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slava [35]
3 years ago
13

Escoge cualquier elemento químico de la tabla periódica perteneciente al periodo 3 y Determina cada uno de sus cuatro números cu

ánticos para cada electrón.
Chemistry
1 answer:
Tju [1.3M]3 years ago
4 0

Answer:

No hoblas espanol :(

Explanation:

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Air bags are activated when a severe impact causes a steel ball to compress a spring and electrically ignite a detonator cap. Th
Zarrin [17]

Answer: 136 g of  NaN_3 must be reacted to inflate an air bag to 70.6 L at STP.

Explanation:

Using ideal gas equation:

PV=nRT

P= pressure of nitrogen gas = 1 atm (at STP)

V =volume of nitrogen gas = 70.6 L

n = number of moles of nitrogen gas = 1 atm (at STP)

R = gas constant = 0.0821 Latm/Kmol

T = temperature of nitrogen gas = 273 K (at STP)

1atm\times 70.6L=n\times 0.0821 L atm/K mol\times 273

n=3.14

For the balanced chemical reaction:

2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)

3 moles of nitrogen are produced by = 2 moles of NaN_3

3.14 moles of nitrogen are produced by =\frac{2}{3}\times 3.14=2.09 moles of NaN_3

Mass of NaN_3 = Moles × Molar mass = 2.09 mole × 65 g/mol = 136 g

7 0
4 years ago
Which of the following is an acid? HF KOH LiF NH3
seraphim [82]

The compound HF is an acid. This is because it dissociates in water to give hydrogen ions (H⁺) as the only positively charged ions.

The others dissociate as follows: NH₃  reacts with water to form an alkaline solution NH₄OH.

LiF dissolves to form an ionic solution

KOH dissociates in water to form an alkaline solution.

3 0
3 years ago
Read 2 more answers
Lab: Applications of electromagnetic​
Pavlova-9 [17]

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.

7 0
2 years ago
Are blue stars young or old? How can you tell?.
andriy [413]
Astronomers measure the brightness of stars using light-years. This means that the light we see now left in that years ago, traveling through space in that distance at kilometer per second.

A star is born, it radiates energy for a long time, toward the end it expands, it may or may not explode, and then it dies. It vary in sizes, masses and surface temperature range. The colors of stars reflect their surface temperature. Their relative brightness is expressed in a scale of six magnitudes. The brightest the stars are first-magnitude stars, while the dimmest are sixth-magnitude stars.
 
The lower the number, the brighter the star.

For example:
                      Star color                        Surface Temperature
                     Blue-white                         around 25 000 K & higher
                     white                                  around 10 000 K
                     yellow                                around 7 000 K
                     Red                                    around 5 000 K & lower

4 0
3 years ago
Butane (C4H10) has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 ∘C. A 250 mL sealed flask contains
erastova [34]

Given that:

  • The heat of vaporization = 22.44 kJ/mol = 22440 J/mol
  • normal boiling point which is the initial temperature = 0.4° C = (273 + (-0.4))K  = 272.6 K
  • volume  = 250 mL = 0.250 L
  • Mass of butane = 0.8 g
  • the final temperature = -22° C = (273 + (-22)) K = 251 K

The first step is to determine the vapor pressure at the final temperature of 251K by using the Clausius-Clapeyron equation. This is following by using the ideal gas equation to determine the numbers of moles of butane gas. After that, the mass of butane present in the liquid is determined by using the relation for the number of moles.

Using Clausius-Clapeyron Equation:

\mathbf{In (\dfrac{P_2}{P_1} )= -\dfrac{\Delta H_{vap}}{R}(\dfrac{1}{T_2} - \dfrac{1}{T_1})}

where;

P1 and P2 correspond to the temperature at T1 and T2.

∴

replacing the values into the given equation, we have;

\mathbf{In \dfrac{P_2}{1\  atm} = -\dfrac{22440 \ J/mol}{8.314 \ J/mol.K}(\dfrac{1}{251 \ K} - \dfrac{1}{272.6 \ K})}

\mathbf{In \dfrac{P_2}{1\  atm} =-(0.852053785)}

\mathbf{P_2=0.427 \ atm}

As such, at -22° C; the vapor pressure = 0.427 atm

Now, using the ideal gas equation:

PV = nRT

where:

  • P = Pressure
  • V = volume
  • n = number of moles of butane
  • R = universal gas constant
  • T = temperature

∴

Making (n) the subject of the formula:

\mathbf{n = \dfrac{PV}{RT}}

\mathbf{n = \dfrac{0.427 atm \times 0.250 L}{(0.08206 \ L.atm/k.mol) \times 251}}

\mathbf{n =0.00518 mol}

We all know that the standard molecular weight of butane = 58.12 g/mol

∴

Using the relation for the number of moles which is:

\mathbf{number \  of \  moles = \dfrac{mass}{molar mass}}

mass = 0.00518 mole × 58.12 g/mol

mass = 0.301 g

∴

The mass of butane in the flask = 0.301 g

But the mass of the butane present as a liquid in the flask is

= 0.8 g - 0.301 g

= 0.499 g

In conclusion, the mass of the butane present as a liquid in the flask is 0.499 g

Learn more about vapourization here:

brainly.com/question/17039550?referrer=searchResults

7 0
3 years ago
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