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SIZIF [17.4K]
2 years ago
6

Consider the reaction. c2h5oh(l)⟶ c2h6(g)+12o2(g). state what further data would be needed to obtain δrh∘ at 500∘c and 1 bar.

Chemistry
1 answer:
UkoKoshka [18]2 years ago
6 0
For this type of problem, it is essential for you to have a data on the standard heats of formation of the substances given. For elemental substances or diatomic gases, the standard heat of formation is 0. Standard means the temperature is at 0°C and pressure at 1 atm. Calculate the standard heat of reaction using:

ΔH°rxn = ∑(Stoichiometric coefficient×ΔHf of products) - ∑(Stoichiometric coefficient×ΔHf of reactants)
Then, use this equation to find the reaction at T = 500°C and P = 1 bar:

ΔHrxn = ΔH°rxn + [∑(Stoichiometric coefficient×Cp of products) - ∑(Stoichiometric coefficient×ΔHf of reactants)]ΔT
So, you also need the Cp or specific heat capacities of the substances.
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If I have 450 grams of iorn how many miles of it do I have
Alja [10]

Answer:

sorry this is a kind of confusing question.

Explanation:

4 0
3 years ago
Is glucose a solution colloid or suspension?
Tju [1.3M]
Glucose is a solution.

5 0
3 years ago
How many moles of glucose does 1.2 x 10^24 molecules represent
sammy [17]
<h3>Answer:</h3>

2.0 mol C₆H₁₂O₆

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

1.2 × 10²⁴ molecules C₆H₁₂O₆ (glucose)

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                               \displaystyle 1.2 \cdot 10^{24} \ molecules \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{6.022 \cdot 10^{23} \ molecules \ C_6H_{12}O_6})
  2. Divide:                                                                                                                      \displaystyle 1.99269 \ mol \ C_6H_{12}O_6

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.99269 mol C₆H₁₂O₆ ≈ 2.0 mol C₆H₁₂O₆

3 0
3 years ago
Amino acids are the building blocks of proteins. The simplest amino acid is glycine (H2NCH2COOH). Draw a Lewis structure for gly
VashaNatasha [74]

Answer:

Explanation:

The lewis structure (indicating all the atoms and patterns provided as hint in the question) of glycine can be seen in the attachment below. While the chemical structure of glycine can be seen below

         H

          |

H₂N - C - C =O

          |      \

         H      OH

The structure (of glycine) above provides a "fair idea" of how the lewis structure will be.

3 0
2 years ago
Instructions
ivann1987 [24]

Answer:

I got a 100 with this, sorry if this is not what you want just trying to help

Explanation:

1. This experiment was to find how mass and speed effect KE. This is important because if you were in a situation where you needed something to go higher, you would know to add more or less of mass/speed.  

To test mass, we filled the bean bag with a certain amount of water, then dropped it. After, you recorded how high it made the bean bag go. The same with speed, but same amount in the bottle, just dropped from different heights.  

My hypothesis is when you have more mass, the KE will be greater. This is also the same with speed, if it is dropped from a higher place, the bean bag will launch farther than the last time.  

2. Data I collected from the lab was like my hypothesis explained. When the height of the bottle increased, it made the bean bag go higher than the last. And I tested 4 different masses, 0.125 kg, 0.250kg, 0.375kg and 0.500kg. Each time the bean bag went higher on a larger mass.  

A lot of times on the speed test, the bean bag would go higher than the bottle drop point, but not every time. Also, when it was dropped from the same height each time, some results varied quite a bit, like when it was dropped from 1.28 the results were 1.14 then 1.30 1.30. Mass on the other hand was all in the same number range, only once the numbers were a bit off from each other.  

3.  Some formulas I used were KE= ½ mv^2 and Ht v^2/2g. The first was to calculate the kinetic energy of an object, m=mass v=speed. Second was for finding out what height I needed to drop something to reach a certain speed, Ht=Height and g= Gravitational Acceleration of 9.8 m/s^2.  

I used these to figure out tables that showed relationships between different things like mass and KE or speed and height. The whole time I was doing the lab, my data was going up, when there was more mass/speed there were higher values in the table.  

This means that my hypothesis at the beginning was correct, more of m/s means KE will increase proportionally because they are all linear. I found it surprising when the bean bag height went over the water bottle drop mark.  

4.     To conclude, my hypothesis matched my data. The data values went up when more mass or speed was added. This means if I were in a situation where I needed more kinetic energy for something, I would know to increase mass or the speed of the object giving it energy.  

The reason that this hypothesis is correct is when you have more mass, you have more energy. So, when you drop let's say a baseball, it isn’t that heavy so it would only launch the bean bag so far. But a bowling ball is very heavy and has lots of energy when falling because of that, it would make the bean bag go very high.  

To make this experiment better, I would use a smoother material for the lever so energy wouldn’t be lost by friction from wood rubbing together. Also, maybe a scanner or video camera to more accurately record how far the bean bag went. All of these would help the lab get more precise results, maybe they could be used in a future lab.

8 0
3 years ago
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