C. Increases. Increasing temperature=Increasing Volume
Answer:
Percent Yield Fe = 82.5%
Explanation:
The actual yield is the value produced after an experiment is conducted. The theoretical yield is the value calculated using the balanced chemical equation and atomic/molar masses.
To find the percent yield of iron (Fe), you need to (1) convert grams Al to moles Al (via atomic mass), then (2) convert moles Al to moles Fe (via mole-to-mole ratio from equation coefficients), then (3) convert moles Fe to grams Fe (via atomic mass), and then (4) calculate the percent yield. It is important to arrange the ratios in a way that allows for the cancellation of units. The final answer should have 3 sig figs to reflect the sig figs of the given values.
Atomic Mass (Mg): 24.305 g/mol
Atomic Mass (Fe): 55.845 g/mol
3 Mg + 2 FeCl₃ -----> 2 Fe + 3 MgCl₂
20.5 g Mg 1 mole 2 moles Fe 55.845 g
----------------- x ----------------- x ---------------------- x ----------------- =
24.305 g 3 moles Mg 1 mole
= 31.4 g Fe
Actual Yield
Percent Yield = ---------------------------------- x 100%
Theoretical Yield
25.9 g Fe
Percent Yield = -------------------- x 100%
31.4 g Fe
Percent Yield = 82.5%
Answer:
Explanation:
In CF4 and NF3, the valence electron groups on the central C and N atoms have a tetrahedral arrangement. The shapes of the molecules are determined by the number of bonding and nonbonding of electrons: since CF4 has four bonded atom(s) and zero lone pair(s) of electrons, the shape is tetrahedral.
Answer:
0.83 mL
Explanation:
Given data
- Initial concentration (C₁): 12 M
- Final concentration (C₂): 1.0 M
- Final volume (V₂): 10.0 mL
We can calculate the initial volume of HCl using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 1.0 M × 10.0 mL / 12 M
V₁ = 0.83 mL
The required volume of the initial solution is 0.83 mL.
Al
Explanation:
The limiting reactant will be Al:
4Al + 3O₂ → 2Al₂O₃
The limiting reactant is the reactant in short supply in a chemical reaction.
Given parameters:
Mass of Al = 30g Molar mass = 27g/mol
Number of moles = 
= 
Number of moles of Al = 1.111 mole
Mass of O₂ = 30g, molar mass = 32g/mol
Number of moles = 
= 0.94mol
In the reaction:
4 moles of Al reacted with 3 moles of O₂
1.11moles of Al will require 
= 0.83mole to react
But we have been given 0.94mole of O₂. This is more than required.
Therefore O₂ is in excess and Al is the limiting reactant.
Learn more:
Limiting reagents brainly.com/question/6078553
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