the cell for 41.0 s

We will use the following relationships.

  • 1 A = 1 C/s
  • 1 mole of electrons has a charge of 96486 C (Faraday's constant)
  • 1 mole of Al is produced when 3 moles of electrons pass through the cell.
  • The molar mass of Al is 26.98 g/mol.

The mass of Al produced is:

41.0s \times \frac{100C}{s} \times \frac{1mole^{-} }{96486C} \times \frac{1molAl}{3mole^{-} } \times \frac{26.98gAl}{1molAl} = 0.382gAl

7 0
3 years ago
Consider the following reaction:
Kitty [74]

Answer : The value of equilibrium constant (Kc) is, 0.0154

Explanation :

The given chemical reaction is:

                        SO_2Cl_2(g)\rightarrow SO_2(g)+Cl_2(g)

Initial conc.    2.4\times 10^{-2}          0             0

At eqm.          (2.4\times 10^{-2}-x)   x              x

As we are given:

Concentration of Cl_2 at equilibrium = 1.3\times 10^{-2}M

That means,

x=1.3\times 10^{-2}M

The expression for equilibrium constant is:

K_c=\frac{[SO_2][Cl_2]}{[SO_2Cl_2]}

Now put all the given values in this expression, we get:

K_c=\frac{(x)\times (x)}{2.4\times 10^{-2}-x}

K_c=\frac{(1.3\times 10^{-2})\times (1.3\times 10^{-2})}{2.4\times 10^{-2}-1.3\times 10^{-2}}

K_c=0.0154

Thus, the value of equilibrium constant (Kc) is, 0.0154

4 0
3 years ago
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