Metallic properties head to the left.
Answer:
Explanation:
2 moles hydrogen reacts with one mole of oxygen to give 2 moles of water.
a ) rate of consumption of hydrogen ( moles per second) is twice the rate of consumption of oxygen .
b ) rate of formation of water ( moles per second ) is twice the rate of consumption of oxygen
c ) rate of formation of water ( moles per second ) is equal to the rate of consumption of hydrogen.
Answer:
15.4%
Explanation:
If Ka = 0.54 mM = 1.51x10⁻⁵
Then;
C₄H₈O₂ --------> C₄H₇O₂⁻ + H⁺
I 0.54x10⁻³ 0 0
E 0.54x10⁻³(1-x) 0.54x10⁻³x 0.54x10⁻³x
Recall that x is the percentage degree of dissociation
From the ICE table;
Ka = [C₄H₇O₂⁻] [ H⁺]/[C₄H₈O₂]
1.51x10⁻⁵=(0.54x10⁻³x) (0.54x10⁻³x)/ 0.54x10⁻³(1-x)
1.51x10⁻⁵ = 0.54x10⁻³x^2/1-x
1.51x10⁻⁵(1-x) = 0.54x10⁻³x^2
1.51x10⁻⁵ - 1.51x10⁻⁵x = 0.54x10⁻³x^2
Hence;
0.54x10⁻³x^2 + 1.51x10⁻⁵x - 1.51x10⁻⁵=0
x^2 + 0.028x - 0.028 = 0
Solving the quadratic equation here;
x = 0.154 or −0.182
Ignoring the negative result, x = 0.154
Hence, fraction of butanoic acid that is in the dissociated form in this solution = 15.4%
Answer:
Average atomic mass = 63.553 amu.
Explanation:
Given data:
Abundance of Y-63 = 69.17%
Abundance of Y-65 = 100 - 69.17 = 30.83%
Atomic mass of Y-63 = 62.940 amu
Atomic mass of Y-65 = 64.928 amu
Atomic mass of Y = ?
Solution:
Average atomic mass= (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass= (62.940×69.17)+(64.928×30.83) /100
Average atomic mass = 4353.560 + 2001.730 / 100
Average atomic mass = 6355.29 / 100
Average atomic mass = 63.553 amu.
