A) 5.2 x 10^2
B) 86.
C) 6.4 x 10^3
D) 5.0
E) 22.
F) 0.89
Answer:
it is an exothermic reaction
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
To answer this question, we will use pressure law which states that:
"At constant volume of a fixed mass of gas, the pressure of the gas is directly proportional to the temperature"
This means that:
Pi / Ti = Pf / Tf where:
Pi is the initial pressure = 1 atm
Ti is the initial temperature = 100 + 273 = 373 degree kelvin
Pf is the final pressure that we want to calculate
Tf is the final temperature = 139 + 273 = 412 degree kelvin
Substitute in the equation with the givens to calculate the pressure as follows:
1 / 373 = Pf / 412
Pf = (1/373) * 412 = 1.104557641 atm
For the answer to the questions above,
a) Ag2CO3(s) => Ag2O(s)+CO2(g)
<span>b) Cl2(g)+2(KI)(aq) => I2(s)+2(KCl)(aq) (coefficients are for balanced equation) </span>
<span>net ionic is Cl2(g)+2I- => I2(s)+2Cl-(aq) </span>
<span>c) I2(s)+3(Cl2)(g)=>2(ICl3)
</span>I hope I helped you with your problem