please help i dont get chemistry as much as I've only just started and I dont know how to separate all these matters

What is the hybridization of the central atom in the trichlorostannanide anion?

Mademuasel [1]

Answer:

sp²

Explanation:

The trhichlorostannanide anion has molecular formula SnCl₃⁻, Sn is the central atom, which is at the group 14 of the periodic table, so it has 4 valence electrons. It must form 4 bonds to be stable.

In the anion, the tin (Sn) forms 3 bonds, so, it has 1 electron that is isolated. According to the VSEPR theory, the geometry of a molecule depends on the bonded electrons and the lone pairs, because the repulsive force between them will form the angles of the molecule.

Thus, only the bonded electrons will contribute, and because of that, the geometry will be trigonal planar.

To form the bonds, the subshells (s, p, d, f) of the atoms in the bond must interact. But, only isolated electrons can bond, so, to be stable and form the compounds, some atoms have hybrids shells, which are formed by the join of subshells.

The hybridization can be found by the determination of how many bonds are done, or by the geometry. For a trigonal planar geometry, the hybridization is sp², which means that the orbital is formed by one orbital s and two orbitals p.

8 0

4 years ago

Muriatic acid, HCl, is often used to remove rust. A solution of muriatic acid, HCl, reacts with Fe2O3 deposits on industrial equ

Alina [70]

Answer:

3L

Explanation:

Step 1:

The balanced equation for the reaction.

Fe2O3(s) + 6HCl(aq) → 2FeCl3(aq) + 3H2O

Step 2 :

Determination of the masses of HCl and Fe2O3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Fe2O3 = 159.70g/mol

Molar mass of HCl = 36.46 g/mol

Mass of HCl from the balanced equation = 6 x 36.46 = 218.76g

From the balanced equation above,

159.70g of Fe2O3 reacted with 218.76g of HCl

Step 3:

Determination of the mass of HCl needed to react with 439g of Fe2O3. This is illustrated below:

From the balanced equation above,

159.70g of Fe2O3 reacted with 218.76g of HCl.

Therefore, 439g of Fe2O3 will react with = (439 x 218.76) /159.70 = 601.35g of HCl.

Step 4:

Conversion of 601.35g of HCl to mole. This is illustrated below:

Molar mass of HCl = 36.46 g/mol

Mass of HCl = 601.35g

Number of mole = Mass/Molar Mass

Number of mole of HCl = 601.35/36.46

Number of mole of HCl = 16.49 moles

Step 5:

Determination of the volume of the HCl that reacted.

This is illustrated below:

Mole of HCl = 16.49 moles

Molarity of HCl = 5.50 M

Volume =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 16.49/5.5

Volume of HCl = 3L

Therefore the volume of HCl needed for the reaction is 3L

5 0

3 years ago

22. A piece of wood board contains approximately 6.76 X 1023 molecules of wood, C2H4920, how

Licemer1 [7]

Answer:

<h2>1.23 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{6.76 \times {10}^{23} }{6.02 \times {10}^{23} } = \frac{6.76}{6.02} \\ = 1.22923...$

We have the final answer as

<h3>1.23 moles</h3>

Hope this helps you

5 0

3 years ago

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans

Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of $H_2SO_4$ (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

$H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)$

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of $H_2SO_4(aq)$ ? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of $H_2SO_4$ (aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_1,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL$

Putting values in above equation, we get:

$2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M$

0.044 M is the molarity of $H_2SO_4$ (aq).

4 0

3 years ago

16. A small piece of silver was placed in a solution containing both magnesium nitrate and copper(II) nitrate.

sashaice [31]

I believe it's about the dislodging response as Ag structures solvent salt with nitrate. For dislodging response, more receptive metal will uproot less receptive metal from the arrangement. In your question, the reactivity of the metal are positioned as takes after: Mg > Cu > Ag. Take note of that more responsive metals, that are Mg and Cu, are in the arrangement.

3 0

3 years ago

please help i dont get chemistry as much as I've only just started and I dont know how to separate all these matters ​

please help i dont get chemistry as much as I've only just started and I dont know how to separate all these matters