Change occurs in the density of the gummy bear to change if we soaked it in isopropanol.
<h3>Effects on density by soaking isopropanol</h3>
We expect the density of the gummy bear to change if we soaked it in isopropanol because the gummy bear absorb the liquid which increase its density due to increasing weight of the gummy bear.
If a substance gains weight then its density also increases and we know that when the gummy bear was soaked in the isopropanol then its weight increases so its density is also changes so we can conclude that change occurs in the density of the gummy bear to change if we soaked it in isopropanol.
Learn more about density here: brainly.com/question/1354972
Answer:
a) 965,1 lbf
b) 4,5 kg
c) 1,33 * 10^6 dynes
Explanation:
Mass of an object refers to the amount of mattter it cotains, it can be expressed it gr, kg, lbm, ton, etc.
Weight of an object refers to a force, and is the measurement of the pull of gravitiy on an object. It may be definide as the mass times the acceleration of gravity.
w=mg
In Planet Earth, the nominal "average" value for gravity is 9,8 m/s² (in the International System) or 32,17 ft/s² (in the FPS system).
To solve this problem we'll use the following conversion factors:
1 lbf = 1 lbm*ft/s²
1 N = 1 kg*m/s²
1 dyne = 1 gr*cm/s² and 1 N =10^5 dynes
1 ton = 907,18 kg
1 k = 1000 gr
a) m = 30 lbm

b) w = 44 N
First, we clear m of the weight equation and then we replace our data.

c) m = 15 ton
Answer:
the final volume of the gas is
= 1311.5 mL
Explanation:
Given that:
a sample gas has an initial volume of 61.5 mL
The workdone = 130.1 J
Pressure = 783 torr
The objective is to determine the final volume of the gas.
Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.
Converting the external pressure to atm ; we have
External Pressure
:


The workdone W =
V
The change in volume ΔV= 
ΔV = 
ΔV = 
ΔV = 1.25 L
ΔV = 1250 mL
Recall that the initial volume = 61.5 mL
The change in volume V is 

multiply through by (-), we have:

= 1250 mL + 61.5 mL
= 1311.5 mL
∴ the final volume of the gas is
= 1311.5 mL