The Fermentation Of Glucose (C6H12O6) Produces Ethyl Alcohol ...

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  • The fermentation of glucose (C6H12O6) produces ethyl alcohol (C2H5OH) and CO2...
The fermentation of glucose (C6H12O6) produces ethyl alcohol (C2H5OH) and CO2... Content Type

User Generated

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gungpuhoolxvq

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Science

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The fermentation of glucose (C6H12O6) produces ethyl alcohol (C2H5OH) and CO2:

C6H12O6(aq) → 2 C2H5OH(aq) + 2 CO2(g)

a) How many moles of CO2 are produced when 0.700 mol of C6H12O6 reacts in this fashion?

b) How many grams of C6H12O6 are needed to form 7.50 g of C2H5OH?

c) How many grams of CO2 form when 7.50 g of C2H5OH are produced?

User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.

Explanation & Answer

Thank you for the opportunity to help you with your question!

To answer these questions you need to use stoichiometry of the balanced equation given.

a) .7 mols of glucose * (2mols CO2/1 mol glucose) = 1.4 mols of CO2

b) for this question we will need the MM of glucose and alcohol. 46.07 g/mol for ethyl al, and 180.16 g/mol for glucose.

7.5 g of ethyl al. * (1 mol ethyl alcohol/46.07) * (1 mol glucose/2 mols ethyl al) * (180.16 g glucose/mol) = 14.66 g of glucose

c) is similar to b, but wants the weight of CO2 instead of glucose.

7.5 g of ethyl al. * (1 mol ethyl alcohol/46.07) * (2 mol CO2/2 mols ethyl al) * (44.01g CO2/mol) = 7.16 g of CO2

Please let me know if you didn't understand something that I did or need any clarification. I'm always happy to answer your questions!

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Get 24/7 Study help Our tutors provide high quality explanations & answers. Question details... Post question Most Popular Content ​WAN Technologies Paper WAN Technologies Paper: Research Point-to-Point (dedicated), Packet Switched, and Circuit Switched WAN protocols/circuits/ ... ​WAN Technologies Paper WAN Technologies Paper: Research Point-to-Point (dedicated), Packet Switched, and Circuit Switched WAN protocols/circuits/types. Define each protocol and describe at least two data transmission technologies associated with the protocol. 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If there are passengers in the elevator then their weight is applied towards the floor of the elevator and the elevator force also exerts a normal force on the passenger. When the elevator is accelerating upward the weight of the passengers appears to be greater than the real weight. When the elevator is coming to a rest at a floor, then it begins to decelerate and acceleration is less than the acceleration due to gravity. The weight of the passenger here appears to be less than the real weight. (8 points) Score 3. Use your data from Part 3 and Newton’s laws to explain why the force meter measures a force if the cart is moving at a constant velocity. 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For Part 4, describe the motion of the ball as it moves from the top of the ramp and comes to rest in the cup. Explain how the motion changes in terms of the velocity, distance, acceleration, and force that you measured. Use your data and Newton’s laws to calculate these values and explain the situation. Include the tables you used to calculate average initial velocity and average displacement. Also show your work for your calculations of acceleration and net force. Here are the kinematic equations:PICTURE 4 Answer: when the ball was rolling down the ramp, the force of gravity was acting on it and therefore it was decelerating. Once the ball reached the end, it collided with the cup and moved a few distance and came to stop. The ball had a constant acceleration which means the velocity was not constant. The acceleration of the ball can be calculate using a kinematic formula vf2 = vi2 + 2aΔd. This formula can be modified to a = vf2 -vi ^2 / 2Δd. Final velocity was zero and the first velocity was 1 m/s. Moreover the change in displacement was 0.36 meters a =( 0m/s)^2 – (1m/s )^2 / 2* 0.36 a= -1.4 m/s^2 Time (s) Distance (m) Velocity (m/s) 0.93 1 1.08 1 1 1 1 1 1 1 1 1 1.07 1 0.93 Average velocity (m/s) = 1. 0 m/s ∆d (cm) ∆d (m) Trial 1 34 0.34 Trial 2 36 0.36 Trail 3 36 0.36 Trial 4 38 0.38 Average ∆d(m) = 0.36 m The net force of the collision of the ball with cup can also be calculated using the formula Fnet= mass x acceleration Fnet = ? a = 1.4 m= 0.67kg Fnet = 0.67kg * 1.4 m/s^2 Fnet = 0.94 N concept map Design and submit a concept map on what you have learned. It should include concepts and connections between concepts inc ... concept map Design and submit a concept map on what you have learned. It should include concepts and connections between concepts including:How genetics affects replication and growth of microbes by type Rasmussen University Anatomy and Physiology Discussion Vertigo often presents as dizziness, which can have many causes. In this discussion, we will examine causes and their rela ... Rasmussen University Anatomy and Physiology Discussion Vertigo often presents as dizziness, which can have many causes. In this discussion, we will examine causes and their related anatomy and physiology. Within the article, The Treatment and Natural Course of Peripheral and Central Vertigo, select one type of vertigo to read about. Focus on the anatomy and physiology, as opposed to the treatments. Initial post: In your initial post, describe in your own words the type of vertigo you chose to read about and explain the related anatomy and physiology. View more Earn money selling your Study Documents Sell Documents

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