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3 years ago

What is the correct formula for potassium sulfite? khso khso4 k2so4 k2so3

Chemistry

2 answers:

Aleonysh [2.5K]3 years ago

7 0

Answer: The chemical formula for potassium sulfite is $K_2SO_3$

Explanation:

We are given a chemical compound having name potassium sulfite. This compound is formed by the combination of potassium ions and sulfite ions.

Potassium is the 19th element of periodic table having electronic configuration of $[Ar]4s^1$ .

This element will loose 1 electron to form $K^{+}$ ion

Sulfite ion is a polyatomic ion having chemical formula of $SO_3^{2-}$

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

Hence, the chemical formula for potassium sulfite is $K_2SO_3$

Verdich [7]3 years ago

3 0

Potassium sulfite (K₂SO₃)

hope this helps!

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3 years ago

A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.

BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

$\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\ &=185.59\text{ g/mol} \end{aligned}$

Dimensional Analysis:

$\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }$

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

8 0

3 years ago

Help pleaseeeeeeeeeee

hichkok12 [17]

Answer:

D 1 and 3 only I am not sure

Explanation:

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3 years ago

The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.

maw [93]

CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X

Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²

Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M

Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L

Molar mass of CaCO₃ = 100 g mol⁻¹

Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L

7 0

3 years ago