Answer:
"gene" is the segment of DNA containg the inheritance information.
Answer:
A. m C5H12 = 108.23 g
B. m F2 = 547.142 g
C. m Ca(CN)2 = 71.85 g
Explanation:
- mass (m) = mol (n) × molecular weigth (Mw)
∴ Mw C5H12 = ((12.011)(5)) + ((1.008)(12)) = 72.151 g/mol C5H12
∴ Mw F2 = (18.998)(2) = 37.996 g/mol F2
∴ Mw = Ca(CN)2 = 40.078+((12.011+14.007)(2)) = 92.114 g/mol Ca(CN)2
A. m C5H12 = ( 1.50 mol)×(72.151 g/mol) = 108.23 g C5H12
B. m F2 = (14.4 mol)×(37.996 g/mol) = 547.142 g F2
C. m Ca(CN)2 = (0.780 mol)×(92.114 g/mol) = 71.85 g Ca(CN)2
Weighs 0.001836 gram per cubic centimeter or 1.836 kilogram per cubic meter
Try to see if this helps
Answer:
The 1st and 4th options are correct
I.the oxidized form has a higher affinity for electrons
IV. the greater the tendency for the oxidized form to accept electrons
Explanation:
Half reaction can be described as the oxidation or reduction reaction in a redox reaction.it is In the redox rection there is a change in the oxidation states of Chemical species involved. the oxidized form in the redox has a higher affinity for electrons and the greater the tendency for the oxidized form to accept electrons.
Standard reduction potential which is also referred to as standard cell potential can be described as the potential difference that exist between cathode and anode of the cell. In the standard reduction potential most times the species will be reduced which is usually analysed in a reduction half reaction.
(Standard Hydrogen Electrode) is utilized when determining the Standard reduction or potentials of a chemical specie. this is because of Hydrogen having zero reduction and oxidation potentials, as a result of this a measured potential of any species is compared with that of Hydrogen, the difference helps to know the potential reduction of that particular specie.
Answer:
N2H2(aq) + 2OH^-(aq) ----------> N2(g) + 2H2O(l) + 2e
Explanation:
Hydrazine is mostly used in thermal engineering as an anticorrosive agent. Hydrazine can be oxidized in aqueous solution as shown in the equation above. Oxidation has to do with loss of electrons and increase in oxidation number.
The oxidation number of nitrogen in the equation increased from -1 in hydrazine on the lefthand side of the reaction equation to zero in nitrogen on the right hand side of the reaction equation. Two electrons were lost in the process as shown.
