Answer:
See explanation
Explanation:
The reason why the droplets are spherical is the surface area to volume ratio of the falling droplet in a gravitational field. Recall that a sphere has a small surface area to volume ratio.
Between X and Y, one key difference that will define the rate at which the two drops of liquid falls is the viscosity of the fluid. Since the images were not attached, I can not really tell what liquid droplet is more flatter than the other.
However, the liquid with a greater surface tension will form larger droplets and experience a greater air resistance as the droplet falls. Hence the less the surface tension, the flatter the droplets. Cohesive forces pull molecules of a liquid droplets inwards leading to a more spherical shape and reducing the surface area. Surface tension is therefore the reason why liquids form droplets.
Answer:
1.99grams
Explanation:
- First, we need to calculate the molar mass of the compound: Ca(HCO3)2
Ca = 40g/mol, H = 1g/mol, C = 12g/mol, O = 16g/mol
Hence, Ca(HCO3)2
= 40 + {1 + 12 + 16(3)}2
= 40 + {13 + 48}2
= 40 + {61}2
= 40 + 122
= 162g/mol
Molar mass of Ca(HCO3)2 = 162g/mol
- Next, we calculate the mass of oxygen in one mole of the compound, Ca(HCO3)2.
Oxygen = {16(3)}2
= 48 × 2
= 96g of Oxygen
- Next, we calculate the percentage composition of oxygen by mass by dividing the mass of oxygen in the compound by the molar mass of the compound i.e.
% composition of O = 96/162 × 100
= 0.5926 × 100
= 59.26%.
- The number of moles of the compound, Ca(HCO3)2, must be converted to mass by using the formula; mole = mass/molar mass
0.0207 = mass/162
Mass = 162 × 0.0207
Mass = 3.353grams
However, in every gram of Ca(HCO3)2, there is 0.5926 g of oxygen
Hence, in 3.353grams of Ca(HCO3)2, there will be;
0.5926 × 3.353
= 1.986
= 1.99grams.
Therefore, there is 1.99grams of Oxygen in 0.0207 moles (3.353g) of Ca(HCO3)2.
Answer:
I don't know because I don't know about chemistry
Answer:
The final temperature of the setup = 36.6°C
Explanation:
Let the final temperature of the setup be T
Heat lost by the copper tubing = Heat gained by water and the vessel
Heat lost by the copper tubing = mC ΔT = 455 × 0.387 × (89.5 - T) = (15759.61 - 176.1T) J
Heat gained by water = mC ΔT = 159 × 4.186 × (T - 22.8) = (665.6T - 15175.1) J
Heat gained by vessel = c ΔT = 10 × (T - 22.8) = (10T - 228) J
Heat lost by the copper tubing = Heat gained by water and the vessel
(15759.61 - 176.1T) = (665.6T - 15175.1) + (10T - 228)
851.7 T = 31162.71
T = 36.6°C
