You can predict it based of the electronegativity
The decomposition time : 7.69 min ≈ 7.7 min
<h3>Further explanation</h3>
Given
rate constant : 0.029/min
a concentration of 0.050 mol L to a concentration of 0.040 mol L
Required
the decomposition time
Solution
The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time
For first-order reaction :
[A]=[Ao]e^(-kt)
or
ln[A]=-kt+ln(A0)
Input the value :
ln(0.040)=-(0.029)t+ln(0.050)
-3.219 = -0.029t -2.996
-0.223 =-0.029t
t=7.69 minutes
Answer:
1.59 x 10⁻²⁵ J.
Explanation:
- The energy of a photon is calculated Planck - Einstein's equation:
E = h ν
, where
E is the energy of the photon,
h is Planck's constant <em>(h = 6.626 x 10
⁻³⁴ J.s)</em>
ν is the frequency of the photon
-
There is a relation between the frequency (ν
) and wave length (λ).
λ.ν = c,
where c is the speed of light in vacuum (c = 3
.0 x 10
⁸ m/s).
λ = 125 cm = 1.25 m.
<em>Now, E = h.c/λ.</em>
∴ E = h.c/λ = (6.626 x 10
⁻³⁴ J.s) (3
.0 x 10
⁸ m/s) / (1.25 m) = 1.59 x 10⁻²⁵ J.
<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask.
1mmol = 10^-3 mol
Therefore 4.10*10^-5mmol = 4.10*10^-8mol
molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol
You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below)
But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g
Mass is = 9.75*10^-7 grams
1µg = 10^-6 g
You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4
(*see below) at this point you could have said:
1µg = 10^-6 g therefore you have a solution of 6.29µg per litre,
155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
Answer:
The isotopic mass of 41K is 40.9574 amu
Explanation:
Step 1: Data given
The isotopes are:
39K with an isotopic mass of 38.963707u and natural abundance of 93.2581%
40K with an isotopic mass of 39.963999u
41K wit natural abundance of 6.7302 %
Average atomic mass =39.098 amu
Step 2: Calculate natural abundance of 40 K
100 % - 93.2581 % - 6.7302 %
100 % = 0.0117 %
Step 3: Calculate isotopic mass of 41K
39.098 = 38.963707 * 0.932581 + 39.963999 * 0.000117 + X * 0.067302
39.098 = 36.33681 + 0.0046758 + X * 2.067302
X = 40.9574 amu
The isotopic mass of 41K is 40.9574 amu
