Answer: The answer is K3PO4(s) → 3K+(aq) + PO43–(aq) since water-soluble ionic tripotassium phosphate dissociates completely into K+ and PO43– ions when dissolved, that is, no K3PO4 remains in the solution. Carbonic acid H2CO3 and acetic acid CH3COOH are weak electrolytes since they are weak acids that do not completely ionize, while nonelectrolyte CH3OH do not dissociate into ions.
Answer: The correct answer is E.
Explanation:
The path for the ejaculation of the sperm starts in the testicles, more exactly in the seminiferous tubules, then the path continues trough the epididymis for reaching vas deferens. Finally, the path is completed by the sperm passing through the ejaculatory duct for being released by the urethra.
Therefore the correct answer is E.
Answer:
The pH of the buffer solution = 8.05
Explanation:
Using the Henderson - Hasselbalch equation;
pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]
where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21
Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)
[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M
[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M
Therefore,
pH = 7.21 + log (0.663 / 0.096)
pH = 7.21 + 0.84
pH = 8.05
Answer:
Explanation:
Azimuthal maps projection is the easiest for plotting travel by airplane because Azimuthal maps make the UV plane to be tangent to the globe and it's make the directions from the center of projections to be correct and accurate to any other points , likewise circles are also projected directly to the straight lines on the plane. And this is because Any line that is normally drawn through the tangent point give distance correctly and accurately.
Answer : The average atomic mass of an element X is, 63.546 amu
Solution : Given,
Mass of isotope X-63 = 62.9296 amu
% abundance of isotope X-63 = 69.15% = 0.6915
Mass of isotope X-64 = 64.9278 amu
% abundance of isotope X-64 = 30.85% = 0.3085
Formula used for average atomic mass of an element X :
![\text{ Average atomic mass of an element X}=\sum[(62.9296\times0.6915)+(64.9278\times 0.3085)]](https://tex.z-dn.net/?f=%5Ctext%7B%20Average%20atomic%20mass%20of%20an%20element%20X%7D%3D%5Csum%5B%2862.9296%5Ctimes0.6915%29%2B%2864.9278%5Ctimes%200.3085%29%5D)
Therefore, the average atomic mass of an element X is, 63.546 amu
