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mrs_skeptik [129]
2 years ago
12

The Ksp of barium sulfate (BaSO4) is 1.1 × 10–10. What is the solubility concentration of sulfate ions in a saturated solution a

t 25°C?
a.1.0 x 10^-5m
b. 1.5 x10^-5m
c. 5.5 x 10^-11m
d.7.4 x 10^-6m
Chemistry
2 answers:
vlada-n [284]2 years ago
6 0
Each mole of barium sulfate releases equal moles of barium ions and sulfate ions. Let the concentration of one substance be x. Thus: 
Ksp = [Ba⁺²][SO₄⁻²]
1.1 x 10⁻¹⁰ = (x)(x)
x = 1.05 x 10⁻⁵

The answer is A.

ch4aika [34]2 years ago
6 0

<u>Given:</u>

Ksp of BaSO4 = 1.1 * 10⁻¹⁰

<u>To determine:</u>

The solubility concentration of SO4²⁻ ions

<u>Explanation:</u>

BaSO4 ↔ Ba²⁺(aq) + SO4²⁻(aq)

Ksp = [Ba²⁺][SO4²⁻]

if 's' is the solubility of the ions then we have

Ksp = s²

s = √ksp = √1.1*10⁻¹⁰ = 1.05*10⁻⁵M

Ans: solubility of sulfate ions is 1.05*10⁻⁵ M

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Anna007 [38]
Molality is the number of moles of solute in 1 kg of solvent
number of moles of sucrose - mass of sucrose / molar mass
number of moles of sucrose - 34.2 g / 342.34 g/mol = 0.0999 mol
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molality of sucrose solution - 0.799 mol/kg
7 0
2 years ago
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You are asked to determine the mass of a piece of copper using its reported density, 8.96 g/ml, and a 150-ml graduated cylinder.
kakasveta [241]

Answer:- Mass of copper piece is 290 gram.

Solution:- We know that, mass = density * volume

density of copper is given as 8.96 gram per mL.

Volume of copper piece is the rise change in volume.

Volume of copper piece = 137 mL - 105 mL = 32 mL

Let's multiply the volume by density to calculate the mass of copper:

mass of copper = 32mL(\frac{8.96g}{mL})

mass of copper = 286.72 g

Volume has two significant figures, so if we round the mass to two significant figures then it becomes 290 g.

7 0
2 years ago
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Two nonpolar organic liquids, hexane (C6H14) and heptane (C7H16), are mixed. Do you expect ΔHsoln to be a large positive number,
Aliun [14]

Answer:

ΔH of solution is expected to be close to zero.

Explanation:

When we mix two non polar organic liquids like hexane and heptane,the resulting mixture formed is an ideal solution.An ideal solution is formed when the force of attraction between the molecules of the two liquids is equal to the force of attraction between the molecules of the same type.

For instance if liquids A and B are mixed,

F_{A-A} = F_{B-B} = F_{A-B}

Hence the condition before and after mixing remains unchanged.

Since enthalpy change is associated with inter molecular force of attraction the enthalpy change for ideal solution is zero.

More examples of ideal solutions are:

1. Ethanol and Methanol

2. Benzene and Toluene

3. Ethyl bromide and Ethyl iodide

5 0
2 years ago
If a bottle of olive oil contains 1.2 kg of olive oil, what is the volume, in milliliters (mL), of the olive oil?
Paul [167]

Answer:

1.3 mL

Explanation:

First, get the density of the olive oil, which is 0.917 kg/mL. Then divide the mass by the density:

1.2kg/0.917kg/mL= 1.3086150491 mL. The kg cancel out, leaving us with mL.

It should have 2 significant figures, because 1.2kg has 2 and we are dividing.

7 0
2 years ago
Be sure to answer all parts. The sulfate ion can be represented with four S―O bonds or with two S―O and two S═O bonds. (a) Which
VMariaS [17]

Answer:

a) Representation - (in attachment)

b) Tetrahedral geometry and

c) sp^{3} hybrid orbitals invovle sigma bonding.

\pi orbitals are formed by the overlapping of d-orbitals of sulfur with p-orbitals of oxygen.

Explanation:

a)

Representation in attachment.

The overall charge in the both structures are -2. The structure (b) is favored by the resonance structures since the formal charge with in the species are remained.

Therefore, structure (b) is the better representation of sulfate ion.

b)

In sulfate ion, sulfur atom is attached with four different oxygen atoms. According to the VSEPR theory , the sufate ion has tetrahedral geometry.

There is four sigma bonds and zero lone pairs present on the central metal atom.

Hence, the hybridization of the sulfur atom is sp^{3}

c)

The s and p orbitals of sulfur invovles hybridization and form sigma bonds.The orbitals of sulfur involves \pi bonding.

Therefore, \pi bonds are formed by the overlapping of d-orbitals of sulfur with  p-orbitals of oxygen.

5 0
2 years ago
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