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steposvetlana [31]
3 years ago
9

What is the solubility product expression for Ag2CO3

Chemistry
1 answer:
Dominik [7]3 years ago
7 0
<span> Ksp = [Ag+]^2[CO32-]that should be it </span>
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Answer:1 2 and 4

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The bottom number on each element of a on the periodic table is called
ruslelena [56]

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On the periodic table, there are two numbers for every element: the atomic number and the atomic weight.

The numbers at the top of the square represents the number of protons present in the atom's nucleus of that element. This is called the atomic number.

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3 years ago
Using the mmoles of (35)-2,2,-dibromo-3,4-dimethylpentane calculated earlier and the molecular weight of the product (962 g/mol)
Sedbober [7]

Answer:

The yield of the product in gram is \mathsf{{w_P}=0.26 \ gram}

Explanation:

Given that:

the molecular mass weight of the product = 96.2 g/mol

the molecular mass of the reagent (3S)-2,2,-dibromo-3,4-dimethylpentane is 257.997 g

given that the millimoles of the reagent = 2,7 millimoles = 2.7 \times 10^{-3} \ moles

We know that:

Number of moles = mass/molar mass

Then:

2.7 \times 10^{-3} = \dfrac{ mass}{257.997}

mass = 2.7 \times 10^{-3} \times 257.997

mass = 0.697

Theoretical yield = (number of moles of the product/ number of moles of reactant) × 100

i.e

Theoretical yield = \dfrac{n_P}{n_R}\times 100\%

where;

n_P = \dfrac{w_P}{m_P}    and n_R = \dfrac{w_R}{m_R}

Theoretical yield = \dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%

Given that the theoretical yield = 100%

Then:

100\% =\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%

\dfrac{w_P}{m_P}=\dfrac{w_R}{m_R}

{w_P}=\dfrac{w_R \times m_P}{m_R}

where,

w_P = derived weight of the product

m_P =the molecular mass of the derived product

m_R = the molecular mass of the reagent

w_R = weight in a gram of the reagent

{w_P}=\dfrac{w_R \times m_P}{m_R}

{w_P}=\dfrac{0.697 \times 96.2}{257.997}

\mathsf{{w_P}=0.26 \ gram}

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3 years ago
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Answer:

Mixture.

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C is the answer for this!
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