Chemitey half equations

What is the difference between ionic formula and the molecular formula?

Kaylis [27]

Answer:

Molecular compounds are pure substances formed when atoms are linked together by sharing of electrons while ionic compounds are formed due to the transfer of electrons. 2. ... Molecular compounds are formed between two non-metals while ionic compounds are formed between metals and non-metals.

3 0

3 years ago

The sp of pbbr2 is 6. 60×10−6. what is the molar solubility of pbbr2 in pure water?

Inessa05 [86]

Ksp of PbBr₂ is 6.60 × 10⁻⁶. The molar solubility of PbBr₂ in pure water is 0.0118M.

Ksp or Solubility Product Constant is an equilibrium constant for the dissociation in an aqueous solution.

Molar solubility (S) is the concentration of the dissolved substance in a solution that is saturated.

Let the molar solubility be S upon dissociation.

PbBr₂ or Lead Bromide dissociates in pure water as follows:

PbBr₂ ----------> Pb⁺² + Br⁻

S 2S

Ksp = [Pb⁺²] [ Br⁻]

Ksp = (S) (2S)²

Ksp = 4S³

6.60 × 10⁻⁶ = 4S³

S = 0.0118M

Hence, the Molar solubility S is 0.0118M.

Learn more about Molar solubility here, brainly.com/question/16243859

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6 0

2 years ago

When three pairs of electrons are shared in a molecule, what shape will the molecule be?

djyliett [7]

The correct answer is B. The shape of a molecule where three pairs of electrons are shared is a trigonal planar. This is characterized by one central atom and three atoms forming an equilateral triangle which is bonded to the central atom.

8 0

3 years ago

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The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 Fe2(SO4)3 + 2Na3PO4 What is the t

slava [35]

<u>Answer:</u> The theoretical yield of iron(III) sulfate is 26.6 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of iron(III) phosphate = 20.00 g

Molar mass of iron(III) phosphate = 150.82 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron(III) phosphate}=\frac{20g}{150.82g/mol}=0.133mol$

The given chemical equation follows:

$2FePO_4+3Na_2SO_4\rightarrow Fe_2(SO_4)_3+2Na_3PO_4$

As, sodium sulfate is present in excess. So, it is considered as an excess reagent.

Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate

So, 0.133 moles of iron(III) phosphate will produce = $\frac{1}{2}\times 0.133=0.0665moles$ of iron(III) sulfate

Now, calculating the mass of iron(III) sulfate from equation 1, we get:

Molar mass of iron(III) sulfate = 399.9 g/mol

Moles of iron(III) sulfate = 0.0665 moles

Putting values in equation 1, we get:

$0.0665mol=\frac{\text{Mass of iron(III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron(III) sulfate}=(0.0665mol\times 399.9g/mol)=26.6g$

Hence, the theoretical yield of iron(III) sulfate is 26.6 grams

8 0

3 years ago

A gas occupies 2.0 m3 at 100.0k and exerts a pressure of 100.0kPa. What volume will the gas occupy if the temperature is increas

Svetradugi [14.3K]

According to ideal gas equation, we know for 1 mole of gas: PV=RT
where P = pressure, T = temperature, R = gas constant, V= volume
If '1' and '2' indicates initial and final experimental conditions, we have
$\frac{P1V1}{P2V2} = \frac{T1}{T2}$

Given that: V1 = 100.0 kPa, T1 = 100.0 K, V1 = 2.0 m3, T2 = 400 K, P2 = 200.0 kPa

∴ on rearranging above eq., we get V2 = $\frac{P1V1T2}{T1} = \frac{100 X 2 X 400}{200X100}$
∴ V2 = 4 m3

7 0

3 years ago