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bearhunter [10]
4 years ago
6

CHEM! Why do we use roman numerals (aka stock system) when you write the name for a compound with (example:) tin in it even thou

gh it’s a main group metal?
Chemistry
1 answer:
mylen [45]4 years ago
7 0

Roman numerals are used in naming ionic compounds when the metal cation forms more than one ion. The metals that form more than one ion are the transition metals, although not all of them do this.

SnBr2 - Tin(II) Bromide


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If 22.99 g of sodium and 35.45 g of chlorine fully react, how much sodium chloride forms?​
matrenka [14]

Answer:

The answer to your question is: HCl = 58.44 g

Explanation:

To solve this problem, we must remember the law of conservation of mass that says that, matter cannot be created or destroyed, only changes its form.

That means that, the mass of the reactants must be equal to the mass of the products.

               Process

                    Sodium (Na)  +    Chlorine (Cl)   ⇒     Sodium chloride (HCl)

                         22.99 g              35.45 g                            x

              22.99 + 35.45 = HCl

              HCl = 58.44 g

3 0
3 years ago
Calculate the enthalpy of the reaction
harkovskaia [24]

Answer : The enthalpy of the reaction is, -2552 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given enthalpy of reaction is,

4B(s)+3O_2(g)\rightarrow 2B_2O_3(s)    \Delta H=?

The intermediate balanced chemical reactions are:

(1) B_2O_3(s)+3H_2O(g)\rightarrow 3O_2(g)+B_2H_6(g)     \Delta H_A=+2035kJ

(2) 2B(s)+3H_2(g)\rightarrow B_2H_6(g)    \Delta H_B=+36kJ

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_C=-285kJ

(4) H_2O(l)\rightarrow H_2O(g)    \Delta H_D=+44kJ

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :

(when we are reversing the reaction then the sign of the enthalpy change will be change.)

The expression for enthalpy of the reaction will be,

\Delta H=-2\times \Delta H_A+2\times \Delta H_B-6\times \Delta H_C-6\times \Delta H_D

\Delta H=-2(+2035kJ)+2(+36kJ)-6(-285kJ)-6(+44)

\Delta H=-2552kJ

Therefore, the enthalpy of the reaction is, -2552 kJ/mole

4 0
3 years ago
Which compound is classified as a hydrocarbon?(1) ethane (3) chloroethane(2) ethanol (4) ethanoic acid
jok3333 [9.3K]
The answer is (1). The hydrocarbon is a compound consists only with carbon and hydrogen. Ethane is C2H6. Chloroethane is C2H5Cl. Ethanol is CH3OH. Ethanoic acid is CH3COOH.
4 0
3 years ago
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3 years ago
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Name the complex ion [cr(cn)6]3−. the oxidation number of chromium is +3. enter the name.
frez [133]
The given complex Ion is,

                                          [Co(CN)₃]³⁻

In complex compounds the Positive part is named first, followed by name of Negative part. In given case there is no Positive part. So, we will name the Negative part.

In Sphere the Ligands are named first. In this case there are 6 CN⁻ ligands. As CN⁻ (cyanide) contains a negative charge so the -ide is replaced by -o. i.e. Cyanide → Cyano. Also, Hexa is placed before Cyano as there are 6 CN⁻ Ligands.
So,
                                           Hexacyano

After that Chromium metal is named. As there is a negative charge on the sphere, so the -ium in Chromium is replaced by -ate.
i.e. Chromium → Chromate. And the oxidation state of Cr is written in Roman. Hence,the name is as,
          
                                Hexacyanochromate (III) Ion
5 0
3 years ago
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