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Musya8 [376]
3 years ago
11

Given the enthalpies of reaction below, determine the enthalpy change for the reaction: N2O (g) + NO2 (g) 3NO (g). N2 (g) + O2 (

g) 2NO (g) ΔH = +180.7 kJ 2NO (g) + O2 (g) 2NO2 (g) ΔH = -113.1 kJ 2N2O (g) 2N2 (g) + O2 (g) ΔH = -163.2 kJ A. 155.6 kJ B. 130.6 kJ C. 311.3 kJ D. 65.3 kJ
Chemistry
1 answer:
lubasha [3.4K]3 years ago
6 0
This problem applies Hess' Law to come up with the final reaction from a series of reactions given. Equation (1) stays as it is. Equation (2) should be inverted and divided by 2. Equation (3) is simply divided by 2. The energies should also correspond to the operations. Adding the net energies of each reaction, the total energy is 155.65 kJ. Hence, the answer is A.
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How many grams of NaF should be added to 500 mL of a 0.100 M solution of HF to make a buffer with a pH of 3.2
Korolek [52]

Answer:

2.25g of NaF are needed to prepare the buffer of pH = 3.2

Explanation:

The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:

pH = pKa + log [A-] / [HA]

<em>Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF). </em>

The moles of HF are:

500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF

Replacing:

3.2 = 3.17 + log [A-] / [0.0500moles]

0.03 = log [A-] / [0.0500moles]

1.017152 = [A-] / [0.0500moles]

[A-] = 0.0500mol * 1.017152

[A-] = 0.0536 moles NaF

The mass could be obtained using the molar mass of NaF (41.99g/mol):

0.0536 moles NaF * (41.99g/mol) =

<h3>2.25g of NaF are needed to prepare the buffer of pH = 3.2</h3>
4 0
3 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
Friction is the force that occurs when: A. one object attracts or repels another. B. one object is rubbing against another. C. o
AlladinOne [14]

Answer:

b i think. hope im not wrong

4 0
3 years ago
Read 2 more answers
How much heat is required to raise the temperature of a 6.21 g sample of iron from 25.0 oC to 79.8 oC?
telo118 [61]

Answer:

151.1J

Explanation:

Given parameters:

Mass of iron  = 6.21g

Initial temperature of iron  = 25°C

Final temperature of iron  = 79.8°C

Unknown:

Amount of heat = ?

Solution:

The amount of heat require to cause this temperature can be determined using the expression below;

    H  = m c (T₂ - T₁)

H is the amount of heat

m is the mass

c is the specific heat capacity

T is the temperature

    Specific heat capacity of iron 0.444J/g°C

Insert the parameters and solve;

     H  = 6.21 x 0.444 x (79.8 - 25)

     H   = 151.1J

5 0
3 years ago
Calculate the theoretical yield of ammonia produced by the reaction of 100 g of H2 gas and 200g of N2 gas? 3H2(g) + N2(g)-----&g
Alex787 [66]
Moles of Hydrogen present: 100 / 2 = 50 moles

Moles of Nitrogen present: 200 / 28 = 7.14 moles

Hydrogen required by given amount of nitrogen = 7.14 x 3 = 21.42 moles

Hydrogen is excess so we will calculate the Ammonia produced using Nitrogen.

Molar ratio of Nitrogen : Ammonia = 1 : 2

Moles of ammonia = 7.14 x 2 = 14.28 moles
8 0
4 years ago
Read 2 more answers
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