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Leviafan [203]
3 years ago
8

List three factors thatinfluence the rate at whick dissolving occurs

Chemistry
1 answer:
lys-0071 [83]3 years ago
5 0
1) Temperature (heat) of the solution
2) Concentration (amount) of both solvent (usually water) and solute (substance being dissolved by solvent)
3) Movement (kinetic energy) of the solution, as in shaking/stirring
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A 10 gram sample of iron reacts with oxygen to form 18.2 grams of ferric oxide. How many grams of oxygen reacted?
RoseWind [281]

Answer:

\boxed {\boxed {\sf 8.2 \ grams}}

Explanation:

According to the Law of Conservation of Mass, the mass of the products must equal the mass of the reactants.

  • mass products = mass reactants

In this problem, the reaction is:

iron + oxygen = ferric \ oxygen

  • The reactants are iron and oxygen. We know the mass of the iron sample is 10 grams.
  • The product is ferric oxide. The mass of the ferric oxide sample is 18.2 grams.

10 \ g + oxygen=18.2 \ g

We want to find how many grams of oxygen reacted. We have to get the oxygen by itself. 10 is being added to oxygen. The inverse of addition is subtraction. Subtract 10 from both sides of the equation.

10 \ g - 10 \ g+ oxygen = 18.2 \ g - 10 \ g

oxygen= 18.2 \ g - 10 \ g

oxygen= 8.2 \ g \\

<u>8.2 grams of oxygen </u>reacted with 10 grams of iron to form 18.2 grams of ferric oxide.

8 0
3 years ago
Cdvshduejsnxhhwjxhxh ydujdbxg
fiasKO [112]

Answer:

This is not a question

Explanation:

5 0
3 years ago
Read 2 more answers
Reactant(s)<br> H<br> 0,<br> Product(s)<br> H,0<br> 3.4g<br> 10g
zalisa [80]

Answer:

H,0

3.4g

Explanation:

H,0

3.4g

8 0
3 years ago
8.0 mol AgNO3 reacts with 5.0 mol Zn in
Yakvenalex [24]

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:

  • AgNO₃: 2 moles
  • Zn: 1 mole
  • Ag: 2 moles
  • Zn(NO₃)₂: 1 mole

<h3>Limiting reagent</h3>

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?

amount of moles of AgNO_{3}= \frac{5 moles of Znx2 moles of AgNO_{3}}{1 mole of Zn}

<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?

amount of moles of Ag=\frac{8 moles of AgNO_{3}x2 moles of Ag }{2 moles of AgNO_{3}}

<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>

#SPJ1

4 0
2 years ago
You are given a solution of HCOOH (formic acid) with an approximate concentration of 0.20 M and you will titrate this with a 0.1
jeka57 [31]

Answer:

\boxed{\text{36 mL}}

Explanation:

1. Write the balanced chemical equation.

\rm HCOOH + NaOH $ \longrightarrow$ HCOONa + H$_{2}$O

2. Calculate the moles of HCOOH

\text{Moles of HCOOH} =\text{20.00 mL HCOOH } \times \dfrac{\text{0.20 mmol HCOOHl}}{\text{1 mL HCOOH}} = \text{4.00 mmol HCOOH}

3. Calculate the moles of NaOH.

\text{Moles of NaOH = 4.00 mmol HCOOH } \times \dfrac{\text{1 mmol NaOH} }{\text{1 mmol HCOOH}} = \text{4.00 mmol NaOH}

4. Calculate the volume of NaOH

c = \text{4.00 mmol NaOH } \times \dfrac{\text{1 mL NaOH }}{\text{0.1105 mmol NaOH }} = \textbf{36 mL NaOH }\\\\\text{The titration will require }\boxed{\textbf{36 mL of NaOH}}

3 0
3 years ago
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