hydrogen is h
The atomic number of hydrogen is 1, indicating that hydrogen has one proton in its nucleus.
In H-N=N-H are 2 sigma's (-) between N and H,and there are 1 pi bond + 1 sigma (=) between N and N
H = hydrogen O = oxygen N = nitrogen Ne = neon Hydrogen is smallest, so H is the answer.
14 n has 5 valence electrons and H has 1 we have 2 N and 4 H so 2*5 + 4*1 10+4 = 14
Hydrogen bonds are formed within molecules. In chemistry, they are the strongest of the 3 types of bonds (London Dispersion, Dipole-Dipole, and Hydrogen Bonding). Molecules that have hydrogen bonds have to have bonds between hydrogen and nitrogen or hydrogen and oxygen or hydrogen and fluorine (N-H, O-H, or F-H).
multiplication is point to point and convolustion is point to multi-point ex multiplication-- s[n]=x[n].h[n] s[0]=[x[0].h[0] s[1]=[x[1].h[1] s[2]=[x[2].h[2] . . . .. s[n-1]=[x[n-1].h[n-1] convollustion s[n]=x[n]*h[n] s[0]=[x[0].h[0]+x[0].h[1]+x[0].h[2]+.......+x[0].h[n-1] s[1]=[x[1].h[0]+x[1].h[1]+x[1].h[2]+.......+x[1].h[n-1] s[2]=[x[2].h[2]+x[2].h[1]+x[2].h[2]+.......+x[2].h[n-1] . . . s[n-1]=[x[n-1].h[0]+x[n-1].h[1]+x[n-1].h[2]+.......+x[n-1].h[n-1].
1 the Atomic Number of Hydrogen
In N2H4, the oxidation number of nitrogen (N) is -2. Each hydrogen (H) has an oxidation number of +1.
The oxidation number of N in N2H4 is -2. In hydrazine (N2H4), each hydrogen atom has an oxidation number of +1, and the overall molecule has a charge of 0. Since there are two hydrogen atoms for each nitrogen atom, the oxidation number of nitrogen must be -2 to balance the charges.
1 the Atomic Number of Hydrogen
The atomic number of hydrogen is 1
The atomic number of hydrogen is 1, indicating that hydrogen has one proton in its nucleus.
In hydrazine (N2H4), the oxidation number of N is -2 because hydrogen (H) is more electronegative than nitrogen (N), so each hydrogen contributes +1 to the overall charge of the molecule. Since hydrazine is a neutral molecule, the two nitrogen atoms must have an oxidation number of -2 each to balance the charges.
Good question. It's because of the Sun's temperature. H-alpha is the line produced from exciting an electron from n=2 to n=3 level. Note that to be able to absorb H-alpha, you would need to have a hydrogen atom with its electron already in the n=2 excited state! Exciting n=1 to n=2 (Lyman-alpha) takes a lot of energy, much more than n=2 to n=3, and the Sun's photosphere is not hot enough for much of its hydrogen to be in the n=2 excited state. Hotter stars have more hydrogen in the n=2 state, so it is then easy to do the (less energetic) n=2 to n=3 H-alpha absorption. That's why hotter stars like A stars have much stronger H-alpha lines. Note that stars hotter than A stars (O,B) again have weak hydrogen absorption lines, because they are so hot most of their hydrogen is ionized, and again there is not much hydrgoen in the n=2 state to excite.
In H-N=N-H are 2 sigma's (-) between N and H,and there are 1 pi bond + 1 sigma (=) between N and N
For nitrogen in HNO3 it is '5'. To calculate oxidation numbers. Use oxygen as the stabndard at '-2' -2 x 3 = -6 is the oxygen moiety. Since hydrogen only oxides to '+1' as H^(+). Create a little sum +1 +N - 6 = 0 To balance then N must be '+5'.
H = hydrogen O = oxygen N = nitrogen Ne = neon Hydrogen is smallest, so H is the answer.