CHEMISTRY HIGH SCHOOL

Hi! could you help with this (Brainlist + 50 points)

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Answer 1

Answer:

1. 44 grams

2. KCIO3

3. the third one

hope this helps :)

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Pls help ASAP I am literally so tired I hate this so much

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a still lake . Is your answer

Does anyone know how to do this? the solubility of barium carbonate, BaCO3, is 0.0100 g/L. Its molar mass is 197.3 g/mol. What is the Ksp of barium carbonate?

Answers

The value of ${{\text{K}}_{{\text{sp}}}}$ for barium carbonate is $\boxed{{\text{2}}{{.6 * 1}}{{\text{0}}^{{\text{ - 9}}}}}$ .

Further explanation:

Solubility product constant:

The equilibrium constant between the compound and its ion, when dissolved in solution, is known as solubility product constant. It is denoted by ${{\text{K}}_{{\text{sp}}}}$ . The solubility product constant is used to calculate the product of the concentration of ions at equilibrium.

Higher the solubility product constant more will be the solubility of the compound.

The general reaction is as follows:

${\text{AB}}\left({aq}\right)\to{{\text{A}}^+}\left({aq}\right)+{{\text{B}}^-}\left({aq}\right)$

The expression to calculate the solubility product for the general reaction is as follows:

$\boxed{{{\text{K}}_{{\text{sp}}}}=\left[{{{\text{A}}^+}}\right]\left[{{{\text{B}}^-}}\right]}$

Here,

${{\text{K}}_{{\text{sp}}}}$ is the solubility product constant.

$\left[ {{{\text{A}}^ + }} \right]$ is the concentration of ${{\text{A}}^ + }$ ions.

$\left[ {{{\text{B}}^ - }} \right]$ is the concentration of ${{\text{B}}^ - }$ ions.

The dissociation of ${\text{BaC}}{{\text{O}}_{\text{3}}}$ occurs as follows:

${\text{BaC}}{{\text{O}}_3}\to{\text{B}}{{\text{a}}^(2+)}+{\text{CO}}_3^(2-)$

The given solubility of ${\text{BaC}}{{\text{O}}_{\text{3}}}$ is 0.0100 g/L. Firstly, it is to be converted in mol/L. So the solubility of ${\text{BaC}}{{\text{O}}_{\text{3}}}$ is calculated as follows:

${\text{Solubility of BaC}}{{\text{O}}_{\text{3}}}\left({{\text{mol/L}}}\right)= \frac{{{\text{Solubility of BaC}}{{\text{O}}_{\text{3}}}\left({{\text{g/L}}}\right)}}{{{\text{Molar mass of BaC}}{{\text{O}}_{\text{3}}}\left({{\text{g/mol}}}\right)}}$ …… (1)

The solubility of ${\text{BaC}}{{\text{O}}_{\text{3}}}$ is 0.0100 g/L.

The molar mass of ${\text{BaC}}{{\text{O}}_{\text{3}}}$ is 197.3 g/mol.

Substitute these values in equation (1).

$\begin{aligned}{\text{Solubility of BaC}}{{\text{O}}_{\text{3}}}&=\left( {\frac{{{\text{0}}{\text{.0100 g}}}}{{{\text{1 L}}}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{197}}{\text{.3 g}}}}} \right)\n&=0.00005068\;{\text{mol/L}}\n\end{aligned}$

It is evident from the chemical equation, one mole of ${\text{BaC}}{{\text{O}}_{\text{3}}}$ dissociates to form one mole of ${\text{B}}{{\text{a}}^(2+)}$ and one mole of ${\text{CO}}_3^(2-)$ . So the solubility of both ${\text{B}}{{\text{a}}^(2+)}$ and ${\text{CO}}_3^(2-)$ is 0.00005068 mol/L.

The formula to calculate the solubility product of is as follows:

${{\text{K}}_{{\text{sp}}}}=\left[{{\text{B}}{{\text{a}}^{{\text{2+}}}}}\right]\left[{{\text{CO}}{{_3^2}^-}}\right]$ …… (2)

Substitute 0.00005068 mol/L for $\left[{{\text{B}}{{\text{a}}^{{\text{2+}}}}}\right]$ and 0.00005068 mol/L for $\left[{{\text{CO}}_3^(2-)}\right]$ in equation (2).

$\begin{aligned}{{\text{K}}_{{\text{sp}}}}&=\left({{\text{0}}{\text{.00005068}}} \right)\left({{\text{0}}{\text{.00005068}}}\right)\n&=2.6*{10^(-9)}\n\end{aligned}$

Therefore, the value of ${{\mathbf{K}}_{{\mathbf{sp}}$ for ${\mathbf{BaC}}{{\mathbf{O}}_{\mathbf{3}$ is ${\mathbf{2}}{\mathbf{.6 * 1}}{{\mathbf{0}}^(-9)}$ .

Learn more:

1. Sort the solubility of gas will increase or decrease: brainly.com/question/2802008.

2. What is the pressure of the gas?: brainly.com/question/6340739.

Answer details:

Grade: School School

Subject: Chemistry

Chapter: Chemical equilibrium

Keywords: solubility, Ba2+, CO32-, BaCO3, Ksp, solubility product, molar mass, 197.3 g/mol, mol/L, g/L, 0.0100 g/L.

It's simple, just follow my steps.

1º - in 1 L we have $0.0100~g$ of $BaCO_3$

2º - let's find the number of moles.

