Isn't this false? For the industrial production of indigo carmine, a blue food colouring additive, a synthetic process with an E-factor of 17.4 produces less waste than a synthetic process with an E-factor of 3.0.

The answer I got was False, is this correct?

Answers

Answer 1

Answer:

Answer: yes it is false

Explanation:

The statement is false. A synthetic process with a lower E-factor produces less waste than a process with a higher E-factor.

The E-factor is a measure of the waste generated during a manufacturing process. It is calculated by dividing the total mass of waste produced by the mass of the desired product. A lower E-factor indicates that less waste is generated per unit of product.

In this case, the synthetic process with an E-factor of 3.0 produces less waste than the process with an E-factor of 17.4. This means that the process with an E-factor of 3.0 is more efficient in terms of waste reduction.

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Complete combustion of 4.20 g of a hydrocarbon produced 12.9 g of CO2 and 6.15 g of H2O. What is the empirical formula for the hydrocarbon?

Answers

We calculate first the number of moles of CO2 and H2O produced by dividing the given masses by the molar masses of CO2 and H2O.
moles CO2 = (12.9 g CO2) x (1 mole CO2 / 12 g CO2) = 1.075 moles.
moles H2O = (6.15 g H2O) x (1 mole H2O / 18 g H2O) = 0.36 moles
Then, we count the number of C, H, and O moles. This gives us 1.075 moles C, 2.5 moles O and 0.72 moles H. The empirical formula is,
C1.075H0.72O2.5
Simplifying,
C4H3O10

A cool, yellow-orange flame is used to heat the crucible. Would this affect the mass of the crucible? If so, how?

Answers

Answer:

yes

Explanation:

Usually, it would not affect the crucible, but depending on the temperature of the flame the enamel of the crucible may begin to melt and stick to the metal object being used to handle the crucible. This tiny amount that is melted off can cause very small changes in the original mass of the crucible, which although it is almost unnoticeable it is still there. Therefore, the answer to this question would be yes.

Final answer:

Using a cool, yellow-orange flame to heat the crucible does not directly affect its mass, but can lead to the burning off or decomposition of any impurities or residues present.

Explanation:

When a cool, yellow-orange flame is used to heat the crucible, it does not directly affect the mass of the crucible. The color of the flame is an indication of the temperature and the type of fuel being burned.

However, if there are impurities or residues in the crucible, the heat from the flame can cause them to burn off or decompose, which may slightly affect the mass of the crucible.

Learn more about Effect of flame color on crucible mass here:

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Calculate the number of C atoms in 9.837 x 1024 molecules of CO2.

Please help

Answers

Answer:

Explanation:

1 molecule contains 1 carbon atom.

9.837 * 10^24 molecules contains 9.837 * 10^24 atom of carbon.

It's a 1 to 1 ratio.

help asap What type of cell does the cheek cell represent, plant cell or animal cell? What did you see that let you know?

Answers

Answer:

The human cheek cell is a good example of a typical animal cell. It has a prominent nucleus and a flexible cell membrane which gives the cell its irregular, soft-looking shape.

The ratio of oxygen-16 and oxygen-18 isotopes in plankton fossils in deep-sea sediments can be used to determine ________.

Answers

The ratio of oxygen-16 and oxygen-18 isotopes in plankton fossils in deep-sea sediments can be used to determine past temperatures.

Answer;

-past temperatures

The ratio of oxygen-16 and oxygen-18 isotopes in plankton fossils in deep-sea sediments can be used to determine past temperatures.

Explanation;

-O-16 will evaporate more readily than O-18 since it is lighter, therefore; during a warm period, the relative amount of O-18 will increase in the ocean waters since more of the O-16 is evaporating.

-Hence, looking at the ratio of O16 to O18 in the past can give clues about global temperatures.

Consider the following reaction at equilibrium. What effect will adding 1.4 mole of He to the reaction mixture have on the system? 2 H2S(g) + 3 O2(g) ⇌ 2 H2O(g) + 2 SO2(g) Consider the following reaction at equilibrium. What effect will adding 1.4 mole of He to the reaction mixture have on the system? 2 H2S(g) + 3 O2(g) ⇌ 2 H2O(g) + 2 SO2(g) The reaction will shift to the left in the direction of reactants. No effect will be observed. The reaction will shift to the right in the direction of products. The equilibrium constant will increase. The equilibrium constant will decrease.

Answers

Answer:

The reaction will shift to the left in the direction of reactants.

Explanation:

According to Le Chatelier's principle, when an external constraint is applied to a chemical system in equilibrium, the system adjust in order to annul the effect impose on it by the external system.

Also, from the principle, the addition of an inert gas can affect the equilbrium of a gaseous system, but only if the volume is allowed to change.

There are two cases on which equilibrium depends. These are:

1. Addition of an inert gas at constant volume:

When an inert gas is added to the system in equilibrium at constant volume, the total pressure will increase. But the concentrations of the products and reactants (i.e. ratio of their moles to the volume of the container) will not change. Hence, there will be no effect on the equilibrium.

2. Addition of an inert gas at constant pressure:

When an inert gas is added to a system in equilibrium at constant pressure, then the total volume will increase(i.e. the number of moles per unit volume of various reactants and products will decrease). Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases.

Considering the given reaction in equilibrium:

2H₂S(g) + 3O₂(g) ⇌ 2H₂O(g) + 2SO₂(g)

The addition of an inert gas at constant pressure to the above reaction will shift the equilibrium towards the backward direction because the number of moles of reactants is more than the number of moles of the products.

Final answer:

Adding 1.4 moles of He to the reaction mixture will have no effect on the equilibrium of the system.

Explanation:

Adding 1.4 moles of He to the reaction mixture will have no effect on the system. The equilibrium of the reaction will not shift to the left or right, and there will be no change in the equilibrium constant. This is because He is considered an inert gas and does not participate in the reaction.