Answer:
John Dalton:
John Dalton was the scientist who introduced atomic theory in the field of chemistry. Dalton worked on different gases and formulated this theory. The main points of Dalton's theory are:
- Every element present is made up of atoms.
- Atoms of an elements are have the same same properties whereas these properties are different for each element.
- According to his theory, an atom could not be broken down.
- Different atoms combine or get separated from each other during a chemical reaction.
Ernest Rutherford:
Ernest Rutherford is known as the father of nuclear physics due to his impressing research work on radioactivity of atoms. Rutherford was the first scientist to discover the nucleus of an atom and prove that the nucleus was charged. He also described that the electrons circle around the nucleus of an atom.
Answer:
The atomic mass of gallium (Ga) = <u>69.723 g/mol</u>
Explanation:
Given: Two isotopes of Gallium (Ga) are Gallium-69 (⁶⁹Ga) and Gallium-71 (⁷¹Ga)
<u>For ⁶⁹Ga: </u>
Relative abundance = 60.12% = 60.12 ÷ 100 = 0.6012; Atomic mass = 68.9257 g/mol
<u>For ⁷¹Ga:</u>
Relative abundance = 39.88% = 39.88 ÷ 100 = 0.3988; Atomic mass = 70.9249 g/mol
∴ The atomic mass of Ga = (Relative abundance of ⁶⁹Ga × Atomic mass of ⁶⁹Ga) + (Relative abundance of ⁷¹Ga × Atomic mass of ⁷¹Ga)
⇒ Atomic mass of Ga = (0.6012 × 68.9257 g/mol) + (0.3988 × 70.9249 g/mol) = <u>69.723 g/mol</u>
<u>Therefore, the atomic mass of gallium (Ga) = 69.723 g/mol</u>
Answer:
0.00316
Explanation:
You have to use the following equation:
![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
You are given the pH and need to find the concentration of H+. Plug in the given components and solve.
![2.5=-log[H^+]\\H^+ = 10^{-2.5}\\H^+=0.00316](https://tex.z-dn.net/?f=2.5%3D-log%5BH%5E%2B%5D%5C%5CH%5E%2B%20%3D%2010%5E%7B-2.5%7D%5C%5CH%5E%2B%3D0.00316)
The concentration of H is 0.00316.
Answer:
0.1988 J/g°C
Explanation:
-Qmetal = Qwater
Q = mc∆T
Where;
Q = amount of heat
m = mass of substance
c = specific heat of substance
∆T = change in temperature
Hence;
-{mc∆T} of metal = {mc∆T} of water
From the information provided in this question, For water; m= 22.0g, ∆T = (24°C-19°C), c = 4.18J/g°C.
For metal; m= 34.0g, ∆T = (24°C-92°C), c = ?
Note that, the final temperature of water and the metal = 24°C
-{34 × c × (24°C-92°C)} = 22 × 4.18 × (24°C-19°C)
-{34 × c × (-68°C)} = 459.8
-{34 × c × -68} = 459.8
-{-2312c} = 459.8
+2312c = 459.8
c = 459.8/2312
c = 0.1988
The specific heat capacity of the metal is 0.1988 J/g°C
