Answer:
Explanation:
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In this case, for this neutralization reaction, it is possible to realize that one the neutralization products is water (pH=7) and the other one is the salt coming up from the cation of the NaOH and the anion of the HI:
Moreover, since the solubility of NaI is large in water, we infer it remains aqueous whereas the water is maintained as liquid:
Which is also balanced as the number of atoms of all the elements is the same at both sides.
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Taking into account the reaction stoichiometry, 8 moles of Ag can be produced from 8 moles of AgNO₃ and 5 moles of Zn.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
2 AgNO₃ + Zn → 2 Ag + Zn(NO₃)₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
<h3>Limiting reagent in this case</h3>To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Zn reacts with 2 moles of AgNO₃, 5 moles of Zn reacts with how many moles of AgNO₃?
<u><em>amount of moles of AgNO₃= 10 moles </em></u>
But 10 moles of AgNO₃ are not available, 8 moles are available. Since you have less moles than you need to react with 5 moles of Zn, AgNO₃ will be the limiting reagent.
<h3>Moles of Ag formed</h3>Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 2 moles of AgNO₃ form 2 moles of Ag, 8 moles of AgNO₃ form how many moles of Ag?
<u><em>amount of moles of Ag= 8 moles</em></u>
Then, 8 moles of Ag can be produced from 8 moles of AgNO₃ and 5 moles of Zn.
Learn more about the reaction stoichiometry:
<u>brainly.com/question/24741074</u>
<u>brainly.com/question/24653699</u>
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Answer:
3.91 minutes
Explanation:
Given that:
Biacetyl breakdown with a half life of 9.0 min after undergoing first-order reaction;
As we known that the half-life for first order is:
where;
k = constant
The formula can be re-written as:
Let the initial amount of butter flavor in the food be = 100%
Also, the amount of butter flavor retained at 200°C = 74%
The rate constant
To determine how long can the food be heated at this temperature and retain 74% of its buttery flavor; we use the formula:
Substituting our values; we have:
t = 3.91 minutes
∵ The time needed for the food to be heated at this temperature and retain 74% of its buttery flavor is 3.91 minutes