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Alik [6]
2 years ago
15

Lab: Applications of electromagnetic​

Chemistry
1 answer:
Pavlova-9 [17]2 years ago
7 0

Answer:Not the same class but it will give u 100%

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.

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Be sure to answer all parts. Write an unbalanced equation to represent each of the following reactions: Do not include phase abb
Eva8 [605]

<u>Answer:</u> The unbalanced chemical equations are written below.

<u>Explanation:</u>

An unbalanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side is not equal to the total number of individual atoms on the product side. These equations does not follow law of conservation of mass.

  • <u>For a:</u>

The chemical equation for the reaction of nitrogen gas and oxygen gas follows:

N_2+O_2\rightarrow NO_2

The product formed is nitrogen dioxide.

  • <u>For b:</u>

The chemical equation for the decomposition of dinitrogen pentaoxide follows:

N_2O_5\rightarrow N_2O_4+O_2

The product formed is dinitrogen tetroxide and oxygen gas.

  • <u>For c:</u>

The chemical equation for the reaction of ozone to oxygen gas follows:

O_3\rightarrow O_2

The product formed is oxygen gas.

  • <u>For d:</u>

The chemical equation for the reaction of chlorine and sodium iodide follows:

Cl_2+NaI\rightarrow NaCl+I_2

The product formed is sodium chloride and iodine gas

  • <u>For e:</u>

The chemical equation for the reaction of magnesium and oxygen gas follows:

Mg+O_2\rightarrow MgO

The product formed is magnesium oxide

3 0
3 years ago
Imagine two solutions with the same concentration and the same boiling point, but one has benzene as the solvent and the other h
11Alexandr11 [23.1K]

Complete question:

Imagine two solutions with the same concentration and the same boiling point, but one has benzene as the solvent and the other has carbon tetrachloride as the solvent. Determine that molal concentration, m (or b), and boiling point, Tb.

benzene boiling point=80.1 Kb=2.53

carbon tetrachloride boiling point=76.8 Kb=5.03

Answer:

m = 1.32 mol/kg

Boiling point: 83.4°C

Explanation:

When a nonvolatile solute is added to a pure solvent, the boiling point of the solvent increases, a phenomenon called ebullioscopy. This happens because of the interactions between the solute and the solvent. The temperature variation (new boiling point - normal boiling point) can be calculated by:

ΔT = m*Kb*i

Where m is the molal concentration (moles o solute/mass of solvent in kg), Kb is the ebullioscopy constant of the solvent, and i is the van't Hoff factor, which indicates how much of the solute dissociates. Let's assume that i is equal in both solvents and equal to 1 (the solvent dissociates completely)

Calling the new boiling point as Tb, for benzene:

Tb - 80.1 = m*2.53*1

Tb = 2.53m + 80.1

For carbon tetrachloride:

Tb - 76.8 = m*5.03*1

Tb = 5.03m + 76.8

Because Tb and m are equal for both:

5.03m + 76.8 = 2.53m + 80.1

2.5m = 3.3

m = 1.32 mol/kg

So, substituting m in any of the equations (choosing the first):

Tb = 2.53 * 1.32 + 80.1

Tb = 83.4°C

3 0
3 years ago
Read 2 more answers
A chemist determines by measurements that moles of nitrogen gas participate in a chemical reaction. Calculate the mass of nitrog
Likurg_2 [28]

Answer:

0.56 g

Explanation:

<em>A chemist determines by measurements that 0.020 moles of nitrogen gas participate in a chemical reaction. Calculate the mass of nitrogen gas that participates.</em>

Step 1: Given data

Moles of nitrogen gas (n): 0.020 mol

Step 2: Calculate the molar mass (M) of nitrogen gas

Molecular nitrogen is a gas formed by diatomic molecules, whose chemical formula is N₂. Its molar mass is:

M(N₂) = 2 × M(N) = 2 × 14.01 g/mol = 28.02 g/mol

Step 3: Calculate the mass (m) corresponding to 0 0.020 moles of nitrogen gas

We will use the following expression.

m = n × M

m = 0.020 mol × 28.02 g/mol

m = 0.56 g

4 0
3 years ago
What is the molar mass of an unknown gas with a density of 2.00 g/L at 1.00 atm and 25.0 °C?
soldier1979 [14.2K]

Answer:

Explanation:Explanation:

Your starting point here will be the ideal gas law equation

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

P

V

=

n

R

T

a

a

∣

∣

−−−−−−−−−−−−−−−

, where

P

- the pressure of the gas

V

- the volume it occupies

n

- the number of moles of gas

R

- the universal gas constant, usually given as

0.0821

atm

⋅

L

mol

⋅

K

T

- the absolute temperature of the gas

Now, you will have to manipulate this equation in order to find a relationship between the density of the gas,

ρ

, under those conditions for pressure and temperature, and its molar mass,

M

M

.

You know that the molar mass of a substance tells you the mass of exactly one mole of that substance. This means that for a given mass

m

of this gas, you can express its molar mass as the ratio between

m

and

n

, the number of moles it contains

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

M

M

=

m

n

a

a

∣

∣

−−−−−−−−−−−−−

(

1

)

Similarly, the density of the substance tells you the mass of exactly one unit of volume of that substance.

This means that for the mass

m

of this gas, you can express its density as the ratio between

m

and the volume it occupies

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

ρ

=

m

V

a

a

∣

∣

−−−−−−−−−−−

(

2

)

Plug equation

(

1

)

into the ideal gas law equation to get

P

V

=

m

M

M

⋅

R

T

Rearrange to get

P

V

⋅

M

M

=

m

⋅

R

T

P

⋅

M

M

=

m

V

⋅

R

T

M

M

=

m

V

⋅

R

T

P

Finally, use equation

(

2

)

to write

M

M

=

ρ

⋅

R

T

P

Convert the temperature of the gas from degrees Celsius to Kelvin then plug in your values to find

M

M

=

1.02

g

L

⋅

0.0821

atm

⋅

L

mol

⋅

K

⋅

(

273.15

+

37

)

K

0.990

atm

M

M

=

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

26.3 g mol

−

1

a

a

∣

∣

−−−−−−−−−−−−−−−−

I'll leave the answer rounded to three

7 0
3 years ago
How many moles of chloride ions are there in 2.5 L of 5 M magnesium chloride?
castortr0y [4]

Answer:

n_{Cl^-}=25molCl^-

Explanation:

Hello,

In this case, since the given 5-M concentration of magnesium chloride is expressed as:

5\frac{molMgCl_2}{L}

We can notice that one mole of salt contains two moles of chloride ions as the subscript of chlorine is two, in such a way, with the volume of solution we obtain the moles of chloride ions as shown below:

n_{Cl^-}=5\frac{molMgCl_2}{L}*\frac{2molCl^-}{1molMgCl_2} *2.5L\\\\n_{Cl^-}=25molCl^-

Best regards.

4 0
3 years ago
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