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Brut [27]
3 years ago
13

Methanol (ch3oh) can be made by the reaction of co with h2: co(g)+2h2(g)⇌ch3oh(g) to maximize the equilibrium yield of methanol,

would you use a high or low temperature?
Chemistry
1 answer:
BlackZzzverrR [31]3 years ago
8 0
The enthalpy of the creation of Methanol is negative. This means heat is released when the reaction proceeds. You would want to use a low temperature.
You might be interested in
Potassium and fluorine are both halogens?​
andreev551 [17]

Answer:

false, Potassium and fluorine are not halogens.

only fluorine here is halogen.

potassium is an alkali earth metal it doesn't comes under category of halogens, but fluorine

is a non metal which comes under halogen family.

7 0
3 years ago
How do you prepare 500 ml of a 1.77M H2SO4 solution from an 18.0 M H2SO4 stock solution?
finlep [7]

Answer:

Molarity Problems Worksheet  

M=nV   n= # moles  

V must be in liters (change if necessary)  

1. What is the molarity of a 0.30 liter solution containing 0.50 moles of NaCl?  

1.7M

2. Calculate the molarity of 0.289 moles of FeCl3 dissolved in 120 ml of solution?  

2.41 M

3. If a 0.075 liter solution contains 0.0877 moles of CuCO4, what is the molarity?  

1.2M

4. How many moles of NaCl are present in 600. ml a 1.55 M NaCl solution?  

.930 moles

5. How many moles of H2SO4 are present in 1.63 liters of a 0.954 M solution?  

1.56 molse

6. How many liters of solution are needed to make a 1.66 M solution containing 2.11 moles of KMnO4?  

1.27 g

7. What volume of a 0.25 M solution can be made using 0.55 moles of Ca(OH)2?  

2.2 L

For all of the problems below you will need to do a mole-mass conversion. Each problem will involve two steps.  

8. What is the molarity in 650. ml of solution containing 63 grams of NaCl?  

1.7 M

9. How many grams of Ca(OH)2 are needed to produce 500. ml of 1.66 M Ca(OH)2 solution?  

61.5 g

10. What volume of a 0.88 M solution can be made using 130. grams of FeCl2?  

1.2 L

Dilution Problems Worksheet  

1. How do you prepare a 250.-ml of a 2.35 M HF dilution from a 15.0 M stock solution?  

39.2 mL

2. If 455-ml of 6.0 M HNO3 is used to make a 2.5 L dilution, what is the molarity of the dilution?  

1.1 M

3. If 65.5 ml of HCl stock solution is used to make 450.-ml of a 0.675 M HCl dilution, what is the molarity of the stock solution?  

4.64 M

4. How do you prepare 500.-ml of a 1.77 M H2SO4 dilution from an 18.0 M H2SO4 stock solution?  

Take 49.2-ml of 18.0 M H2SO4 stock solution and pour it into a 500-ml volumetric flask. Fill to the 500-ml line with distilled water to make 1.77M H2SO4 solution.

Extra Molarity Problems for Practice  

1. How many moles of LiF would be required to produce a 2.5 M solution with a volume of 1.5 L?  

3.75 M

2. How many moles of Sr(NO3)2 would be used in the preparation of 2.50 L of a 3.5 M solution?  

8.75 M

3. What is the molarity of a 500-ml solution containing 249 g of KI?  

3.00 M

4. How many grams of CaCl2 would be required to produce a 3.5 M solution with a volume of 2.0 L?  

777 g

Explanation:

I've done this before

6 0
3 years ago
How to do limiting reagents for chemistry?
gavmur [86]
Balance the chemical equation for the chemical reaction.
Convert the given information into moles.
Use stoichiometry for each individual reactant to find the mass of product produced.
The reactant that produces a lesser amount of product is the limiting reagent.
The reactant that produces a larger amount of product is the excess reagent.
To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.
4 0
3 years ago
Draw the major addition product of the acid-catalyzed hydration of 4-ethyl-3,3-dimethyl-1-hexene.
kipiarov [429]

Here we have to draw the major product in the acid catalysed hydration reaction of 4-ethyl-3,3-dimethyl-1-hexene.

The 4-ethyl-3,3-dimethyl-1-hexene converts to 2-hydroxy-4-ethyl-3,3-dimethyl-1-hexane as a major product by acid catalyzed hydration reaction.  

The acid catalyzed hydration of an alkene is the Sn¹ reaction. Where in the first step a carbocation is generated. The stability of the carbocation depends upon the position of the neighboring group having +I inductive effect.

In the next step the water molecule attack the carbocation and the corresponding alcohol is produced.

In 4-ethyl-3,3-dimethyl-1-hexene the carbocation formed in the C₂ position which is more stable than the C₁ position due to presence of the dimethyl and ethyl group in the neighboring position which have strong +I inductive effect. This is absence in C₁ position.

In the next step the water molecule attack the C₂ position to form the alcohol.

4-ethyl-3,3-dimethyl-1-hexene converts to 2-hydroxy-4-ethyl-3,3-dimethyl-1-hexane by acid catalyzed hydration reaction which is the major product along with 1-hydroxy-4-ethyl-3,3-dimethyl-1-hexane as a minor product.

The reaction mechanism is shown in the image.          

7 0
3 years ago
At 460 K, the rate constant for this reaction is k = 5.8 × 10–6 s–1 and the activation energy is Ea = 265 kJ/mol. What is the fr
Misha Larkins [42]

The frequency factor from the calculation is 7.1 × 10^25.

<h3>What is the Arrhenius Equation?</h3>

The Arrhenius theory holds that the reaction between molecules owes to the collision between the reactant species. The reacting molecules must posses energy in excess of a minimum amount called the activation energy.

From the theory;

k = A e^-Ea/RT

k = rate constant

A = collision factor

Ea = activation energy

R = gas constant

T = Absolute temperature

5.8 × 10–6 = Ae^-(265 × 10^3/8.314 × 460)

A = 5.8 × 10^–6/8.1 × 10^-31

A = 7.1 × 10^25

Learn more about Arrhenius theory: brainly.com/question/3920636

5 0
2 years ago
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