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ra1l [238]
2 years ago
10

A 250. cm3 sample of neon is collected at standard temperature and pressure. Assuming the volume remains constant, what would be

the pressure of the neon at 44.0°C?
a) 118 b)87.2 c) 0.00 d)290
Chemistry
1 answer:
Olenka [21]2 years ago
3 0

Answer:

The answer is 118 kPa.

Explanation:

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A balloon is filled with 12 L of air at a pressure of 2 atm. What is the volume of the balloon if the pressure is changed to 3 a
mihalych1998 [28]

Answer:

8L

Explanation:

Using Boyle's law which states that the volume of a given mass of gas is inversely proportional to the pressure, provided temperature remains constant

P1V1= P2V2

P1 =  2atm, V1 = 12L ,

P2 = 3atm , V2 =

12 × 2 = V2 × 3

Divide both sides by 3

V2 = 24 ÷ 3

V2 = 8L

I hope this was helpful, please mark as brainliest

3 0
3 years ago
Metallic crystals are excellent conductors of electricity due to the existence of positive and negative ions.
My name is Ann [436]

the answer is FALSE


4 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
6.02 kJ>mol. When a small ice cube at -10°C is put into a cup of water at room temperature, which of the following plays a gr
9966 [12]

Answer:

Heat transfer during melting of ice plays greater role in cooling of liquid water.

Explanation:

Temperature of ice = -10 °c

Temperature of water = 0 °c

When ice cube is dipped in to the water.the heat transfer

Q = m c ΔT

⇒ Q = 1 × 2.01 × 10

⇒ Q = 20.1 KJ

Heat transfer during melting of ice Q_{melt} = latent heat of ice

Latent heat of ice = 334 KJ

⇒ Q_{melt} = 334 KJ

Heat transfer during melting of ice is greater value than heat transfer during warming of ice from -10°C to 0°C.

Thus heat transfer during melting of ice plays greater role in cooling of liquid water.

8 0
3 years ago
Find the molar mass of s6
Mrrafil [7]

Moles: 1

Weight, g : 192.3900

8 0
3 years ago
Read 2 more answers
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