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Fiesta28 [93]
4 years ago
9

A water bottling company is looking to achieve maximum production from the following reversible reaction. If the reaction has re

ached a point of equilibrium, what would be the best way to maximize production of water? 2 H2+O2⇔2 H2O Remove hydrogen and allow the reaction to reach equilibrium again. Stop of the process because no more water can be made after equilibrium is reached. Remove the already produced water and allow the reaction to reach equilibrium again. Remove oxygen and allow the reaction to reach equilibrium again.
Chemistry
1 answer:
dybincka [34]4 years ago
7 0

Answer:

Remove the already produced water and allow the reaction to reach equilibrium again.

Explanation:

<em>According to Le Chatelier principle, when a reaction is in equilibrium and one of the factors that influence the rate of reaction is altered, the equilibrium will shift so as to annul the effects of the change.</em>

If the product is continuously being removed from a reaction that is in equilibrium, more product will continued to be formed in another to annul the effect of reduction in the concentration of product.

Hence, in order to maximize production of water in the reaction, the product (water) needs to be removed and the reaction allowed to reach equilibrium again.

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Another chem question I need help with (multiple choice)
Andreyy89

Answer:

The answer to your question is letter B, 2-methylhexane.

Explanation:

Remember that for naming organic compounds first, we need to look for the largest chain of carbons.

In your example, the largest chain is horizontal and has 6 carbons.

Later, we need to circle all the branches, in your example there is only one branch located close to the left side

After that, we number the carbons of the main chain, starting in the corner with more branches, in your example we start from the first carbon on the left.

Finally, start naming the number of the carbon branch, later hte name of the branch and finally the name of the main chain.

7 0
3 years ago
...............………...mkmlkml;m
bagirrra123 [75]

Answer:

390

Explanation:

Specific heat capacity= heat/mass × temperature

x = 2925 \div 0.5 \times 15 = 390

Remember you convert gram into kilogram and 1 gram =0.001 kilogram

5 0
3 years ago
If 20 grams of Zinc phosphate reacts with excess hydrochloric acid and produces 18 grams of Zinc chloride what is the percent yi
fenix001 [56]

Answer:

Y=85\%

Explanation:

Hello!

In this case, since we know the balanced chemical reaction, we are first able to realize there is a 1:3 mole ratio between zinc phosphate and zinc chloride; it means that we can first compute the moles of the desired product via stoichiometry:

n_{ZnCl_2}=20gZn_3(PO_4)_2*\frac{1molZn_3(PO_4)_2}{386.11gZn_3(PO_4)_2}*\frac{3molZnCl_2}{1molZn_3(PO_4)_2}=0.16gZnCl_2

Next, since those moles are associated with the theoretical yield of zinc chloride, we obtain the corresponding mass:

m_{ZnCl_2}^{theoretical}=0.16molZnCl_2*\frac{136.29gZnCl_2}{1molZnCl_2} =21gZnCl_2

Finally, we compute the percent yield by diving the actual yield (18 g) by the theoretical yield:

Y=\frac{18g}{21g}*100\%\\\\Y=85\%

Best regards!

4 0
3 years ago
A cylinder with a moveable piston contains 92g of Nitrogen. The external pressure is constant at 1.00 atm. The initial temperatu
Jobisdone [24]

Answer:

Work done in this process = 4053 J

Explanation:

Mass of the gas = 0.092 kg

Pressure is constant = 1 atm = 101325 pa

Initial temperature T_{1} = 200 K

Final temperature T_{2} = 200 - 85 = 115 K

Gas constant for nitrogen = 297 \frac{J}{kg k}

When pressure of a gas is constant, volume of the gas is directly proportional to its temperature.

⇒ V ∝ T

⇒ \frac{V_{2} }{V_{1} } = \frac{T_{2} }{T_{1} } ------------ ( 1 )

From ideal gas equation P_{1} V_{1} = m R T_{1} ------ (2)

⇒ 101325 × V_{1} = 0.092 × 297 × 200

⇒ V_{1} = 0.054 m^{3}

This is the volume at initial condition.

From equation 1

⇒ \frac{V_{2} }{0.054} = \frac{200}{115}

⇒ V_{2} = 0.094 m^{3}

This is the volume at final condition.

Thus the work done is given by W = P [V_{2} - V_{1} ]

⇒ W = 101325 × [ 0.094 - 0.054]

⇒ W = 4053 J

This is the work done in that process.

7 0
3 years ago
If you mix a 25.0mL sample of a 1.20m potassium sulfide solution with 15.0 mL of a 0.900m basium nitrate solution and the reacti
ddd [48]

Answer:

Limiting reagent: barium nitrate

Theoretical yield: 2.29 g BaS

Percent yield: 87%

Explanation:

The corrected balanced reaction equation is:

K₂S + Ba(NO₃)₂ ⇒ 2KNO₃ + BaS

The amount of potassium sulfide added is:

(25.0 mL)(1.20mol/L) = 30 mmol

The amount of barium nitrate added is:

(15.0mL)(0.900mol/L = 13.5 mmol

Since the reactant react in a 1:1 molar ratio, barium nitrate is the limiting reagent. If all of the limiting reagent reacts, the amount of barium sulfide produced is:

(13.5 mmol Ba(NO₃)₂)(BaS/Ba(NO₃)₂ ) = 13.5 mmol BaS

Converting this amount to grams gives the theoretical yield of BaS (molar mass 169.39 g/mol).

(13.5 mmol)(169.39 g/mol)(1g/1000mg) = 2.29 g BaS

The percent yield is calculated as follows:

(actual yield) / (theoretical yield) x 100%

(2.0 g) / (2.29 g) x 100% = 87%

6 0
3 years ago
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