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 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.
To find the H ion concentration in a KOH solution, first determine the hydroxide ion (OH-) concentration. Since KOH is a strong base, it fully dissociates in water. Therefore, the OH- concentration is equal to the KOH concentration, which is 4.8x10^-2 M. The H ion concentration can be calculated using the equation Kw = [H+][OH-], where Kw is the ion product of water (1.0x10^-14 at 25°C). From there, you can determine the H ion concentration by dividing Kw by the OH- concentration.