1. A student tried to solve the following problem by selecting the tile as shown. What, if anything, did the student do wrong?
1 answer:
Answer:
B. The student chose the correct tile, but needs to flip the tile to make the units cancel
Explanation:
Based on the reaction:
2AgNO₃(aq) + Cu(s) → 2Ag(s) + Cu(NO₃)₂ (aq)
<em>2 moles of AgNO₃ react per mole of Cu producing 2 moles of Ag and 1 mole of Cu(NO₃)₂</em>
Thus, if you want to produce 6.75moles of Cu(NO₃)₂ you need:
= 13.50 moles of AgNO₃ are needed
Thus, if you analize the tile shown by the student:
<em>B. The student chose the correct tile, but needs to flip the tile to make the units cancel</em>
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Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.
First of all, the STP conditions refer to the standard temperature and pressure, where the values used are: pressure at 1 atmosphere and temperature at 0°C. These values are reference values for gases.
On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P×V = n×R×T
where:
- P is the gas pressure.
- V is the volume that occupies.
- T is its temperature.
- R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
- n is the number of moles of the gas.
Then, in this case:
- P= 1 atm
- V= 44.1 L
- n= ?
- R= 0.082
- T= 0°C =273 K
Replacing in the expression for the ideal gas law:
1 atm× 44.1 L= n× 0.082 × 273 K
Solving:
n=1.97 moles
Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:
= 63.04 g ≈ <u><em>63 g</em></u>
Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.
Learn more about the ideal gas law: