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sesenic [268]
3 years ago
15

How does an atom of one element differ from the atom of another element?

Chemistry
1 answer:
Temka [501]3 years ago
8 0

Explanation:

The atoms of one element differs from the atoms of other elements in terms of the number of protons they contain. This is often taken as the atomic number of such an atom.

  • The number of proton is the best indicator of the atom one is dealing with.
  • Based on this number, elements are categorized into distinct columns and rows on the periodic table.
  • The atomic number is the number of protons or positively charge particles in the atom.

II.

It is possible to change the identity of an atom. This is only possible by altering the atomic number of the atom.

Only nuclear reactions have this capability.

When an atom undergoes nuclear reaction that involves change in number of protons, transmutation occurs and a new atom forms.

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Las nubes el agua y el Hielo<br> son lo mismo?
xenn [34]

Answer:

Si,las mubes el agua y el hielo son lo mismo

7 0
3 years ago
Need help real quick with these questions
Yuliya22 [10]

Answer:

what are the questions???

8 0
3 years ago
A. Match each type of titration to its pH at the equivalence point.
maks197457 [2]

Answer:answers are in the explanation

Explanation:

(a). pH less than 7 between 1 - 3.5 are strong acid, and between 4.5-6.9 weak acid.

pH greater than 7; between 10-14 is a strong base, and between 7.1 - 9, it is weakly basic.

(b). Equation of reaction;

HBr + KOH ---------> KBr + H2O

One mole of HBr reacts with one mole of KOH to give one Mole of KBr and one mole of H2O

Calculating the mmol, we have;

mmol KOH = 28.0 ml × 0.50 M

mmol KOH= 14 mmol

mmol of HBr= 56 ml × 0.25M

mmol of HBr= 14 mmol

Both HBr and KOH are used up in the reaction, which leaves only the product,KBr and H2O.

The pH here is greater than 7

(C). [NH4^+] = 0.20 mol L^-1 × 50 ml. L^-1 ÷ 50 mL + 50mL

= 0.10 M

Ka=Kw/kb

10^-14/ 1.8× 10^-5

Ka= 5.56 ×10^-10

Therefore, ka= x^2 / 0.20

5.56e-10 = x^2/0.20

x= (0.20 × 5.56e-10)^2

x= 1.05 × 10^-5

pH = -log [H+]

pH= - log[1.05 × 10^-5]

pH = 4.98

Acidic(less than 7)

(c). 0.5 × 20/40

= 0.25 M

Ka= Kw/kb

kb= 10^-14/1.8× 10^-5

Kb = 5.56×10^-10

x= (5.56×10^-10 × 0.5)^2

x= 1.667×10^-5 M

pH will be basic

3 0
3 years ago
The following questions are about ethyl bromide,n-propyl bromide, isopropyl bromide,t-butyl bromide, &amp; neopentyl bromide
mrs_skeptik [129]

Answer:

a) Watch the attaccment

b) Ethyl bromide is more reactive than n-propyl bromid, and this more than neopentyl bromide. Ethyl bromide has less steric hindrance than the others, to SN2 reactions.

c) t-butyl bromide is more reactive than  isopropyl bromide, and this more than ethyl bromide. t-butyl bromide structure stabilize the carbocation, better than the others.

Explanation:

Speed of SN2 reactions depends on steric hindrance, the less hindrance, the most reaction speed, meaning more reactivity. Then, those linear structures are more reactive to SN2 reactions.

In the other hand, speed of SN1 reactions depends on the stability of the carbocation formed. Structure with ramifications can stabilize better the carbocation, these structures are more reactive to SN1 reactions.

Download pdf
7 0
3 years ago
When 1.14 g of octane (molar mass = 114 g/mol) reacts with excess oxygen in a constant volume calorimeter, the temperature of th
maxonik [38]
I can't answer this question without knowing what the specific heat capacity of the calorimeter is. Luckily, I found a similar problem from another website which is shown in the attached picture. 

Q = nCpΔT
Q = (1.14 g)(1 mol/114 g)(6.97 kJ/kmol·°C)(10°C)(1000 mol/1 kmol)
<em>Q = +6970 kJ</em>

8 0
3 years ago
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