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Vadim26 [7]
3 years ago
7

In 1808, John Dalton proposed a series of statements regarding the atom that became known as Dalton's postulates or Dalton's ato

mic theory. Since Dalton's original proposal, some of the statements have been updated or changed due to new discoveries. Which of the statements was not a part of Dalton's original atomic theory? 1) Different elements have atoms of different masses and properties. 2) All four statements are part of Dalton's original atomic theory. 3) Matter is comprised of small particles called atoms. 4) Atoms cannot be created or destroyed in a chemical reaction. 5) Atoms of the same element have the same size, mass, and structure.
Chemistry
1 answer:
Elden [556K]3 years ago
8 0

Answer:

2) All four statements are part of Dalton's original atomic theory.

Explanation:

The postulates of the Dalton’s atomic theory are:

  • The matter is comprised of small units which are known as atoms.
  • The atoms are indivisible which means that they can not be further divided.
  • Atoms of the matter can neither be created nor be destroyed in the chemical reaction.  
  • All atoms of the particular element are identical.
  • The atoms of the different elements vary in the size and the mass. and posses different properties.
  • The compounds are formed by the different whole number combinations of the atoms.

<u>Hence, all 4 are correct.</u>

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Classify each substance as a strong acid, strong base, weak acid, or weak base. drag each item into the appropriate bin.
Zepler [3.9K]
Since you forgot to include the choices for classification, I would just define each of these and tell you the hints that would help you classify them.

Among these acids and bases, its is the strong acids and strong bases that are easily classified. You should note that there are only 7 strong acids existing. All the rest are weak acids. These 7 acids are: HCl, HBr, HI, HClO₃, HClO₄, HNO₃ and H₂SO₄. On the other hand, there are only 8 strong bases; the rest are weak bases. These are the hydroxides of the Group ! and !! metals: LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)₂, Sr(OH)₂, and Br(OH)₂.

For the weak acids and weak bases, just remember the definitions of Arrhenius, Lewis and Bronsted-Lowry. A weak base are those compounds that accept H⁺ protons, produce OH⁻ ions when solvated and an electron donor. A weak acid are those compounds that donate H⁺ protons, produce H⁺ ions when solvated and an electron acceptor.
6 0
3 years ago
A drop of water with a mass of 0.48 g is vaporized at 100 ∘C and condenses on the surface of a 55- g block of aluminum that is i
bagirrra123 [75]

The final temperature in Celsius of the metal block is 49°C.

<h3>How to find the number of moles ?</h3>

Moles water = \frac{\text{Given mass}}{\text{Molar Mass}}

                     = \frac{0.48\ g}{18\ \text{g/mol}}

                     = 0.0266 moles  

                   

Heat lost by water = 0.0266 mol x 44.0 kJ/mol

                                = 1.17 kJ

                                = 1170 J           [1 kJ = 1000 J]

Heat lost = Heat gained

Heat gained by aluminum = 1170 J  

1170 = 55 x 0.903 (T - 25) = 49.7 T - 1242  

1170 + 1242 = 49.7 T  

T = 48.5°C (49°C at two significant figures)

Thus from the above conclusion we can say that The final temperature in Celsius of the metal block is 49°C.

Learn more about the Moles here: brainly.com/question/15356425

#SPJ1

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Answer:c the correct technology cannot support this mission

Explanation:

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A 50.0g g sample of 16n decays to 12.5g in 14.4 seconds. What is its half life
mixas84 [53]
T is amount after time t 
<span>Ao is initial amount </span>
<span>t is time </span>
<span>HL is half life </span>

<span>log (At) = log [ Ao x (1/2)^(t/HL) ] </span>
<span>log (At) = log Ao + log (1/2)^(t/HL) </span>
<span>log (At) = log Ao + (t/HL) x log (1/2) </span>

<span>( log At - log Ao) / log (1/2) = t / HL </span>
<span>log (At/Ao) / log (1/2) = t / HL </span>

<span>HL = t / [( log (At / Ao)) / log (1/2) ] </span>

<span>HL = 14.4 s / [ ( log (12.5 / 50) / log (1/2) ] </span>

<span>HL = 14.4 s / 2 = 7.2 seconds </span>
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