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3 years ago

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Chromium (Cr) can combine with chlorine (Cl2) to form chromium chloride (CrCl3). Which equation is a correct representation of t

his reaction?

2 answers:

scoray [572]3 years ago

7 0

The balanced equation is 2Cr + 3Cl2 = 2CrCl3

Alisiya [41]3 years ago

3 0

<u>Answer:</u> The chemical equation is written below.

<u>Explanation:</u>

When chromium metal reacts with chlorine gas, it leads to the production of chromium chloride.

The chemical equation for the formation of chromium chloride follows:

$2Cr(s)+3Cl_2(g)\rightarrow 2CrCl_3(s)$

This is a type of combination reaction because two small substances react to form one large substance.

Hence, the chemical equation is written above.

You might be interested in

Determine what is missing from this neutralization reaction: AgNO3+KCl→AgCl+−−−−

lapo4ka [179]

Answer:

The answer to your question is: KNO₃

Explanation:

AgNO3 + KCl → AgCl + −−−−

A. KNO3 this option is correct because it is a double replacement reaction then potassium must attached to NO₃.

B. KOH this product is not possible because there is no water to form OH⁻ ions.

C. Ag2K this product is not possible because both Ag and K are metals, then it is difficult that they attach.

D. KN2O This product is imposible to form, this option is wrong.

7 0

2 years ago

Determine the number of atoms of O in 92.3 moles of Cr3(PO4)2

Irina18 [472]

739 is the number of atoms of O in 92.3 moles of Cr3(PO4)2.

Explanation:

Molecular formula given is = Cr3(PO4)2

number of moles of the compound is 92.3 moles

number of 0 atoms in 92.3 moles =?

From the chemical formula 1 mole of the compound has 8 atoms of oxygen

So, it can be written as

1 mole Cr3(PO4)2 has 8 atoms of oxygen

92.3 moles of Cr3(PO4)2 has x atoms of oxygen

$\frac{8}{1}$ = $\frac{x}{92.3}$

x = 8 x 92.3

x = 738.4 atoms

There will be 739 oxygen atoms in the 92.3 moles of Cr3(PO4)2.

4 0

3 years ago

A ________ reaction occurs when one compound reacts and is broken down into different elements or simpler compounds?

faust18 [17]

Answer: decomposition

Explanation:

1 . Combustion is a type of chemical reaction in which a hydrocarbon reacts with oxygen to form carbon dioxide ans water along with liberation of large amount of energy.

2. Decomposition is a type of chemical reaction in which a single reactant gives two or more than two products.

$CaCO_3\rightarrow CaO+CO_2$

3. Single replacement is a type of chemical reaction in which a more reactive element displaces the less reactive element from its slat solution.

4. Synthesis is a type of chemical reaction in which two or more than two reactants combine together to give a single product.

5 0

3 years ago

Read 2 more answers

Can some atoms exceed the limits of the octet rule in bonding? If so, give an example.

harkovskaia [24]

Answer:

Yes. Example: <u>Sulfur hexafluoride (SF₆) molecule</u>

Explanation:

According to the octet rule, elements tend to form chemical bonds in order to have <u>8 electrons in their valence shell</u> and gain the stable s²p⁶ electronic configuration.

However, this rule is generally followed by main group elements only.

Exception: <u>SF₆ molecule</u>

In this molecule, six fluorine atoms are attached to the central sulfur atom by single covalent bonds.

<u>Each fluorine atom has 8 electrons in their valence shells</u>. Thus, it <u>follows the octet rule.</u>

Whereas, there are <u>12 electrons around the central sulfur atom</u> in the SF₆ molecule. Therefore, <u>sulfur does not follow the octet rule.</u>

<u>Therefore, the SF₆ molecule is known as a </u><u>hypervalent molecule</u><u> or expanded-valence molecule.</u>

6 0

3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3.

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago