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Blizzard [7]
1 year ago
13

Write the chemical formula for tin(ii)sulfite.. Please show working thank you

Chemistry
1 answer:
Alex17521 [72]1 year ago
6 0

Accordingly, the majority of the bonds in the molecule, tin (ii) sulfide, are covalent, accounting for around 90% of the bonds. An extremely thin anode interlayer is used in organic photovoltaics by the compound tin(II) thiocyanate, Sn(SCN)2.

Stannic sulfide, tin disulfide, and tin bis(sulfanylidene) are examples. EC number: 215-252-9; CAS number: 1315-01-1. A chemical compound is tin(II) sulfate. SnSO4 is its atomic number. Tin and sulfate ions can be found in it. The chemical element tin has the atomic number 50 and the letter Sn for its symbol. At room temperature, tin is solid and is categorized as a post-transition metal. In order to create Tin dioxide and two molecules of hydrogen gas, Tin combines with two molecules of water, which are really in the form of steam. A reminder: Acids and alkalis corrode tin.

To know more about bonds.,

brainly.com/question/977324

#SPJ4

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If a substance has a half-life of 55.6 s, and if 230.0 g of the substance are present initially, how many grams will remain afte
Assoli18 [71]

Answer:

m=0.127g

Explanation:

Hello,

In this case, for a first-order reaction, we can firstly compute the rate constant from the given half-life:

k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{55.6s}=0.0125s^{-1}

In such a way, the integrated first-order law, allows us to compute the final mass of the substance once 10.0 minutes (600 seconds) have passed:

m=m_0*exp(-kt)=230.0g*exp(-0.0125s^{-1}*600s)\\\\m=0.127g

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a solution is made by dissolving cadmium fluoride (CdF2 Ksp =6.44×10 −3 ) in 100.0 mL of water until excess solid is present. So
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Answer:

Answers are in the explanation

Explanation:

Ksp of CdF₂ is:

CdF₂(s) ⇄ Cd²⁺(aq) + 2F⁻(aq)

Ksp = 6.44x10⁻³ = [Cd²⁺] [F⁻]²

When an excess of solid is present, the solution is saturated, the molarity of Cd²⁺ is X and F⁻ 2X:

6.44x10⁻³ = [X] [2X]²

6.44x10⁻³ = 4X³

X = 0.1172M

<h3>[F⁻] = 0.2344M</h3><h3 />

Ksp of LiF is:

LiF(s) ⇄ Li⁺(aq) + F⁻(aq)

Ksp = 1.84x10⁻³ = [Li⁺] [F⁻]

When an excess of solid is present, the solution is saturated, the molarity of Li⁺ and F⁻ is XX:

1.84x10⁻³ = [X] [X]

1.84x10⁻³ = X²

X = 0.0429

<h3>[F⁻] = 0.0429M</h3><h3 /><h3>The solution of CdF₂ has the higher fluoride ion concentration</h3>
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