C
2 to 3
2 of al3-
And 3 of s2+
Equals 0
Answer:
2Na2O2+2H2O⟶O2+4NaOH
2×78g 32g
156g of Na2O2 produces 32g of O2,
12g of Na2O2 produces =15632×12=10.66g.
Density of O2 at NTP=1.428g/mL.
DensityMass=Vol.
1.42810.66=7.46mL
Vol. of O2 at NTP is 7.46mL.
Explanation:
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0.32000616 miles equal 515 meters.
Answer:
See the answer below
Explanation:
From the original equation in the image, the mole ratio of C:CO2:CO is 1:1:2. This means that for every 1 mole of C and CO2, 2 moles of CO would be produced.
Now, looking at the simulation below the equation of the reaction, 3 moles of C and 8 moles of CO2 were supplied as input. Applying this to the original equation of reaction, C seems to be a limiting reagent for the reaction because the ratio of C to CO2 should 1:1.
Hence, taking all the 3 moles of C available means that only 3 moles out of the available 8 for CO2 would be needed. 3 moles c and 3 moles CO2 means that 6 moles CO would be produced (remember that the ratio remains 1:1:3 for C, CO2, and CO). This means that 5 moles CO2 would be leftover.
<em>In other words, all the 3 moles C would be consumed, 3 out of 8 moles CO2 would be consumed, and 6 moles CO would be produced while 5 moles CO2 would be leftover. </em>
Answer:
0.5 M
Explanation:
First, let us look at the balanced equation of the reaction.
The solute formed is .
Recall that: mole = molarity x volume
Hence,
50 ml, 1.00 M H2SO4 = 0.05 x 1 = 0.05 mole
50 ml, 2.0 M KOH = 0.05 x 2 = 0.1 mole
From the equation
<em>1 mole of H2SO4 reacts with 2 moles of KOH to give 1 mole of K2SO4.</em>
Hence,
<em>0.05 mole H2SO4 reacting with 0.1 mole KOH will give 0.05 mole </em><em>.</em>
Also recall that: concentration = mole/volume
Total volume of resulting solution = 50 ml + 50 ml = 100 ml or 0.1 liter
Concentration of = mole of /volume of resulting solution
= 0.05/0.1 = 0.5 M
The concentration of the resulting solute = 0.5 M