Hey there!:
Given the reaction:
2 C2H2 + 5 O2 → 4 CO2 + 2 H2O
5 moles O2 ------------- 4 moles CO2
3.00 moles O2 ---------- ( moles of CO2 ?? )
moles of CO2 = 3.00 * 4 / 5
moles of CO2 = 12 / 5
moles of CO2 = 2.4 moles
So, molar mass CO2 = 44.01 g/mol
Therefore:
1 mole CO2 -------------- 44.01 g
2.4 moles CO2 ---------- ( mass of CO2 )
mass of CO2 = 2.4 * 44.01 / 1
mass of CO2 = 106 g
Answer A
Hope that helps!
Yes because molecules is solid
Answer:
The unknown solution had the higher concentration.
Explanation:
When two solutions are separated by a semi-permeable membrane, depending on the concentration gradient between the two solutions, there is a tendency for water molecules to move across the semi-permeable in order to establish an equilibrium concentration between the two solutions. This movement of water molecules across a semi-permeable membrane in response to a concentration gradient is known as osmosis. In osmosis, water molecules moves from a region of lower solute concentration or higher water molecules concentration to a region of higher solute concentration or lower water molecules concentration until equilibrium concentration is attained.
Based on the observation that when the glucose solution described in part A is connected to an unknown solution via a semipermeable membrane, the unknown solution level rises, it means that water molecules have passed from the glucose solution through the semipermeable membrane into the unknown solution. Therefore, the solution has a higher solute concentration than the glucose solution.
Answer:
The reaction will proceed to the left to attain equilibrium.
Explanation:
The question is missing but I guess it must be about <em>how the reaction will proceed to attain equilibrium.</em>
First, we have to calculate the partial pressures using the ideal gas equation.
![pP_{4}=\frac{2.50mol\times (0.08206atm.L/mol.K)\times 673K}{25.0L} =5.52atm](https://tex.z-dn.net/?f=pP_%7B4%7D%3D%5Cfrac%7B2.50mol%5Ctimes%20%280.08206atm.L%2Fmol.K%29%5Ctimes%20673K%7D%7B25.0L%7D%20%3D5.52atm)
![pP_{2}=\frac{1.50mol\times (0.08206atm.L/mol.K)\times 673K}{25.0L}=3.31atm](https://tex.z-dn.net/?f=pP_%7B2%7D%3D%5Cfrac%7B1.50mol%5Ctimes%20%280.08206atm.L%2Fmol.K%29%5Ctimes%20673K%7D%7B25.0L%7D%3D3.31atm)
Now, we have to calculate the reaction quotient (Qp).
![Qp=\frac{pP_{2}^{2}}{pP_{4}} =\frac{3.31^{2} }{5.52} =1.98](https://tex.z-dn.net/?f=Qp%3D%5Cfrac%7BpP_%7B2%7D%5E%7B2%7D%7D%7BpP_%7B4%7D%7D%20%3D%5Cfrac%7B3.31%5E%7B2%7D%20%7D%7B5.52%7D%20%3D1.98)
Since Qp > Kp, the reaction will proceed to the left to attain equilibrium.
Answer:
It indicates that the ratio of the particles in the composition of the sample will always be 2:1 with Hydrogen: Sulfur