Answer:
None is wrinkled and waxy
Explanation:
Spliting the alleles and pairing them usng the two characters listed, The male pea plant will have four possible alleles: RS, Rs, rS, and rs. The female also we have the same. Using punnet's square, we are to cross the two parents, Using gentotypes obtained from this cross; RRSS, RRSs, RrSS, RrSs, RRSs, RRss, RrSs, Rrss, RrSS, RrSc, rrSS, rrSs, RrSs, Rrss, rrSs, rrss. The phenotupic ration will be 9 Round starchy: 3 Round waxy: 3 Wrinkled starchy: 1 wrinkled waxy. It cannot be said that none of the progeny will be wrinkled waxy.
It tends to happen in capillaries
for example in the lungs avioli are coverd in capilarries so to facilitate the diffusion (exchange) of oxygen in to the blood stream and carbon dioxide out of the blood stream
Answer:
H2S.
Explanation:
Ions may be defined as the element that contains either positive or negative charge over them. Two types of ions are cations and anions. The outermost electrons are involved in the formation of ions.
The atomic number of sulfur is 16. Its outermost electronic configuration is K=2, L= 8, M= 6. The sulfur requres two more electrons to complete its orbit and accquire -2 charge.
H S
+1 -2
Then, the formula will be H2S.
Thus, the correct answer is option (c).
Answer:
It's a duplication........