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IgorLugansk [536]
4 years ago
9

A chemical engineer is developing a process for producing a new chemical. One step in the process involves allowing a solution o

f potassium hydroxide to react with a solid. Which action would most likely increase the reaction rate for this step? using larger pieces of the solid using a more concentrated potassium hydroxide solution adding water to the system
Chemistry
1 answer:
g100num [7]4 years ago
4 0

Answer:

using a more concentrated potassium hydroxide

Explanation:

<em>The option that would likely increase the rate of reaction would be to use a more concentrated potassium hydroxide.</em>

<u>The concentration of reactants is one of the factors that affect the rate of reaction. The more the concentration of the reactants, the faster the rate of reaction. </u>

Granted that there are enough of the other reactants, increasing the concentration of one of the reactants will lead to an increased rate of reaction.

Hence, using a more concentrated potassium hydroxide which happens to be one of the reactants would likely increase the rate of reaction.

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Magnesium salt is the type of substance
7 0
3 years ago
Read 2 more answers
Please help my teacher has like given up
NikAS [45]

Answer:

24.47 L

Explanation:

Using the general gas law equation:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = 0.0821 Latm/molK

T = temperature (K)

According to the provided information in this question,

P = 1.0 atm

V = ?

n = 1 mol

T = 25°C = 25 + 273 = 298K

Using PV = nRT

V = nRT ÷ P

V = 1 × 0.0821 × 298 ÷ 1

V = 24.465 ÷ 1

V = 24.465

V = 24.47 L

5 0
3 years ago
Which of Graphs 1 correctly represents the relationship between the volume and Kelvin temperature of a gas?
pav-90 [236]

Answer:

B

Explanation:

Pressure is directly proportional to temperature

8 0
3 years ago
how many ml of 0.032 molar kmno4 are required to react with 50.0 ml of 0.100 molar h2c2o4 in the presence of excess h2so4
BartSMP [9]

Answer:

62.5 ml of 0.032 M  KMnO₄ are required to react with 50.0 ml of 0.100 molar H₂C₂O₄ in the presence of excess H₂SO₄

Explanation:

The balanced reaction is:

2 KMnO₄ + 5 H₂C₂O₄ + 3 H₂SO₄ → K₂SO₄ + 2 MnSO₄ + 8 H₂O + 10 CO₂

By stoichiometry of the reaction (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:

  • KMnO₄: 2 moles
  • H₂C₂O₄: 5 moles
  • H₂SO₄: 3 moles
  • K₂SO₄: 1 mole
  • MnSO₄: 2 moles
  • H₂O: 8 moles
  • CO₂: 10 moles

Molarity or Molar Concentration is the number of moles of solute that are dissolved in a certain volume.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

Molarity=\frac{number of moles of solute}{volume}

Molarity is expressed in units \frac{moles}{liter}

In this case, 50 mL (0.05 L) of 0.1 M H₂C₂O₄ react. So, replacing the data in the definition of molarity:

0.1 M=\frac{number of moles of solute}{0.05 L}

Solving:

number of moles of solute= 0.1 M*0.05 L

number of moles of solute= 0.005 moles

So, 0.005 moles of H₂C₂O₄ react.  Then you can apply the following rule of three: if by stoichiometry 5 moles of H₂C₂O₄ react with 2 moles of KMnO₄, 0.005 moles of H₂C₂O₄ react with how many moles of KMnO₄?

moles of KMnO_{4} =\frac{0.005moles of H_{2} C_{2} O_{4}* 2moles of KMnO_{4} }{5moles of H_{2} C_{2} O_{4} }

moles of KMnO₄= 0.002 moles

Knowing that the molarity of KMnO₄ is 0.032 M, replacing in its definition and solving:

0.032 M=\frac{0.002 moles}{volume}

volume=\frac{0.002 moles}{0.032 M}

volume= 0.0625 L= 62.5 mL

<u><em>62.5 ml of 0.032 M  KMnO₄ are required to react with 50.0 ml of 0.100 molar H₂C₂O₄ in the presence of excess H₂SO₄</em></u>

6 0
3 years ago
If nitrogen gas is collected over water at 20 o C and the total pressure of the gas in the collection vessel is 200 kPa, what is
Citrus2011 [14]

Answer:

Explanation:

pressure of dry nitrogen gas at 20° = Partial pressure of nitrogen and partial pressure of water at that temperature .

The pressure of dry nitrogen gas= total pressure - vapour pressure of water at 20°C

= 200 k Pa - 2.34 k Pa

= 197.66 kPa.

8 0
3 years ago
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