Answer:
80cm3 of water, and 60cm3 carbon IV oxide is formed while 20cm3 of oxygen is left unreacted.
Explanation:
From Gay-Lussac's law, there are five volumes of oxygen, 1 volume if propane, 4 volumes of water and three volumes of CO2. Applying this shows the reacting volumes as we have in the image attached, hence the volumes left after reaction.
Answer:
its not cear to answer this question
Answer: look at the close because that is the answer
Explanation:
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>
The balanced equation for the given reaction:
C₆H₁₂O₆ (glucose) + 6O₂→ 6CO₂ + 6H₂O
So in the balanced equation the coefficients before glucose, oxygen, water and carbon dioxide are 1, 6, 6 and 6 respectively.
Therefore, the sum of the coefficients in the balanced equation
= 1 + 6 + 6 + 6
= 19
The correct answer is 19.
