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:
Explanation:
The first one is CrO. The Chromium has the same charge as the oxygen so mol numbers are dropped.
The Second one is CrO2 The two oxygens have a charge of 2(-2) = -4. To balance this, the Chromium must have a charge of +4 Cr(Iv)O2
The third one is can be set up like this
Cr + 3(-2) = 0
Cr - 6 = 0
Cr = 6
Therefore the formula is Cr(vi)O3
The last one is a bit tricky. Follow this carefully. There are 2 Crs and 3Os.
The formula looks like this
2Cr + 3(-2) = 0
2Cr - 6 = 0
2Cr = 6
Cr = 3
The formula is Cr(iii)2 O3
Answer:
3.2×10^-3 mol
Explanation:
The equation for molarity is M= n/L. Where "M" is Molarity, "n" is the number of moles of solute, and "L" is the total liters in solution.
The question gives you the volume in mL, so to convert "mL" to "L" you need to divide by 1000. (6.70mL/ 1000L)= 0.0067L.
Now you can plug the numbers into the equation. 0.480M= n/ 0.0067L), multiply (0.480M×0.0067L)= 0.003216 mol. The scientific notation is 3.2×10^-3, 10^-3 because you move the decimal back three times and 3.2 because there are 2 sig figs.
Correct option is B ethyl ethanoate
The following compound is ester R−COO−R ′ and the naming of ester are done on the basis of alkanoate group (R−COO−) and alkyl (R ′ ) combined to form ester.The suffix for ester is "-oate". The full name of ester is like alkyl alkanoate. Thus the name of given compound is ethyl ethanoate.
<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em><em> ❤️</em>
Stirring and agitating chemical reactions is desirable and stirring reflux systems or any system under heating is necessary to a distribute the heat evenly throughout the system and b) to prevent splashing and boilovers.
