Answer:
d properties
i took a test on this before
Answer:
38.75 L
Explanation:
From the question,
Applying Boyles Law,
PV = P'V'....................... Equation 1
Where P = Original pressure of the Argon gas, V = Original Volume of Argon gas, P' = Final pressure of Argon gas, V' = Final Volume of Argon gas.
make V the subject of the equation
V = P'V'/P.................... Equation 2
Given: P = 34.6 atm, V' = 456 L, P' = 2.94 atm.
Substitute these values into equation 2
V = (456×2.94)/34.6
V = 38.75 L
Answer:
223 g O₂
Explanation:
To find the mass of oxygen gas needed, you need to (1) convert moles Al to moles O₂ (via the mole-to-mole ratio from reaction coefficients) and then (2) convert moles O₂ to grams O₂ (via the molar mass). When writing your ratios/conversions, the desired unit should be in the numerator in order to allow for the cancellation of the previous unit. The final answer should have 3 sig figs because the given value (9.30 moles) has 3 sig figs.
4 Al + 3 O₂ ----> 2 Al₂O₃
^ ^
Molar Mass (O₂): 32.0 g/mol
9.3 moles Al 3 moles O₂ 32.0 g
------------------- x --------------------- x -------------------- = 223 g O₂
4 moles Al 1 mole
A: Trial 1, because the average rate of the reaction is lower.
The rate of reaction is the speed with which reactants are converted into products. It is also the rate at which reactants disappear and products appear. The higher the rate of reaction, the greater the amount of product formed in a reaction.
If we look at the graph, we will realize that trial 1 produces a lesser amount of product than trial 2. This implies that the average rate of the reaction in trial 1 is lower than in trial 2.
Lower average rate of reaction implies lower concentration of the reactants since the rate of reaction depends on the concentration of reactants.
Hence trial 1 has a lower concentration of reactants because the average rate of the reaction is lower.
