Answer:
more electron deficient
Explanation:
The nitro group is an electron withdrawing group. It withdraws electrons from the pyridine ring by resonance.
This electron withdrawal by resonance makes the pyridine ring less electron rich or more electron deficient.
Hence, the nitro group makes the pyrinde ring more electron deficient
Based on the incomplete neutralization reaction above, the corresponding salts that each reaction would form are as follows:
1. H2SO4 + 2NH4OH----(NH4)2SO4 with water as a byproduct
2. 2NaOH + H2CO3 ----Na2CO3 with water as a byproduct
3. HNO3 + KOH ----KNO3 with water as a byproduct
Answer:
(a) r = 6.26 * 10⁻⁷cm
(b) r₂ = 6.05 * 10⁻⁷cm
Explanation:
Using the sedimentation coefficient formula;
s = M(1-Vρ) / Nf ; where s is sedimentation coefficient, M is molecular weight, V is specific volume of protein, p is density of the solvent, N is Avogadro number, f if frictional force = 6πnr, n is viscosity of the medium, r is radius of particle
s = M ( 1 - Vρ) / N*6πnr
making r sbjct of formula, r = M (1 - Vρ) / N*6πnrs
Note: S = 10⁻¹³ sec, 1 KDalton = 1 *10³ g/mol, I cP = 0.01 g/cm/s
r = {(3.1 * 10⁵ g/mol)(1 - (0.732 cm³/g)(1 g/cm³)} / { (6.02 * 10²³)(6π)(0.01 g/cm/s)(11.7 * 10⁻¹³ sec)
r = 6.26 * 10⁻⁷cm
b. Using the formula r₂/r₁ = s₁/s₂
s₂ = 0.035 + 1s₁ = 1.035s₁
making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁
r₂ = 6.3 * 10⁻⁷cm / 1.035
r₂ = 6.05 * 10⁻⁷cm
A bond is a force of attraction between atoms. They are mainly fthree types of bonds namely; ionic bond, which involves transfer of electrons between a metal and a non metal, covalent bond which occurs between non metal atoms by sharing of electrons, metallic bond which is a bond in the metal structure between metal atoms and the sea of electrons. in this case carbon and hydrogen are non metals hence they will have a covalent bond between their atoms.
Answer:
There are 3, 64 moles of NaCl.
Explanation:
First we calculate the mass of 1 mol of NaCl, starting from the atomic weights of Na and Cl obtained from the periodic table. Then we calculate themoles in 213 grams of NaCl, making a simple rule of three:
Weight NaCl= Weight Na + Weight Cl = 23 g + 35, 5 g= 58, 5 g/ mol
58,5 g ------1 mol NaCl
213 g---------x= (213 g x 1 mol NaCl)/ 58, 5 g= <em>3, 64 mol NaCl</em>