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Murljashka [212]
4 years ago
12

Identify the name of the compound Cr2(CO3)3.

Chemistry
2 answers:
Sonja [21]4 years ago
8 0

The name of the compound {\text{C}}{{\text{r}}_{\text{2}}}{\left( {{\text{C}}{{\text{O}}_{\text{3}}}} \right)_3} is \boxed{{\text{chromium }}\left( {{\text{III}}} \right){\text{ carbonate}}}.

Further explanation:

An ionic compound is formed when there occurs complete transfer of electrons between the bonded atoms. Atoms that lose electrons form cations whereas that gain electrons forms anions. Cations are positively charged ions and anions are negatively charged ions.ns.

An ionic compound is named by naming its cation first and then by its anion.

Naming of cations:

1. The main group cations are named the same as the name of their corresponding elements. For example, {\text{N}}{{\text{a}}^ + } is named sodium cation.

2. The transition metals cations are named similar to the main group elements. The only difference is that the charge of transition metal cations is written in parenthesis after the name of the cation. For example, {\text{C}}{{\text{u}}^{2 + }} is named copper (II) cation.

3. The polyatomic cation is named by adding suffix “ium”.

Naming of anions:

1. The suffix “ide” is added when a monoatomic anion is formed by an element. For example, {{\text{H}}^ - } is named as a hydride.

2. When two oxyanions are formed by a single element, the oxyanion with fewer oxygen atoms is written with suffix “ite” and that with more oxygen is named with suffix “ate”. For example, {\text{NO}}_2^ - is named nitrite while {\text{NO}}_3^ - is named nitrate.

3. If a series of oxyanions are formed by an element, prefix “hypo” is used when fewer oxygen atoms are present and prefix “per” is used if more oxygen atoms are present. For example, {\text{Cl}}{{\text{O}}^ - } is named hypochlorite while {\text{ClO}}_4^ - is named as perchlorate.

The given compound is {\text{C}}{{\text{r}}_{\text{2}}}{\left( {{\text{C}}{{\text{O}}_{\text{3}}}} \right)_{\text{3}}}. Since Cr is a transition metal so its oxidation state has to be written parantheses.

The expression to calculate the oxidation state in {\text{C}}{{\text{r}}_{\text{2}}}{\left( {{\text{C}}{{\text{O}}_{\text{3}}}} \right)_{\text{3}}} is:

\left[ {2\left( {{\text{oxidation state of Cr}}} \right) + 3\left( {{\text{oxidation state of C}}} \right) + 9\left( {{\text{oxidation state of O}}} \right)} \right] = 0     … (1)

Rearrange equation (1) for the oxidation state of Cr.

{\text{Oxidation state of Cr}} = \dfrac{{\left[ { - 3\left( {{\text{oxidation state of C}}} \right) - 9\left( {{\text{oxidation state of O}}} \right)} \right]}}{2}        …… (2)  

Substitute +4 for the oxidation {\text{state}} of C and -2 for the oxidation state of O in equation (2).

\begin{aligned}{\text{Oxidation state of Cr}} &= \frac{{\left[ { - 3\left( { + {\text{4}}} \right) - 9\left({ - {\text{2}}} \right)} \right]}}{2}\\&= \frac{{ - 12 + 18}}{2}\\&=+ 3\\\end{aligned}  

Therefore the oxidation state of Cr in {\text{C}}{{\text{r}}_{\text{2}}}{\left( {{\text{C}}{{\text{O}}_{\text{3}}}} \right)_{\text{3}}} is +3.

The formula of any compound is formed by crisscross rule. So the charge on the anion will be -2 and the corresponding anion is {\text{CO}}_3^{2 - } whose name is carbonate. Therefore the name of the given compound is chromium (III) carbonate.

Learn more:

  1. Determine the correct name and formula of acid: brainly.com/question/7072854
  2. Identify the precipitate in the reaction: brainly.com/question/8896163

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Ionic compounds

Keywords: ions, cations, anions, electron, oxidation state, Cr, Cr2(CO3)3, +3, +4, -2, O, C, naming of cations, naming of anions.

kompoz [17]4 years ago
7 0
Cr2(CO3)3 is Chromium (III) Carbonate or Chromic Carbonate. 

It is composed of Chromium, Carbon, and Oxygen. Chromium has 2 atoms, Carbon has 3 atoms, and Oxygen has 9 atoms. 

This compound is composed of 36.6% Chromium, 12.7% Carbon, and 50.7% Oxygen.
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0.2640 g of sodium oxalate is dissolved in a flask and requires 30.74 mL of potassium
Free_Kalibri [48]

Moles of potassium permanganate = 0.0008

<h3>Further explanation  </h3>

Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution). Determination of the endpoint/equivalence point of the reaction can use indicators according to the appropriate pH range  

Reaction

5Na2C2O4(aq) + 2KMnO4(aq) + 8H2SO4(aq) ---> 2MnSO4(aq) + K2SO4(aq) + 5Na2SO4(aq) +  10CO2(g) + 8H2O(1)

The end point ⇒titrant and analyte moles equal

titrant : potassium  permanganate-KMnO4

analyte : sodium oxalate - Na2C2O4

so moles of KMnO4 = moles of Na2C2O4

moles of Na2C2O4(mass = 0.2640 g, MW=134 g/mol) :

\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{0.264}{134 g/mol}\\\\mol=0.002

From equation, mol ratio  Na2C2O4 : KMnO4 = 5 : 2, so mol KMnO4 :

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6 0
3 years ago
The disaccharide lactose can be decomposed into its constituent sugars galactose and glucose. This decomposition can be accompli
n200080 [17]

Answer:

Rate  = 116m⁻¹s⁻¹[lactose][H]⁺

Explanation:

the formula for rate of reaction is given as

Rate = k[lactose]∧α[H]⁺∧β

we solve for the value of α and β

([lactose]₁/[lactose]₂)∧α

α = \frac{ln\frac{0.00116}{0.00232} }{ln\frac{0.01}{0.02} }

when we divide this equation

α = \frac{ln0.5}{ln0.5}

α = 1

we find β

R₁/R₂ = 0.01/0.02(0.001/0.001)∧β

0.00116/0.00232 = 0.5(1)∧β

β = 1

Rate = k[lactose]∧α[H]⁺∧β

we have to find the value for k

k = 0.00116/0.01(0.001)

k = 0.00116/0.00001

= 116m⁻¹s⁻¹

<u>Rate = 116m⁻¹s⁻¹[lactose][H]⁺</u>

3 0
3 years ago
What is the density of a liquid if 12.5 ml of the liquid has a density of 9.80 g?
blagie [28]

Answer:

The answer is

<h2>0.784 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

<h3>density =  \frac{mass}{volume}</h3>

From the question

mass of liquid = 9.8 g

volume = 12.5 mL

The density is

density =  \frac{9.8}{12.5}

We have the final answer as

<h3>0.784 g/mL</h3>

Hope this helps you

3 0
3 years ago
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