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Virty [35]
2 years ago
14

Notice that "SO4" appears in two different places in this chemical equation. SO42− is a polyatomic ion called "sulfate." What nu

mber should be placed in front of CaSO4 to give the same total number of sulfate ions on each side of the equation? ?CaSO4+AlCl3→CaCl2+Al2(SO4)3
Chemistry
1 answer:
masya89 [10]2 years ago
5 0

Answer : The number placed in front of CuSO_4 should be, three (3).

Explanation :

Balanced chemical reaction : It is defined as the number of atoms of individual elements present on reactant side must be equal to the product side.

The given unbalanced chemical reaction is,

CaSO_4+AlCl_3\rightarrow CaCl_2+Al_2(SO_4)_3

This chemical reaction is an unbalanced reaction because in this reaction, the number of atoms of chloride and sulfate ion are not balanced.

In order to balanced the chemical reaction, the coefficient 3 is put before the CuSO_4, the coefficient 2 is put before the AlCl_3 and the coefficient 3 is put before the CaCl_2.

Thus, the balanced chemical reaction will be,

3CaSO_4+2AlCl_3\rightarrow 3CaCl_2+Al_2(SO_4)_3

Therefore, the number placed in front of CuSO_4 should be, three (3).

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2 years ago
Which of the following compounds has the highest pKa value? A)CCl3CH2NH2
Rudik [331]
The lower the pKa<span> of a Bronsted acid, the more easily it gives up its proton. The </span>higher<span> the </span>pKa<span> of a Bronsted acid, the more tightly the proton is held, and the less easily the proton is given up.

Here we need the highest pKa, so we need to see which compound  will less likely to give proton or hydrogen ion.

</span><span>Now, all Nitrogen contains a lone pair. But HALOGEN groups( F, Cl, only) being electronegative than NITROGEN [electronegativity of N=3, F=4 and Cl=3], pulls electron pair towards itself. 
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The more the lone pair of nitrogen is pulled, the more strong bond between N and H will become, which means less likely to give hydrogen ion.
means high Pka

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3 0
2 years ago
In which orbital does an electron in a bromine atom experience the greatest effective nuclear charge?
asambeis [7]

First let us determine the electronic configuration of Bromine (Br). This is written as:

Br = [Ar] 3d10 4s2 4p5

 

Then we must recall that the greatest effective nuclear charge (also referred to as shielding) greatly increases as distance of the orbital to the nucleus also increases. So therefore the electron in the farthest shell will experience the greatest nuclear charge hence the answer is:

<span>4p orbital</span>

4 0
2 years ago
Read 2 more answers
Electroplating is a way to coat a complex metal object with a very thin (and hence inexpensive) layer of a precious metal, such
8_murik_8 [283]

Answer:

0.0164 g

Explanation:

Let's consider the reduction of silver (I) to silver that occurs in the cathode during the electroplating.

Ag⁺(aq) + 1 e⁻ → Ag(s)

We can establish the following relations.

  • 1 A = 1 C/s
  • The charge of 1 mole of electrons is 96,468 C (Faraday's constant)
  • 1 mole of Ag(s) is deposited when 1 mole of electrons circulate.
  • The molar mass of silver is 107.87 g/mol

The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:

19.0s \times \frac{0.770c}{s} \times \frac{1mole^{-} }{96,468C} \times \frac{1molAg}{1mole^{-}} \times \frac{107.87g}{1molAg} = 0.0164 g

5 0
2 years ago
The solubility of KCl is 3.7 M at 20 °C. Two beakers each contain 100. mL of saturated KCl solution: 100. mL of 4.0 M HCl is add
JulijaS [17]

Answer:

a)The Ksp was found to be equal to 13.69

Explanation:

Terminology

Qsp of a dissolving ionic solid — is the solubility product of the concentration of ions in solution.

Ksp however, is the solubility product of the concentration of ions in solution at EQUILIBRIUM with the dissolving ionic solid.

Note that if Qsp > Ksp , the solid at a certain temperature, will precipitate and form solid. That means the equilibrium will shift to the left in order to attain or reach equilibrium (Ksp).

Step-by-step solution:

To solve this: 

#./ Substitute the molar solubility of KCl as given into the ion-product equation to find the Ksp of KCl.

#./ Find the total concentration of ionic chloride in each beaker after the addition of HCl. We pay attention to the amount moles present at the beginning and the moles added.

#./ Find the Qsp value to to know if Ksp is exceeded. If Qsp < Ksp, nothing will precipitate.

a) The equation of solubility equilibrium for KCL is thus;

KCL_(s) ---> K+(aq) + Cl- (aq)

The solubility of KCl given is 3.7 M.

Ksp= [K+][Cl-] = (3.7)(3.7) =13.69

The Ksp was found to be equal to 14.

In pure water KCl

Ksp =13.69 KCl =[K+][Cl-]

Let x= molar solubility [K+],/[Cl-] :. × , x

Ksp =13.69 = [K+][Cl-] = (x)(x) = x²

x= √ 13.69 = 3.7 M moles of KCl requires to make 100mL saturated solutio

37M moles/L

The Ksp was found to be equal to 14.

4.0 M HCl = KCl =[K+][Cl-]

Let y= molar solubility :. y, y+4

Ksp =13.69= [K+][Cl-] = (y)(y*+4)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (y)(y*+4)= y(4)

13.69=4y:. y= 3.42 moles/100mL

y= 34.2moles/L

8 M HCl = KCl =[K+][Cl-]

Let b= molar solubility :. B, b+8

Ksp =13.69= [K+][Cl-] = (b)(b*+8)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (b)(b*+8)= b(8)

13.69=8b:. b= 1.71 moles/100mL

17.1 moles/L

Therefore in a solution with a common ion, the solubility of the compound reduces dramatically.

8 0
2 years ago
Read 2 more answers
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