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natima [27]
2 years ago
12

Which of the following is non polar covalent bond?

Chemistry
1 answer:
vladimir2022 [97]2 years ago
5 0

Answer:

N-Cl

Explanation:

Look at the chart below. Since N-Cl bond has a electronegativity difference of (3.0-3.0) zero, they are non-polar.

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MakcuM [25]
A Wooden Spoon is your answer because metal attracts heat more, so it would get hotter.

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6 0
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konstantin123 [22]

Answer:false

Explanation:

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3 years ago
Name: _________________ Temperature o fwater_25_degreecent.YOU MUST SHOW ALL CALCULATIONS TO RECEIVE CREDIT FOR THEM! DATA ANALY
Anton [14]

Answer:

1. 0.02 M

2. 0.01 M

3. 4×10⁻⁶

Explanation:

We know that V₁S₁ = V₂S₂

1.

Concentration of HCl = 0.05 M

end point comes at = 10 ml

So, concentration of OH⁻(aq) = [OH⁻(aq)] ⇒ (0.05 × 10) ÷ 25 ⇒ 0.02 M

2.

2mol of OH⁻(aq) ≡ 1 mole of Ca²⁺(aq)

[Ca²⁺] = 0.02 ÷ 2 = 0.01 M

3.

K_{sp} = [Ca²⁺(aq)] [OH⁻(aq)]²

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K_{sp} = [0.01 × (0.02)²] = 4×10⁻⁶

4.

If reaction is exothermic which means heat energy will get evolved as a result temperature of the reaction media will get increased during the course of the reaction. If temperature is externally increased, the reaction will go backward to accumulate extra heat energy.

5.

K_{sp} value describes the solubility of a particular ionic compound. The higher the K_{sp} value, the higher the Solubility will be.

6.

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4 0
3 years ago
Define position and reference point in your own words. Dont copy and paste!
irga5000 [103]

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4 0
4 years ago
A compound is 80.0% carbon and 20.0% hydrogen by mass. assume you have a 100.-g sample of this compound. the molar mass of the c
ch4aika [34]
Basis of the calculation: 100g
 
For Carbon: 
 Mass of carbon = (100 g)(0.80) = 80 g
  Number of moles of carbon = (80 g)(1 mole / 12g) = 20/3

For Hydrogen:
  Mass of hydrogen = (100 g)(0.20) = 20 g
     Number of moles of hydrogen = (20 g)(1 mole / 1 g) = 20

Translating the answer to the formula of the substance,
     C20/3H20

Dividing the answer,
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The molar mass of the empirical formula is:
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ANSWER: C2H6

 
4 0
3 years ago
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