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faltersainse [42]
4 years ago
7

The first ionization energies (kl/mol) of hydrogen (H), nitrogen (N), fluroine (F), and oxygen (O) are 1,312, 1,402, 1,742, and

1,314, respectively. From which electrically neutral atom does it take the least energy to remove the first electron?
• Hydrogen
• Fluorine
• Nitrogen
• Oxygen
Chemistry
1 answer:
Oksanka [162]4 years ago
5 0
Fluorine because of the spin pair repulsion fron the p orbitals
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Please answer asap!
kaheart [24]
200 ml is 1/5 of a liter, so the answer is five times the number of moles present in the solution. 0.6 moles/0.2 liter = x moles/1.0 liter. Solving for x gives 0.2 x = 0.6 or x = 3.0 M

so the answer is c
3 0
3 years ago
Breaking bonds requires what type of energy flow?
Rufina [12.5K]

Explanation:

the energy that that is needed to break a bond is called the bond energy or dissciation energy

3 0
3 years ago
If you purchased 2.31 μCi of sulfur-35, how many disintegrations per second does the sample undergo when it is brand new?
nikklg [1K]

Answer:

8.547 x 10⁴disintegrations per second

Explanation:

To calculate the disintegrations per second as -

Given ,

2.31 μCi of sulfur  -35 .

Since ,

1 Ci = 3.7 * 10 ¹⁰ Bq

1 μCi = 10 ⁻⁶ Ci

Hence ,

conversation is done as follows -

2.31 ( 1 * 10⁻⁶) * ( 3.7 * 10¹⁰)

= 8.547 x 10⁴

Hence ,

8.547 x 10⁴disintegrations per second , the sample undergo for it to be brand new .

3 0
3 years ago
A student needs to prepare 100. mL of 0.612 M Cu(NO3)2 solution. What mass, in grams, of copper(II) nitrate should the student u
Temka [501]

Answer: 11.5 grams

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution

Molarity=\frac{n\times 1000}{V_s}

where,

Morality = 0.612 M

n= moles of solute  

V_s = volume of solution in ml = 100 ml

Now put all the given values in the formula of molarity, we get

0.612=\frac{n\times 1000}{100ml}

n=0.0612moles

Mass={\text {moles of solute }}{\times {\text {molar mass}}=0.0612moles\times 187.56g/mol=11.5g

Therefore, the mass of copper (II)nitrate required is 11.5 grams

3 0
3 years ago
The pH of a 0.65M solution of hydrofluoric acid HF is measured to be 1.68. Calculate the acid dissociation constant Ka of hydrof
sergey [27]

Answer:

Kₐ = 6.7 x 10⁻⁴

Explanation:

First lets write the equilibrium expression, Ka , for the dissociation of hydrofluoric acid:

HF + H₂O     ⇄   H₃O⁺ +   F⁻

Kₐ = [ H₃O⁺ ] [ F⁻ ] /[ [ HF ]

Since we are given the pH we can calculate the  [ H₃O⁺ ]  ( pH = - log [ H₃O⁺ ] , and because the acid dissociates into a 1: 1  relation , we will also have [F⁻ ]. The  [ HF ] is given in the question so we have all the information that is needed to  compute Kₐ.

pH = -log [ H₃O⁺ ]

1.68 = - log [ H₃O⁺ ]

Taking antilog to both sides of this equation:

10^-1.68 = [ H₃O⁺ ] ⇒ 2.1 X 10⁻²  M= [ H₃O⁺ ]

[ F⁻ ] = 2.1 X 10⁻² M

Solving for Kₐ :

Kₐ = ( 2.1 X 10⁻² ) x  ( 2.1 X 10⁻² ) / 0.65 = 6.7 x 10⁻⁴  

(Rounded to two significant figures, the powers of 10 have infinite precision )

4 0
3 years ago
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