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4 years ago

Use the molar solubility, 1.08×10−5m, in pure water to calculate ksp for bacro4.

1 answer:

8 0

The solubility constant or Ksp is calculated by the product of the concentration of the ions raise to the stoichiometric coefficient. We calculate as follows:

BaCrO4 = Ba2+ + CrO42-

Ksp = [Ba2+][CrO4]
Ksp = [1.08×10−5<span> ] [1.08×10−5] = 1.1664x10^-10</span>

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What is the pressure in a gas container that is connected to an open-end U-tube manometer if the pressure of the atmosphere is 7

Whitepunk [10]

Answer:

The pressure in the gas is 656mmHg

Explanation:

In calculating the pressure of the gas;

step 1: convert the height of the mercury arm to mmHg

9.60cm = 96.0 mmHg

step 2: convert 752 torr to mmHg

I torr is 1 mmHg

752 torr = 752mmHg

Step 3: since the level of mercury in the container is higher than the level of mercury exposed to the atmosphere, we substrate the values to obtain our pressure.

So, 752mmHg - 96mmHg = 656mmHg

The pressure in the gas container is therefore 656mmHg.

N. B : if the mercury arm is in lower position, you add.

5 0

3 years ago

describe how 250 cm³ of 0.2 mol/dm³ H2SO4 could be prepared from 150 cm³ of 1.0mol/dm³ stock solution of the acid

Reptile [31]

Answer:

Explanation:

250 cm^3 of 0.2 moldm-3 H2SO4 can be prepared from 150cm^3 of 1.0 moldm^-3 by dilution.

150cm^3 of the 1.0 moldm^-3 stock solution is measured out using a measuring cylinder and transferred into a 250 cm^3 standard volumetric flask and made up to mark. The resulting solution is now 250cm^3 of 0.2 moldm-3 H2SO4.

5 0

3 years ago

Including all coefficients, charges, and phases, what is the correct net ionic equation for the following set of reactants? Assu

Musya8 [376]

The <u>correct net ionic equation</u> for the chemical equation involving the reactants Ba(OH)2(aq) and H2SO4(aq) is Ba²⁺ (aq) + 2OH⁻ (aq) + 2H⁺(aq) + SO4²⁻ (aq) --> BaSO₄(s) + H₂O(l)

<h3>Further Explanation </h3>

The chemical reaction between Ba(OH)2(aq) and H2SO4(aq) will be given by;

Ba(OH)₂(aq) + H₂SO4(aq) --> BaSO₄(aq) + 2H₂O(l)

<h3>Net ionic equations </h3>

Net ionic equations that do not include all ions that are present in the reactants and products.
The equations only includes the ions that are involved in the reaction.

The ions that are not involved in a chemical reaction are called spectator ions and are cancelled out while writing ionic equations.
Reaction between Ba(OH)2(aq) and H2SO4(aq is an example of a neutralization reaction or a double displacement reaction.
During the reaction Barium ion bonds with sulfate ions and forms barium sulfate which is a solid, while hydrogen ion bonds with hydroxide ion to form water which is a liquid

That is:

Ba(OH)₂(aq) + H₂SO4(aq) --> BaSO₄(s) + 2H₂O(l)

Barium ion changes from aqueous to solid state barium sulfate there it is an ion involved in the reaction.
Additionally, all the other ions are also involved in the reaction since they have changed their initial state. The chemical reaction has no spectator ions.

Therefore; the ionic equation required will be;

Ba²⁺ (aq) + 2OH⁻ (aq) + 2H⁺(aq) + SO4²⁻ (aq) --> BaSO₄(s) + H₂O(l)

Keywords: Net ionic equations

<h3>Learn more about; </h3>

Net ionic equation: brainly.com/question/5552670
Spectator ions: brainly.com/question/5552670
Example of net ionic equations: brainly.com/question/5552670

Level: High school

Subject: Chemistry

Topic: Chemical equation

Sub-topic: Net ionic equations

7 0

3 years ago

Read 2 more answers

Aluminum oxide (Al2O3) occurs in nature as a mineral called corundum, which is noted for its hardness and resis- tance to attack

Neko [114]

Answer:

7.03×10²³ atoms of Al

Explanation:

Corundum density = 3.97g/cm³

Corundum = Al₂O₃. Then, you have 2 moles of Al and 3 moles of O in 1 mol of the compound.

Let's determine the mass of corundum, by density

Corundum density = Corundum mass / Corundum volume

3.97 g/cm³ = Corundum mass / 15cm³

3.97 g/cm³ . 15cm³ = Corundum mass →59.55 g

Now, let's find out how many moles are in that mass of corundum

1 mol of Al₂O₃ weighs 101.96 g/mol

Then, 101.96 grams of oxide have 2 moles of Al

59.55 g of oxide, will have (59.55 .2)/101.96 = 1.17 moles

Now, we can know the quantity of atoms, by this rule of three

1 mol has NA atoms (6.02×10²³)

1.17 moles will have (1.17 . NA) = 7.03×10²³ atoms of Al

5 0

3 years ago

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Verdich [7]

Answer:

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6 0

3 years ago