$\eta=(m)/(MM)$

$\eta=(0.0100)/(197.3)$

$\boxed{\boxed{\eta=5.07*10^(-5)~mol}}$

3º - The concentration will be

$C=5.07*10^(-5)~mol/L$

But we have this reaction

$BaCO_3\rightleftharpoons Ba^(2+)+CO_3^(2-)$

This concentration will be the concentration of $Ba^(2+)~~and~~CO_3^(2-)$

$K_(sp)=([Ba^(2+)][CO_3^(2-)])/([BaCO_3])$

considering $[BaCO_3]=1~mol/L$

$K_(sp)=[Ba^(2+)][CO_3^(2-)]$

and

$[Ba^(2+)]=[CO_3^(2-)]=5.07*10^(-5)~mol/L$

We can replace it

$K_(sp)=(5.07*10^(-5))*(5.07*10^(-5))$

$K_(sp)\approx25.70*10^(-10)$

Therefore the $K_(sp)$ is:

$\boxed{\boxed{\boxed{K_(sp)\approx2.57*10^(-11)}}}$

Which electron configuration represents the electrons in an atom of Ga in an excited state?

Answers

The ground state is the lowest energy state of an atom while the excited state is any state that is higher than the ground state. Since the ground state configuration of Ga is 2-8-18-3 then the electron configuration of Ga in an excited state is represented by 2-8-17-4. In order to achieve the excited state an electron has to be promoted.

Final answer:

An excited state electron configuration of Gallium (Ga) might be [Ar] 4s2 3d9 4p2, as this configuration represents an electron having absorbed energy and moved from the 3d to the 4p level.

Explanation:

The electron configuration of an atom in its ground state is the arrangement of electrons in the lowest possible energy levels (the 'normal' arrangement). For the element Gallium (Ga), the ground state electron configuration is [Ar] 4s2 3d10 4p1. However, when an atom is in an excited state, one or more electrons have absorbed energy and moved to a higher energy level. An electron configuration for an excited state of Gallium might look like [Ar] 4s2 3d9 4p2 as an electron from the 3d level has absorbed energy and moved to the higher 4p level.

Learn more about Excited State Electron Configuration here:

brainly.com/question/32047683

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What is the charge on a fluorine ion? explain answer plz

Answers

-1

An ion is an atom which has gained or lost electrons.
Atoms want to have full outer shells of electrons.
Electrons are negative.
Fluorine has 7 electrons in its outer shell. When it gains one to fill its shell it will be made negative.
Thus the charge of a fluorine ion is -1

How do you find a Oxidation number

Answers

An oxidation number is the charge of the ion that would form if a covalent molecule was ionic. Oxidation numbers exist for covalent compounds, as ions are not formed.

The oxidation number is determined by the respective electronegativity of the atoms involved. A more electronegative element will always have a negative oxidation number. In the compound, all the oxidation numbers must add up to 0. Oxidation numbers of elements can vary between compunds.

A few rules to follow about oxidation numbers
Fluorine will always be -1
Oxygen will be -2 unless in a compound with fluorine
Hydrogen is usually +1

A few examples
HF (Hydrogen flouride) Oxy number of H= +1 and F = -1

H2O Oxy number of H = +1 and O= -2

NH3 Oxy of H= +1 and since there are 3 Hydrogens each with an oxidation number of +1 N must have an oxy number of -3 for the overall oxidation number of the compound to be 0.

H2SO4 Oxy number of H= +1 O=-2 When adding up the hydrogen and oxygen oxidation numbers ( 1 + 1 -2 - 2 -2 -2) we are left with -6 thus S must have an oxidation number of +6 for the overall number to be 0.

NO2 Oxidation number of O= -2 there are 2 oxygens leaving us with -4 thus N must have an Oxy number of +4 for the overall number to be 0.

What is the net ionic equation for the reaction of solid iron with aqueous copper sulfate? (Hint: First use the activity series to write the formula unit equation!)

Answers

For the answer to the question above,
The reaction of solid iron with aqueous copper sulfate is an example of a single displacement type of reaction which exchanges only the cations of the element and the compound. In this case, the equation is

Fe (s) + CuSO4 (aq) = FeSO4 + Cu (s)

Answer:

The net ionic equation for the reaction of solid iron with aqueous copper sulfate:

$Fe(s)+Cu^(2+)(aq)\rightarrow Fe^(2+)+(aq)+Cu(s)$

Explanation:

Solid iron = Fe

Copper sulfate = $CuSO_4$

$Fe(s)+CuSO_4(aq)\rightarrow FeSO_4(aq)+Cu(s)$

In an aqueous solution of copper sulfate we have copper (II) ions and sulfate ions.

$CuSO_4(aq)\rightarrow Cu^(2+)(aq)+SO_4^(2-)(aq)$

In an aqueous solution of ferrous sulfate we have ion (II) ions and sulfate ions.

$FeSO_4(aq)\rightarrow Fe^(2+)+(aq)+SO_4^(2-)(aq)$

$Fe(s)+Cu^(2+)(aq)+SO_4^(2-)(aq)\rightarrow Fe^(2+)+(aq)+SO_4^(2-)(aq)+Fe^(2+)+(aq)$

Cancelling out the common ions from both sides, we get the net ionic equation:

$Fe(s)+Cu^(2+)(aq)\rightarrow Fe^(2+)+(aq)+Cu(s)$