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>
It is <span>a substance produced by a living organism that acts as a catalyst to bring about a specific biochemical reaction.
So I think C matches the discription </span>
Answer:
There is 541.6 grams of calcium carbonate produced from 5.11 moles
Explanation:
Step 1: Data given
Number of moles of sodium carbonate ( Na2CO3) = 5.11 moles
Molar mass of sodium Carbonate:
⇒ Molar mass of Sodium = 22.99 g/mole
⇒ Molar mass of Carbon = 12.01 g/mole
⇒ Molar mass of Oxygen = 16 g/mole
Molar mass of Na2CO3 = 2* 22.99 + 12.01 + 3*16 = 105.99 g/mole
Step 2: Calculate mass of 5.11 moles of sodium carbonate
mass = number of moles of sodium carbonate * Molar mass of sodium carbonate
mass = 5.11 moles * 105.99 g/moles = 541.6 grams ≈ 542 grams
There is 541.6 grams of calcium carbonate produced from 5.11 moles Na2CO3
