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Dovator [93]
3 years ago
14

Which atom would it be most difficult to remove an electron from

Chemistry
1 answer:
Norma-Jean [14]3 years ago
6 0
I would be difficult to remove an electron from a Noble or Inert Gas (also known as the group 8 or 0 elements).  This is because they all have filled outermost shells and as such the outermost shell would be held tightly to the nucleus and as such make it difficult to remove.  Examples Helium, Neon, Argon, Xenon, Krypton and Radon 
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Explain how neutralization reactions are used in acid-base titrations.
nadya68 [22]
Acid-base tiltrations has DNA that can help in the process of neutralization. not so sure about it
8 0
3 years ago
A chemistry student needs of isopropenylbenzene for an experiment. He has available of a w/w solution of isopropenylbenzene in a
Lera25 [3.4K]

Question:

A chemistry student needs of 10 g isopropenylbenzene for an experiment. He has available 120 g of a 42.7% w/w solution of isopropenylbenzene in acetone. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button.

Answer:

The answer to the question is as follows

The mass of solution the student should use is 23.42 g.

Explanation:

To solve the question we note the following

A solution containing 42.7 % w/w of isopropenylbenzene in acetone  has 42.7 g of isopropenylbenzene in 100 grams of the solution

Therefore we have 10 g of isopropenylbenzene contained in

100 g * 10 g/ 42.7 g = 23.42 g of solution

Available solution = 120 g

Therefore the quantity to used from the available solution = 23.42 g of the isopropenylbenzene in acetone solution.

8 0
3 years ago
0.53g of acetanilide was subjected to kjeldahl determination and the ammonia produced was collected in 50cm3 of 0.50M of h2so4.o
Lady bird [3.3K]

Answer:

10.57% of N in acetanilide

Explanation:

All nitrogen in the sample is converted in NH₃ in the Kjeldahl determination. The NH₃ reacts with H₂SO₄ as follows:

2NH₃ + H₂SO₄ → 2NH₄⁺ + SO₄²⁻

The acid in excess in titrated with Na₂CO₃ as follows:

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

To solve this question we must find the moles of sodium carbonate = Moles of H₂SO₄ in excess. The added moles - Moles in excess = Moles of sulfuric acid that reacts:

<em>Moles Na₂CO₃ anf Moles H₂SO₄ in excess:</em>

0.025L * (0.05mol / L) = 1.25x10⁻³ moles Na₂CO₃ / 0.01360L =

0.09191M * 0.250L = 0.0230 moles H₂SO₄ in excess.

<em>Moles H₂SO₄ added:</em>

0.050L * (0.50mol / L) = 0.0250 moles H₂SO₄ added

<em>Moles that react:</em>

0.0250 moles - 0.0230 moles = 0.0020 moles H₂SO₄

<em>Moles of NH₃ = Moles N:</em>

0.0020 moles H₂SO₄ * (2mol NH₃ / 1mol H₂SO₄) = 0.0040 moles NH₃ = Moles N

<em>mass N and mass percent:</em>

0.0040 moles N * (14g / mol) = 0.056gN / 0.53g * 100 =

<h3>10.57% of N in acetanilide</h3>
7 0
2 years ago
Which of the following mixture types canNOT be filtered to remove solutes?
iVinArrow [24]
1. A) Colloids only

2. C) M<span>olecules of the dispersion medium colliding with dispered phase particles 

Hope this helps!</span>
3 0
2 years ago
Balance the following reaction. A coefficient of "1" is understood. Choose option "blank" for the correct answer if the
Vera_Pavlovna [14]

2NH₂ + O₂ → N₂ + 2H₂O

<u>Explanation:</u>

Balancing the equation means, the number of atoms on both sides of the equation must be the same.

In the case of the given equation, we have to find out whether it is balanced or not.

2NH₂ + O₂ →  N₂ + 2H₂O

Atoms       Number of atoms before balancing     after balancing

                 LHS                 RHS                                    LHS         RHS

N               1                        2                                            2              2

H               2                       2                                            4              4                        

O               2                       1                                            2                2

To balance the N atoms, we have to put 2 in front of NH₂, and then to balance the H, O atoms, we have to put 2 in front of H₂O, so that each atom in left hand as well as right hand side of the equation was balanced.

8 0
3 years ago
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