No se :/, a ti si te paso??
Answer:
Final Volume = 0.31L
Explanation:
Initial volume V1 = 1L
Initial pressure P1 = 150kPa
Initial temperature, T1 = 25'C + 273 = 298K
Final Pressure P2 = 600kPa
Final Temperature T2 = 100'C + 273 = 373K
Final Volume V2 = ?
The combined gas law is given as;
P1V1 / T1 = V2P2 / T2
150 * 1 / 298 = V2 * 600 / 373
V2 = 55950 / 178800 = 0.31L
Answer:
12.07 g.
Explanation:
- The balanced equation for the mentioned reaction is:
<em>2NH₃(g) + CO₂(g) → H₂NCONH₂(g) + H₂O(g),</em>
It is clear that 2.0 moles of NH₃ react with 1.0 mole of CO₂ to produce 1.0 mole of H₂NCONH₂ and 1.0 moles of H₂O.
- Consider the reaction proceeds at STP conditions:
At STP, 9.0 L of NH₃ react with an excess of CO₂ gas:
It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.
<u><em>using cross multiplication:</em></u>
1.0 mol of NH₃ represents → 22.4 L.
??? mol of NH₃ represents → 9.0 L.
∴ 9.0 L of NH₃ represents = (1.0 mol)(9.0 L)/(22.4 L) = 0.4018 mol.
- To find the no. of moles of urea (H₂NCONH₂) produced:
<u><em>Using cross multiplication:</em></u>
2.0 mol of NH₃ produce → 1.0 mol of H₂NCONH₂, from stichiometry.
0.4018 mol of NH₃ produce → ??? mol of H₂NCONH₂.
∴ The no. of moles of H₂NCONH₂ = (1.0 mol)(0.4018 mol)/(2.0 mol) = 0.201 mol.
- Now, we can find the mass of H₂NCONH₂ produced:
<em>mass = n * molar mass</em> = (0.201 mol) * (60.06 g/mol) = <em>12.07 g.</em>
Ok I know this did it last year let's see chemical because you can't fix it back to its normal self