One of the results is that the moon is near the earth and the other one, the oceans tide. Even though the earth can hold any object within
ts proximity, the ocean is partly attracted due to its liquid property. At night, the ocean tends to be attracted to the moon by creating a bulge and assigning it as ‘high tide’. This is due to the strong gravitational pull of th moon to the earth.
I hope this helps!
This might be right..
The experimental density of CO2 at STP is 0.10/0.056=1.78 g/L. The percent error equals to (1.96-1.78)/1.96*100%=9.18%. So the answer is 9.18%.
Answer:
Density of concentrated H2SO4 = 1.99g/cm^3 = 1991.79Kg/m^3
Explanation:
mass of empty flask = 78.23g mass of flask filled when with water = 593.63g.
mass of flask filled when with concentrateds sulfuric acid, H2SO4 = 1026.57g
Mass of water = (mass of flask filled when with water) -
(mass of empty flask) = 593.63g - 78.23g = 515.4g
Volume of flask = volume of water = volume of concentrateds sulfuric acid, H2SO4 =
(mass of water)/ density of water) = 515.4g/1.00g/cm^3 = 515.4cm^3
The density of concentrated sulfuric acid is given by
Density of concentrated H2SO4 = (mass of H2SO4) ÷ (volume of H2SO4) = 1026.57g/515.4cm^3 = 1.99g/cm^3 = 1991.79Kg/m^3
Answer is: D. Cl (chlorine).
The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).
Barium, potassium and arsenic are metals (easily lost valence electrons), chlorine is nonmetal (easily gain electrons).
Alkaline metals (in this example, potassium) have lowest ionizations energy and easy remove valence electrons (one electron), earth alkaline metals (in this example, barium) have higher ionization energy than alkaline metals, because they have two valence electrons.
Nonmetals (in this example chlorine) are far right in the main group and they have highest ionization energy, because they have many valence electrons.
0.600
D) 0.600 is the final concentration of the solution of KCl.
V1 = 50.0 mL.
