Answer: He did not discuss about any of these.
Explanation: Dalton proposed some of the postulates for his atomic theory. They are:
1) Matter is made up of atoms which are not divisible.
2) Atoms of different elements combine in a fixed ratio to form compounds.
3) The atomic properties of given element are same including mass. This states that all the atoms of an element have same mass but the atoms of different elements have different masses.
4) No atoms are either created or destroyed during a chemical reaction.
5) Atoms of an element are identical in mass, size and other chemical and physical properties.
As it is visible from the postulates, he only discussed only about the atoms but not subatomic particles or isotopes.
Answer:
V¹N²= V²N²
here V¹= ?
N¹= 6.00
V²= 175ml
M²= 0.2M
So V¹= (V²N²)/N² = (175 x 0.2)/6
V¹ = 5.83 ml
Explanation:
Therefore diluting 5.83 ml of 6.00M NaOH to 175 m l ,we get 0.2M Solution.
Given parameters:
Mass of sucrose = 5g
Density of sucrose = 1.12g/mL
Percentage of sucrose per liter of cane juice = 12%
Unknown:
Volume of cane juice needed = ?
We need to establish the volume - density relationship. Density is the mass of a substance per unit volume.
Mathematically;
Density = 
Now solve for the volume of sucrose;
1.12g/mL = 
Volume = 
= 4.46mL = 4.46 x 10⁻³L since 1000mL = 1L
Since 12% of 1 liter of cane juice is sucrose;
12% of x liter of cane juice = 4.46 x 10⁻³L
Volume of cane juice = 4.46 x 10⁻³ x 
= 0.037L
Volume of cane juice is 0.037L
Answer:
son 12.6 gramos de HF
Explanation:
Tienes que saber qual es el reactor limitante en este caso es fluoruro con los 20 gramos puedes producer .631 mol qual son 12.6 gramos
Answer:
volume in L = 0.25 L
Explanation:
Given data:
Mass of Cu(NO₃)₂ = 2.43 g
Volume of KI = ?
Solution:
Balanced chemical equation:
2Cu(NO₃)₂ + 4KI → 2CuI + I₂ + 4KNO₃
Moles of Cu(NO₃)₂:
Number of moles = mass/ molar mass
Number of moles = 2.43 g/ 187.56 g/mol
Number of moles = 0.013 mol
Now we will compare the moles of Cu(NO₃)₂ with KI.
Cu(NO₃)₂ : KI
2 : 4
0.013 : 4 × 0.013=0.052 mol
Volume of KI:
<em>Molarity = moles of solute / volume in L</em>
volume in L = moles of solute /Molarity
volume in L = 0.052 mol / 0.209 mol/L
volume in L = 0.25 L
