Answer:
1.5057×10^22 atom
Explanation:
As we
1 mole of argon = 40 g of argon
i.e 40 g of argon = 1 mole of argon
1 g of argon = 1/40 mole of argon
1 mole of argon = 6.023×10^23 atom of argon
1/40 mole if argon = 1/40 ×6.023×10^23
= 1.5057×10^22
Answer:
substance
Explanation:
A pure substance is a form of matter that has a constant composition (meaning it's the same everywhere) and properties that are constant throughout the sample (meaning there is only one set of properties such as melting point, color, boiling point, etc
Answer:
A) CH3CH2SH
Explanation:
Dispersion forces are weak attractions found between non-polar and polar molecules. The attractions here can be attributed to the fact that a non-polar molecule sometimes become polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant. If this happens, the molecule has a temporary dipole. This dipole can induce the neighbouring molecules to be distorted and form dipoles as well. The attractions between these dipoles constitute the Dispersion Forces.
Therefore; the greater the molar mass of a compound or molecule, the higher the Dispersion Force. This implies that the compound or molecule with the highest molar mass have the largest dispersion forces.
Now; for option (A)
CH3CH2SH
The molar mass is :
= (12 + (1 × 3 ) +12 + (1 ×2) + 32+1)
= (12 + 3+ 12 + 2 + 32 + 1)
= 62 g/mol
For option (B)
CH3NH2
The molar mass is:
= (12 + (1 × 3 ) +14 + (1 × 2)
= (12 + 3 + 14 + 2)
= 31 g/mol
For option (C)
CH4
The molar mass is :
= 12 + (1 × 4)
= 12 + 4
= 16 g/mol
For option (D)
CH3CH3
The molar mass is :
= 12 + ( 1 × 3 ) + 12 + ( 1 × 3)
= 12 + 3 + 12 + 3
= 30 g/mol
Thus ; option (A) has the highest molar mass, as such the largest dispersion force is A) CH3CH2SH
Answer: product
Explanation:
Each substance written to the right of the arrow in a chemical equation is referred to as a product.
When writing a chemical equation, the substance that's written to the left of arrow in the equation is the reactants.
On the other hand which is the right side is the product.
Answer:
The section of the bar is 2.92 inches.
Explanation:
Mass of the steel cut ,m = 1.00 kg = 1000 g
Volume of the steel bar = V = Area × height
Height of the of the section of bar = h
Area of Equilateral triangular = 
a = 2.50 inches
Cross sectional area of the steel mass = A
Density of the steel = d =
h = 2.92 inches
The section of the bar is 2.92 inches.
