1*10^9 molecules* (1 mol/ (6.02*10^23 molecules))= 2*10^(-15) mol.
The final answer is 2*10^(-15) mol O2~
Answer:
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Answer: 24.6kpa
Explanation:
This is a classic case where Boyle's Law can be applied.
The equation for Boyle's Law is given as: P1V1 = P2V2
Where P1 = initial pressure
P2 = final pressure
V1 = initial volume
V2 = final volume
From the question, P1 = ?, P2 = 98.2kpa , V1 =4L , V2 =1L
P1V1 = P2V2
P1 x 4L = 98.2 kpa x 1L
Make P1 subject of formula we then have:
P1 = 98.2kpa X 1L / 4L
= 24.6kpa
Explanation:
COMPONENTS OF ENVIRONMENT
Environment mainly consists of atmosphere, hydrosphere, lithosphere and biosphere. But it can be roughly divided into two types such as (a) Micro environment and (b) Macro environment. (a) Micro environment refers to the immediate local surrounding of the organism.
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
