Answer:
c. The atoms of one element can be identical to the atoms of another element.
Explanation:
<em>Which of the following is not a statement of Dalton's atomic theory of matter?</em>
<em>a. Elements are made of atoms.</em> TRUE. An atom is the smallest particle of a chemical element that can exist.
<em>b. Atoms of a given element are identical.</em> TRUE. The only slight difference is in the mass of isotopes.
<em>c. The atoms of one element can be identical to the atoms of another element.</em> FALSE. The atoms of different elements are different from one to another.
<em>d. A given compound always has the same number and kinds of atoms. </em>TRUE. This is known as Dalton's law of constant composition.
Barium fluoride (BaF2) - Also known as Barium(II) fluoride - because it's a combination of two different kinds of ions (binary = two).
Ionic bond involves electrostatic attraction between oppositely charged ions.
The ions are atoms that have gained 1 or more electrons and atoms that have lost 1 or more electrons.
Answer: The type of bond that requires the give and take of electrons is
A ) ionic bond.
Answer:
a) distance is 4+7+1+8=20 blocks
b) displacement is 10 blocks
Explanation:
find displacement: x and y
x axis displacement = 4-1 = 3 blocks
y axis displacement = -7+8= 1 block
displacement = the square root of 3^2 + 1^2
= 9+1 = 10 blocks.
You can find the angle of displacement with respect to the initial position using trig identities, if you wish.
Answer:
(a). 4°C, (b). 2.4M, (c). 11.1 g, (d). 89.01 g, (e). 139.2 g and (f). 58 g/mol.
Explanation:
Without mincing words let's dive straight into the solution to the question.
(a). The freezing point depression can be Determine by subtracting the value of the initial temperature from the final temperature. Therefore;
The freezing point depression = [ 1 - (-3)]° C = 4°C.
(b). The molality can be Determine by using the formula below;
Molality = the number of moles found in the solute/ solvent's weight(kg).
Molality = ( 11.1 / 58) × (1000)/ ( 90.4 - 11.1) = 2.4 M.
(c). The mass of acetone that was in the decanted solution = 11.1 g.
(d). The mass of water that was in the decanted solution = 89.01 g.
(e). 2.4 = x/ 58 × (1000/1000).
x = 2.4 × 58 = 139.2 g.
(f). The molar mass of acetone = (12) + (1 × 3) + 12 + 16 + 12 + (1 x 3) = 58 g/mol.