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Makovka662 [10]
4 years ago
7

In the formula for the compound XCl4, the X could represent (1) C (3) Mg (2) Η (4) Zn

Chemistry
2 answers:
Anvisha [2.4K]4 years ago
7 0
Correct Answer: option 1 i.e. C

Reason: 
The the compound of interest i.e.  XCl4, since there are 4 Cl atoms bonded to X. This signifies that the valency of X is 4.

There atomic number of C is 6. It's electronic configuration is giving by 1s2 2s2 2p2. Thus, there are 4 electrons in valence shell of C. This signifies that valency of C is 4. Hence the compound present in present case is CCl4.
Genrish500 [490]4 years ago
3 0
(1) C.

The name of the compound is Carbon Tetra Chloride. None of the other elements clearly forms a compound with Cl in that form.
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Answer:

Temperature and volume are directly related.

The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure. :)
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2 years ago
You need to prepare .200M solution of hydrochloric acid (HCl). If you took out .830ml of the 12.0M stock solution. How much wate
Alenkinab [10]

Answer:

You must add 48.97 mL of water to make the 0.200 M diluted solution.

Explanation:

In chemistry, dilution is the reduction in concentration of a chemical in a solution. In other words, it is the process of reducing the concentration of solute in solution, simply adding more solvent to the solution.

In a dilution, the quantity or mass of the solute is not changed but only that of the solvent. As only solvent is being added, by not increasing the amount of solute the concentration of the solute decreases.

The expression for the dilution calculations is:

Cinitial* Vinitial = Cfinal* Vfinal

In this case:

  • Cinitial= 12 M
  • Vinitial= 0.830 mL
  • Cfinal= 0.200 M
  • Vfinal= ?

Replacing:

12 M*0.830 mL= 0.200 M*Vfinal

Solving:

Vfinal=\frac{12 M*0.830 mL}{0.200 M}

Vfinal= 49.8 mL

Since 0.830 mL is the volume you initially have of HCl, the amount of water you must add is:

49.8 mL - 0.830 mL= 48.97 mL

<u><em>You must add 48.97 mL of water to make the 0.200 M diluted solution.</em></u>

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A site in Pennsylvania receives a total annual deposition of 2.688 g/mof sulfate from fertilizer and acid rain. The ratio by mas
sertanlavr [38]

According to the statement

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid

<h3>What is neutralization?</h3>

A chemical reaction in which an acid and a base react quantitatively with each other is known as neutralization or neutralization. In a water reaction, neutralization ensures that there is no excess of hydrogen or hydroxide ions in the solution.

<h3>According to the given information:</h3>

The equation of the neutralization reaction between H2SO4 and CaCO3.

CaCO3 + H2SO4 → CaSO4 + H2CO3

H2CO3 dissociate to water and carbon dioxide.

        CaCO3 + H2SO4 → CaSO4  + H2O + CO2

Now solving for the mass of CaCO3 needed to neutralize the acid.

mass of CaCO3 = 9460 Kg H2SO4  × \frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} \times \frac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO} 4}{98.1 \mathrm{gH}_2 \mathrm{SO}_4} \times \frac{1 \mathrm{~mol} \mathrm{CaCO}\left(\mathrm{O}_3\right.}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO}_4}\times \frac{100.1 \mathrm{~g} \mathrm{CaCO}_3}{1 \mathrm{~mol} \mathrm{CaCO}_3} \times \frac{2.205 \mathrm{lb}}{1000 \mathrm{~g}}

= 21284.56606

mass of CaCO3 =  2.12 x 10^4 lbs

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid.

To know more about neutralization visit:

brainly.com/question/12498769

#SPJ4

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