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VashaNatasha [74]
3 years ago
12

How is a compound formed differently than a mixture, and how can each be broken down into their original substances?

Chemistry
2 answers:
Ksivusya [100]3 years ago
7 0
A compound is a pure substance
A mixture is a impure substance
Compounds can be broken into simpler substances only by chemical methods
Mixtures can be separated by physical and chemical methods
saveliy_v [14]3 years ago
5 0

Answer: Pure substances are further broken down into elements and compounds. Mixtures are physically combined structures that can be separated into their original components. A mixture is composed of different types of atoms or molecules that are not chemically bonded.

Explanation:

You might be interested in
ΔG o for the reaction H2(g) + I2(g) ⇌ 2HI(g) is 2.60 kJ/mol at 25°C. Calculate ΔG o , and predict the direction in which the rea
kondaur [170]

Answer:

The reaction is not spontaneous in the forward direction, but in the reverse direction.

Explanation:

<u>Step 1: </u>Data given

H2(g) + I2(g) ⇌ 2HI(g)     ΔG° = 2.60 kJ/mol

Temperature = 25°C = 25+273 = 298 Kelvin

The initial pressures are:

pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

<u>Step 2</u>: Calculate ΔG

ΔG = ΔG° + RTln Q  

with ΔG° = 2.60 kJ/mol

with R = 8.3145 J/K*mol

with T = 298 Kelvin

Q = the reaction quotient → has the same expression as equilibrium constant → in this case Kp = [p(HI)]²/ [p(H2)] [p(I2)]

with pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

Q = (3.10²)/(1.5*1.75)

Q = 3.661

ΔG = ΔG° + RTln Q  

ΔG = 2600 J/mol + 8.3145 J/K*mol * 298 K * ln(3.661)  

ΔG =5815.43 J/mol = 5.815 kJ/mol

To be spontaneous, ΔG should be <0.

ΔG >>0 so the reaction is not spontaneous in the forward direction, but in the reverse direction.

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
Explain how you were able to use your knowledge of how different types of blood react with anti A, anti B, and anti Rh antibodie
dybincka [34]

Depending upon the clumping reaction with anti A , anti B and anti Rh antibodies the blood types are determined.

Explanation:

Agglutination (clumping) will occur when blood that contains the particular antigen is mixed with the particular antibody.

A+ have Agglutination with Anti-A ,Anti-Rh and No agglutination with Anti-B.

A- have Agglutination with Anti-A and No agglutination with Anti-B and Anti-Rh.

B+ have Agglutination with Anti-B Anti-Rh and No agglutination with Anti-A.

B- have Agglutination with Anti-B and No agglutination with Anti-B and Anti-Rh.

Rh+ have Agglutination with Anti-A and Anti-Rh and No agglutination with Anti-B.

Rh- have No Agglutination with Anti-A and Anti-B and Anti-Rh.

3 0
3 years ago
F liquid water runs into an open space within a rock, like a crack, and the temperature drops to the freezing point of water, wh
vodomira [7]

Answer:

b or c is the answer

Explanation:

brainlest plz

4 0
3 years ago
Read 2 more answers
What is the final concentration of cl- ion when 250 ml of 0.20 m cacl2 solution is mixed with 250 ml of 0.40 m kcl solution? (as
serious [3.7K]

CaCl2 and KCl are both salts which dissociate in water when dissolved. Assuming that the dissolution of the two salts are 100 percent, the half reactions are:

<span>CaCl2 ---> Ca2+  +  2 Cl-</span>

KCl ---> K+ + Cl-

Therefore the total Cl- ion concentration would be coming from both salts. First, we calculate the Cl- from each salt by using stoichiometric ratio:

Cl- from CaCl2 = (0.2 moles CaCl2/ L) (0.25 L) (2 moles Cl / 1 mole CaCl2)

Cl- from CaCl2 = 0.1 moles

 

Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1 mole KCl)

Cl- from KCl = 0.1 moles

 

Therefore the final concentration of Cl- in the solution mixture is:

Cl- = (0.1 moles + 0.1 moles) / (0.25 L + 0.25 L)

Cl- = 0.2 moles / 0.5 moles

<span>Cl- = 0.4 moles             (ANSWER)</span>

6 0
3 years ago
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