This gives us the equation: A = hb/2 + c To solve for c, we simply subtract hb/2 from each side! A - hb/2 = c c = A - hb/2
SA = 2B + Ph SA = 2(1/2ab) + (b + c + d)h SA = ab + (b + c + d)h
7g-10h
1.a 2.i 3.c 4.u 5.u 6.m 7.h 8.b 9.q 10.m 11.b 12.a 13.i 14.t 15.p 16.t 17.c 18.h 19.t
v = pi*r^2*h 342 = pi * 3^2 * h 342 = 9h*pi 38 = pi*h h = 38 / pi
Pentanoic acid has 4 carbon atoms with 7 H's bonded to them and 1 carbon atom with a carboxyl ion bonded to it. In the drawings below, I have left off all the single bonded H's to simplify the drawing. . Go to this web site to see a carboxylic group or do a google search for carboxylic group. http://www.ch.ic.ac.uk/vchemlib/mol/glossary/carboxyl.gif . A carboxyl ion is a carbon atom with one oxygen atom double bonded to it, and an O-H single bonded to it as shown below. It has a -1 charge as shown by the - mark to the left of the C. The - ark actually represents one unpaired electron desiring to find another atom with one unpaired electron. …..H …..! ….O ….! ...-C=O . Isotope #1 Draw a row of 4 carbon atoms. Attach a carboxyl ion to the #1 C atom of the 4 you drew. Now the structure has 5 C atoms in a row. . …...H …...! O=C-C-C-C-C . Attaching a carboxyl ion to #4 C is the same as attaching a carboxyl ion to the #1 C atom. (depending on what side of the table you are sitting on, right becomes left) . ………….H ……….…! C-C-C-C=O-H Isotope #2 Attaching a carboxyl ion to the #2 C atom is the same as attaching a carboxyl ion to the #3 C atom. . …....H ....….! ....…O ….....! C-C-C-C . …….H …….! ……O ……! C-C-C-C Isotope #3 Draw a row of 3 carbon atoms, with 1carbon atom bonded below #2 C atom. Attach the carboxyl ion above the #2 C atom. I see 3 carbon atoms in a horizontal line or 3 carbon atoms in a vertical line. There are 4 carbon atoms attached to a central carbon atom. No matter where I attach the double bond O, single bond H, it is attached to one of these 4 carbon atoms. . …H ….! ….C=O-H ….! C-C-C …! ...C . There only 3 isotopes Draw a line of 5 C's …..2..3..4 -C-C-C-C-C- . You might wonder why you can not attach a double bond O, single bond H to Carbon atoms #2, #3,or #4. Looking at the carboxyl ion below, how many bonds does the carbon have around it? I see 3 bonds. (double bond O (2), single bond H (1)) A carbon can only make 4 bonds. If you try to attach a carboxyl ion to Carbon atoms #2, #3, or #4, you still need 2 bonds to attach to the Carbon atoms on the left and right side of that carbon. …..H …..! ….O ….! ...-C=O If you can get a molecular model you can build these isotopes and, determine how many different isotopes you can make.
In even moderately strong acid conditions in aqueous solution, a polyatomic phosphate ion with a charge of -3 will add an hydrogen ion with a charge of +1 to form a polyatomic acid phosphate anion with a charge of -2. In equation form, PO4-3 + H+1 -> HPO4-2.
COOH | H-C-OH | OH-C-H | COOH L(+) Tartaric acid-naturally occurring form with molecular formula C4H6O6
The oxidation number for HCO3, which is the bicarbonate ion, is -1. In this ion, the oxidation numbers of hydrogen, carbon, and oxygen are +1, +4, and -2 respectively.
This is the chemical formula for the ion acetate.
When you make solution of ammonia. Following reaction occurs-NH(3) + H(2)O NH(4)(+) + OH(-).Hence water gives an H(+) ion to ammonia to form ammonium ion.
H H H| | |H-C-C-C-H| | | H I HA complete structural formula will show all bonds
The volume of a cone is equal to 1/3 pi*r2*h. C=2*pi*r, so r=C/(2*pi) and V=1/3*[C/(2*pi)]2*h
The conjugate base of the H2PO4 ion is the HPO4^2- ion. This is formed when H2PO4 loses a hydrogen ion (H+).
The atomicity of HCO3 (hydrogen carbonate or bicarbonate ion) is 3. This means that in one molecule of HCO3, there are a total of 3 atoms - one hydrogen (H), one carbon (C), and three oxygen (O) atoms. This can be determined by counting the individual atoms present in the chemical formula of HCO3.
The standard molar entropy of an H+ ion at 25°C is approximately 0 J/mol-K. Since the H+ ion is a single charged particle with no internal degrees of freedom, its entropy is considered to be close to zero.
The answer is Q. b+1=c, c+2=e, e+3=h, h+4=L, L+5=Q.