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2 years ago

How to balance an ionic equation BaCl2 + 2NH4CH3COO

1 answer:

4 0

Assuming that the reaction proceeds, the reaction would be a double displacement reaction. BaCl2 + 2NH4CH3COO -> Ba(CH3COO)2 + 2NH4Cl If the reaction occurs in an aqueous solution, the products are soluble in water, so the reaction is probably not that interesting to look at.

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Calculate the cell potential for the reaction as written at 25.00 °C 25.00 °C , given that

Taya2010 [7]

Answer:

E = 2.02 V

Explanation:

In order to do this, we need to apply the Nernst equation which is:

E = E° - RT/nF lnQ

The value of RT/F can be simplified to just 0.059 because we are doing this experiment at 25 °C, and R and F are constants. so we need the value of Q which in this case is:

Q = [Mg²⁺] / [Ni²⁺]

We already have the concentrations, so, all we have left is the standard reduction potential, which are:

E° Mg = -2.38 V

E° Ni = -0.25 V

According to the overall reaction:

Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s)

we can see that one element is reducting and the other is oxidizing, so we need to write the semi equation of reduction for each element:

Mg(s) ---------> Mg²⁺ + 2e⁻ E° = 2.38 V oxidizing (Value of E° inverted)

Ni²⁺ + 2e⁻ -----------> Ni(s) E° = -0.25 V reducting

------------------------------------------------------------

Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s) E° = 2.13 V

We have the value of the standard potential, now we need to replace all given data into the nernst equation to solve for the cell potential:

E = 2.13 - 0.059/2 ln(0.757/0.0160)

E = 2.13 - 0.0295 ln(47.3125)

E = 2.13 - 0.11

E = 2.02 V

This is the cell potential

3 0

3 years ago

What were the limition of doberiner classifacation

alexgriva [62]

Answer: I don’t kno

Explanation:

5 0

2 years ago

The solubility of oxygen gas in water at 40 ∘c is 1.0 mmol/l of solution. What is this concentration in units of mole fraction?

juin [17]

The formula for mole fraction is:

$mole fraction of solute = \frac{number of moles of solute}{total number of moles of solution}$ -(1)

The solubility of oxygen gas = 1.0 mmol/L (given)

1.0 mmol/L means 1.0 mmol are present in 1 L.

Converting mmol to mol:

$1.00 mmol\times \frac{1 mol}{1000 mmol} = 0.001 mol$

So, moles of oxygen = 0.001 mol

For moles of water:

1 L of water = 1000 mL of water

Since, the density of water is 1.0 g/mL.

$Density = \frac{mass}{volume}$

$Mass = 1.0 g/ml\times 1000 mL = 1000 g$

So, the mass of water is 1000 g.

Molar mass of water = 18 g/mol.

Number of moles of water = $\frac{1000 g}{18 g/mol} = 55.55 mol$

Substituting the values in formula (1):

$mole fraction = \frac{0.001}{55.55+0.001}$

$mole fraction = 1.8\times 10^{-5}$

Hence, the mole fraction is $1.8\times 10^{-5}$ .

7 0

2 years ago

What mass of H2 is needed to react with 8.75 g of O2 according to the following equation: O2(g) + H2(g) → H2O(g)?

alina1380 [7]

Mass of H₂ needed to react with O₂ : 1.092 g

<h3>Further explanation</h3>

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight / volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.

Reaction

O₂(g) + 2H₂(g) → 2H₂O(g)

mass of O₂ : 8.75 g

mol O₂(MW=32 g/mol) :

$\tt \dfrac{8.75}{32}=0.273$