Answer:
The possible genotypes of a man with blood type B are BB or BO and the genotype of a woman with blood type AB is AB. The child would receive an A allele or a B allele from the mother and a B allele or an O allele from the father. Therefore, the child could not possibly be of blood type O.
Explanation:
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If that is your picture, then the answer is that
Geometry A is trigonal planarand Geometry B is trigonal pyramidal
Answer:
Name Atomic Number Electron Configuration Period 1 Hydrogen 1 1s1 Helium 2 1s2 Period 2 Lithium 3 1s2 2s1 Beryllium 4 1s2 2s2 Boron 5 1s2 2s22p1 Carbon 6 1s2 2s22p2 Nitrogen 7 1s2 2s22p3 Oxygen 8 1s2 2s22p4 Fluorine 9 1s2 2s22p5 Neon 10 1s2 2s22p6 Period 3 Sodium 11 1s2 2s22p63s1 Magnesium 12 1s2 2s22p63s2 Aluminum 13
Answer:
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Explanation:
Determine the [OH−] , pH, and pOH of a solution with a [H+] of 9.5×10−13 M at 25 °C.
Given [H⁺] = 9.5 x 10⁻¹³M => [H⁺][OH⁻] = 1.0 x 10⁻¹⁴ => [OH⁻] = 1.0 x 10⁻¹⁴/9.5 x 10⁻¹³ = 0.0105M
pH = -log[H⁺] = -log(9.5 x 10⁻¹³) = - (-1202) = 12.02.
pOH = -log[OH⁻] = -log(0.0105) = -(-1.98) = 1.98
Now you use the same sequence in the remaining problems.