I believe it’s A, they form whole-number ratios in the compound
Answer:
No
Explanation:
Petroleum cant be mixed with water coz petroleum is less dense than water so petroleum will float on water.
The naturally occurring isotopes of Li are Li-6 of mass 6.015121 amu and Li-7 of mass 7.016003 amu. The atomic mass of Li is 6.9409 amu, the percent abundance can be calculated using the following relation.
Atomic mass=m(Li-6 )×%(Li-6 )+m(Li-7 )×%(Li-7 )
Let the percent abundance of Li-6 be X thus, that of Li-7 will be 1-X, putting the values,
Or,
Or,
X=0.075
Thus, 
Thus, percent abundance of Li-6 is 0.075 or 7.5 % and that of Li-7 is 0.925 or 92.5%.
Answer:
Mass of Rb-87 is 86.913 amu.
Explanation:
Given data:
Average mass of rubidium = 85.4678 amu
Mass of Rb-85 = 84.9117
Ratio of 85Rb/87Rb in natural rubidium = 2.591
Mass of Rb = ?
Solution:
The ration of both isotope is 2.591 to 1. Which means that for 2.591 atoms of Rb-85 there is one Rb-87.
For 100% naturally occurring Rb = 2.591 + 1 = 3.591
% abundance of Rb-85 = 2.591/ 3.591 = 0.722
% abundance of Rb-87 = 1 - 0.722= 0.278
84.9117 × 0.722 + X × 0.278 = 85.4678
61.306 + X × 0.278 = 85.4678
X × 0.278 = 85.4678 - 61.306
X × 0.278 = 24.1618
X = 24.1618 / 0.278
X = 86.913 amu
Answer:
105 grams PbI₂
Explanation:
Pb(NO₃)₂ + 2KI => 2KNO₃ + PbI₂(s)
moles Pb(NO₃)₂ = 0.265L(1.2M) = 0.318 mole
moles KI = 0.293(1.55M) = 0.454 mole => Limiting Reactant
moles PbI₂ from mole KI in excess Pb(NO₃)₂ = 1/2(0.454 mole) = 0.227 mol PbI₂
grams PbI₂ = 0.227 mol PbI₂ x 461 g/mole = 104.68 g ≈ 105 g PbI₂(s)
