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igor_vitrenko [27]
3 years ago
14

Does lithium or potassium have a strong nuclear pull

Chemistry
1 answer:
saveliy_v [14]3 years ago
7 0

Answer:

i believe it is lithium cause its is used for batteries and potassium is in like foods right? lmk if im wrong

and potassium helps to protect from radiation

Explanation:

You might be interested in
Cuando un gas que se encuentra a 20°c se calienta hasta los 40°c sin que varie su presion, su volumen se duplica
kati45 [8]

Answer:

Not doubled

Explanation:

The equation below represent the ideal gases relationship

PV ÷ T = constant

Here

P denotes pressure,

V denotes volume,

T denotes temperature in degrees Kelvin

Now

20 ° c = 273 + 20

= 293 K

And,

40 ° c = 313 K

So,

V = Vo. 313 K ÷  293 K = 1.07 Vo

So,  the volume is NOT doubled.

In the case when the temperature would be determined in degrees celsius at 0 degrees so the volume would be zero

4 0
3 years ago
How many grams are in 1.946 moles of nacl
Ilia_Sergeevich [38]

Answer:

113.8g

Explanation:

Statement of problem: mass of 1.946mole of NaCl

Given parameters:

Number of moles of NaCl = 1.946mole

Unknown: mass of NaCl

Solution

To find the mass of NaCl, we apply the concept of moles which expresses the relationship between number of moles and mass according to the equation below:

                        Number of moles = \frac{mass}{molar mass}

To find the molar mass of NaCl:

                         the atomic mass of Na = 23g

                                atomic mass of Cl = 35.5g

                Molar mass of NaCl = (23 + 35.5) = 58.5gmol⁻¹

Mass of NaCl = Number of moles x molar mass of NaCl

Mass of NaCl = 1.946 x 58.5 = 113.8g

7 0
3 years ago
A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o
Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

4 0
3 years ago
Four animals were fleeing from a wildfire. All of them were running at a speed of 30 miles per hour. Which one has the greatest
RUDIKE [14]

Answer: I don’t know lol

Explanation:  I am so sorry I thought this was easy

6 0
3 years ago
Read 2 more answers
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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