Answer:
1281.25 Moles of glucose
Explanation:
Multiply 1.25x1025= 1281.25
Container must be made up of non metallic elements which can typically covalently bond and must have lone pairs of electrons dative bonds in order to allow further stability.
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:
130 Liters
Explanation:
if 1 mol is 22.4 L, then 5.8 mol is 130 L (129.92 but use sig figs)
Answer:
There was 450.068g of water in the pot.
Explanation:
Latent heat of vaporisation = 2260 kJ/kg = 2260 J/g = L
Specific Heat of Steam = 2.010 kJ/kg C = 2.010 J/g = s
Let m = x g be the weight of water in the pot.
Energy required to vaporise water = mL = 2260x
Energy required to raise the temperature of water from 100 C to 135 C = msΔT = 70.35x
Total energy required = 
Hence, there was 450.068g of water in the pot.
