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3 years ago

Write a balanced net ionic equation for the following double replacement reaction

Chemistry

1 answer:

shtirl [24]3 years ago

3 0

Answer:

Ba²⁺(aq) + SO₄²⁻(aq) ⟶ BaSO₄(s)

Explanation:

There are three steps you must follow. You must write the:

Molecular equation
Ionic equation
Net ionic equation

A. Molecular equation

BaCl₂(aq) + Na₂SO₄(aq) ⟶ BaSO₄(s) + 2NaCl(aq)

B. Ionic equation

You write all the soluble substances as ions.

Ba²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + SO₄²⁻(aq) ⟶ BaSO₄(s) + 2Na⁺(aq) + 2Cl⁻(aq)

C. Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

Ba²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + SO₄²⁻(aq) ⟶ BaSO₄(s) + 2Na⁺(aq) + 2Cl⁻(aq)

The net ionic equation is

Ba²⁺(aq) + SO₄²⁻(aq) ⟶ BaSO₄(s)

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A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

4 0

3 years ago

You have to prepare some 2 M solutions, with 10 g of solute in each. What volume of solution will you prepare, for each solute b

alexdok [17]

Answer:

0.045 L or 45 mL

Explanation:

Moles = Mass/M.Mass

Moles = 10 g / 109.94 g/mol

Moles = 0.09 moles

Also,

Molarity = Moles / Vol in L

Or,

Vol in L = Moles / Molarity

Vol in L = 0.09 mol / 2 mol/L

Vol in L = 0.045 L

8 0

1 year ago

Can one atom in this type of reaction win the "tug of war"? What might happen if it did?

Luden [163]

Answer:

POLAR BONDS will form.

A polar bond is a covalent bond between two atoms where the electrons forming the bond are unequally distributed. This causes the molecule to have a slight electrical dipole moment where one end is slightly positive and the other is slightly negative.

7 0

2 years ago

What does population density indicate? ______

ohaa [14]

Answer:

number of individuals

Explanation:

calculated by dividing the number of people by area

3 0

3 years ago

Read 2 more answers

Methane gas and oxygen gas react to form water vapor and carbon dioxide gas. What volume of water would be produced by this reac

yanalaym [24]

volume of H₂O = 7.2 L

Explanation:

The combustion reaction of methane (CH₄):

CH₄ + 2 O₂ → CO₂ + 2 H₂O

Now we calculate the number of moles of methane using the following formula:

number of moles = volume / 22.4 (L/mole)

number of moles of CH₄ = 3.6 / 22.4

number of moles of CH₄ = 0.16 moles

Taking in account the chemical reaction, we devise the following reasoning:

if 1 mole of CH₄ produce 2 moles of H₂O

then 0.16 moles of CH₄ produce X moles of H₂O

X = (0.16 × 2) / 1 = 0.32 moles of H₂O

And now we can calculate the volume of water (H₂O) produced by the reaction:

number of moles = volume / 22.4 (L/mole)

volume = number of moles × 22.4 (L/mole)

volume of H₂O = 0.32 × 22.4

volume of H₂O = 7.2 L

Learn more about:

combustion reaction

brainly.com/question/14122510

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8 0

2 years ago