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kow [346]
2 years ago
12

Wat two common uses for Zinc???

Chemistry
1 answer:
trapecia [35]2 years ago
5 0

Answer:

Zinc oxide is  used in the manufacture of  many products such as paints, rubber, cosmetics, pharmaceuticals, plastics, inks, soaps, batteries, textiles and electrical equipment.

Explanation:

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What is the hybridization and shape for an XeF6 2+ molecule?
Anna11 [10]

Answer:

Xe:[Kr]4d¹⁰5(sp³d³)₆⁺² => Octahedral Geometry (AX₆)⁺²

Explanation:

Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆²

Ca. #Valence e⁻ = Xe + 6F - 2e⁻ = 1(8) + 6(7) - 2 = 48

Ca. #Substrate e⁻ = 6F = 6(8) = 48

#Nonbonded free pairs e⁻ = (V - S)/2 = (48 - 48)/2 = 0 free pairs

#Bonded pairs e⁻ = 6F substrates = 6 bonded pairs

BPr + NBPr = 6 + 0 = 6 e⁻ pairs => Geometry => [AX₆]⁺² => Octahedron

Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆⁺²    

XeF₆⁺² => 6(sp³d³) hybrid orbitals => Octahedral Geometry (AX₆)      

8 0
2 years ago
36.4 g of SnCl2 is dissolved in 135.7 mL of water.
Yuki888 [10]

Answers:

1st: 189.6 g/mol

2nd: 0.1357 L

3rd: 1.41 M

Explanation:

Finding Molar Mass:

SnCl2 = <u>Tin(II) Chloride</u>

Tin has a molar mass of <u>118.71 g/mol</u>

Chloride has a molar mass of <u>35.453 g/mol</u>

Chloride*2 = <u>70.906</u>

<u>118.71 + 70.906 ≈ 189.6 g/mol</u>

Finding Liters of Solution:

L = mL/1000

135.7 mL / 1000 = <u>0.1357</u>

Finding Molarity:

molarity = <u>moles of solute / liters of solution</u>

M = (36.4g /  189.6g) / 0.1357 L = <u>1.41 M</u>

Hope this helped ;)

6 0
1 year ago
A gas sample enclosed in a rigid metal container at room temperature (20.0∘C∘C) has an absolute pressure p1p1p_1. The container
choli [55]

Answer: p2 = 1.06p1

Explanation: pressure increases with temperature increase.

According to Gass law

P1/T1 = P2/T2

T1 = 20°c = 20 +273 = 293k

T2 = 40°c = 40 +373 = 313k

Therefore

P2 = P1T2/T1 = 313P2/293

P2 = 1.06P1

3 0
3 years ago
Nitrogen has two isotopes. One has an atomic mass of 14.003 amu and a relative abundance of 99.63% while the other isotope has a
Slav-nsk [51]

Answer:

Average atomic mass = 14.0067  amu.

Explanation:

Given data:

Abundance of 1st isotope  = 99.63%

Atomic mass of 1st isotope = 14.003 amu

Abundance of 2nd isotope  = 0.37%

Atomic mass of 2nd isotope = 15.000 amu

Solution:  

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass = (99.63×14.003)+(0.37×15.000) /100

Average atomic mass =  1395.119 + 5.55 / 100

Average atomic mass  = 1400.67 / 100

Average atomic mass = 14.0067  amu.

3 0
3 years ago
How many liters of peroxide should be added to 12 liters of an 8% peroxide solution so that the remaining solution contains 16%
Nastasia [14]
The question above may be solved using the overall mass and component balances. If we let x be the  number of liters of peroxide that should be added, the final mixture will have a quantity of x+12 L. The peroxide balance before and after mixing is,
                            x(1) + 12(0.08) = (x + 12)(0.16)
The value of x is 1.143. Thus, approximately 1.143 L of peroxide should be added to the original solution. 
4 0
3 years ago
Read 2 more answers
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