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3 years ago

What is the net ionic equation for barium sulphate

Chemistry

1 answer:

alexandr1967 [171]3 years ago

4 0

The net ionic equation : Ba²⁺ + SO₄²⁻ ⇒ BaSO₄ (s)

<h3>Further explanation </h3>

The electrolyte in the solution produces ions.

The equation of a chemical reaction can be expressed in the equation of the ions

In the ion equation, there is a spectator ion that is the ion which does not react because it is present before and after the reaction

When these ions are removed, the ionic equation is called the net ionic equation

For gases and solids including water (H₂O) can be written as an ionized molecule

So only the dissolved compound is ionized ((expressed in symbol aq)

Barium sulfate can be formed from the reaction:

Ba(NO₃)₂(aq) + Na₂SO₄(aq)⇒BaSO₄(s)+2NaNO₃(aq)

For full ionic equation :

Ba²⁺ + 2NO₃⁻ + 2 Na⁺ + SO₄²⁻ ⇒ BaSO₄ (s) + 2 Na⁺ + 2 NO³⁻

by removing spectator ions (2NO₃⁻ and 2 Na⁺), the net ionic equation :

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Suppose a 500.mL flask is filled with 1.0mol of CO, 1.5mol of H2O and 0.70mol of CO2. The following reaction becomes possible: +

valentina_108 [34]

Answer:

[CO] = 0.62 M

Explanation:

Step 1: Data given

Volume of the flask = 500 mL

Number of moles CO = 1.0 moles

Number of moles H2O = 1.5 moles

Number of moles CO2 = 0.70 moles

The equilibrium constant K for this reaction is 3.80

Step 2: The balanced equation

CO(g) + H2O(g) ⇆ CO2(g) +H2(g)

Step 3: Calculate the initial concentrations

Concentration = moles / volume

[CO] = 1.0 moles / 0.500 L = 2.0 M

[H2O] = 1.5 moles / 0.500 L = 3.0 M

[CO2] = 0.70 moles / 0.500 L = 1.4 M

[H2] = 0M

Step 4: The concentration at the equilibrium

For 1 mol CO we have 1 mol H2O to produce 1 mol CO2 and 1 mol H2

[CO] = 2.0 -X M

[H2O] = 3.0 - X M

[CO2] =1.4 + X M

[H2] = X M

Step 5: Define Kc

Kc = [CO2][H2]/ [CO][H2O]

3.80 = (1.4 + X) * X / ((2.0 - X)*3.0 -X))

X = 1.38

[CO] = 2.0 -1.38 = 0.62 M

[H2O] = 3.0 - 1.38 = 1.62 M

[CO2] =1.4 + 1.38 M = 2.78

[H2] = 1.38 M

Kc = (2.78*1.38) / (0.62*1.62)

Kc = 3.8

[CO] = 0.62 M

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3 years ago

A 1 kg block of copper is hammered into a thin sheet. The length of the copper sheet is 30 cm. What is the mass of the copper sh

charle [14.2K]

The answer is <span>B) 30 kg ..............................</span>

6 0

4 years ago

Read 2 more answers

An unknown compound has the following chemical formula: where stands for a whole number. Measurements also show that a certain s

anygoal [31]

The question is incomplete, the complete question is;

An unknown compound has the following chemical formula:

N2Ox

where x stands for a whole number.

Measurements also show that a certain sample of the unknown compound contains 3.7 mol of oxygen and 2.45 mol of nitrogen.

Write the complete chemical formula for the unknown compound.

Answer:

The complete formula of the compound is N2O3.

Explanation:

Given;

Number of moles of N = 2.45 moles

Number of moles of O = 3.7 moles

Chemical formula of unknown compound = N2Ox

From the ratio of the number of moles of N to O;

2/x = 2.45/3.7

2 * 3.7 = x * 2.45

x = 2 * 3.7/ 2.45

x =3

Hence the complete formula of the compound is N2O3.

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3 years ago

How many protons are in Cesium-135 (Cs)? 35 55 80 195

sergey [27]

protons of cesium = 55

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3 years ago

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Calculate the [OH-] and the pH for a solution of 0.24M methylamine, CH3NH2. Kb = 3.7 X 10-4.

Naddika [18.5K]

Answer:

$[OH^-]=9.24x10^{-3}M$ .

$pH=11.97$ .

Explanation:

Hello,

In this case, since the ionization of methylamine is:

$CH_3NH_2(aq)+H_2O(l)\rightleftharpoons CH_3NH_3^+(aq)+OH^-(aq)$

The equilibrium expression is:

$Kb=\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}$

And in terms of the reaction extent $x$ which is equal to the concentration of OH⁻ as well as that of CH₃NH₃⁺ via ice procedure we can write:

$3.7x10^{-4}=\frac{x*x}{024-x}$

Whose solution for $x$ via quadratic equation is 9.24x10⁻³ M since the other solution is negative so it is avoided. Therefore, the concentration of OH⁻ is:

$[OH^-]=x=9.24x10^{-3}M$

With which we can compute the pOH at first:

$pOH=-log([OH^-])=-log(9.24x10^{-3})=2.034$

Then, since pH and pOH are related via:

$pH+pOH=14$

The pH turns out:

$pH=14-pOH=14-2.034\\\\pH=11.97$

Best regards.

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3 years ago