Answer:
=1.666 liters
Explanation:
1 mole of a has at standard temperature and pressure occupies a volume of 22.4 liters.
0.5 moles of nitrogen occupy a volume of (0.5 moles×22.4 dm³/mol)/ 1
=11.2 liters.
Standard pressure= 1 atmosphere (Atm)
Standard temperature = 273.15 Kelvin
According to Combined gas equation, P₁V₁/T₁=P₂V₂/T₂
Let us take the conditions under standard conditions as the reference, with the subscript 1 and the conditions under the 5L container to be scenario 2 with subscript 2.
Therefore P₂ =P₁V₁T₂/T₁V₂
Substituting for the values we get:
P₂= (1 atm× 11.2L ×203K)/ (273K×5L)
=1.666 atm
2 cm east is the answer because you go east 5.5 cm and then go back in a sense 3.5 so its basically 5.5-3.5 because its backwards while still facing eat
Well your mass wont change...so it would be 35 grams of a liquid
Answer:
63.05% of MgCO3.3H2O by mass
Explanation:
<em>of MgCO3.3H2O in the mixture?</em>
The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:
<em>Mass water:</em>
3.883g - 2.927g = 0.956g water
<em>Moles water -18.01g/mol-</em>
0.956g water * (1mol/18.01g) = 0.05308 moles H2O.
<em>Moles MgCO3.3H2O:</em>
0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =
0.01769 moles MgCO3.3H2O
<em>Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-</em>
0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O
<em>Mass percent:</em>
2.448g MgCO3.3H2O / 3.883g Mixture * 100 =
<h3>63.05% of MgCO3.3H2O by mass</h3>
