Carbon = 12.010. Oxygen = 15.999 x 2 15.999 x 2 = 31.998 + 12.010 = 44.008 \frac{37.15 grams * 1 mole CO2}{44.008 grams}
First find the oxidation states of the various atoms:
<span>in Cr2O2 2- Cr @ +1; In NH3 N @ +3; in CrO3 Cr @ +3, N2 N @ 0 </span>
<span>Note that N gained electrons, ie, was reduced; Cr was oxidized </span>
<span>Now there is a problem, because B has NH4+ which the problem did not, and is not balanced, showing e- in/out </span>
<span>B.NH4+ → N2 </span>
<span>Which of the following is an oxidation half-reaction? </span>
<span>A.Sn 2+ →Sn 4+ + 2e- </span>
<span>Sn lost electrons so it got oxidized</span>
Answer:
Correct answer is B.
Explanation:
Took the test and got this right. :)
Because the proton has a mass of 1 and a neutron has a mass of 1, we know that there is exactly one proton in the nucleus (because of the atomic number) which therefore tells us there are no neutrons as adding one would make the mass more than one.
Antifreeze is an additive in water-based liquid to lower down the freezing point of such liquid. It is used to make use of the colligative properties of solutions specifically freezing-point depression for cold climate and boiling-point elevation to allow higher coolant temperature.
