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maxonik [38]
3 years ago
14

Use the image above and describe the

Chemistry
1 answer:
Lina20 [59]3 years ago
4 0

Answer:

If you see in the image above, there is an unbalance force applied while playing tug of war. Since it is 1 vs 2, there is a greater net force in the right side then the left side. If it was 2 vs 2 or 1 vs 1, then they are appling balance force. You can also see in the picture that the arrows are pointing outwards (--->) rather then inwards (<---) because you are pulling the rope not pushing the rope. If you add one person on the left side, then the newtons which is 20N will become to 35N and will be balanced, but since there in only 1 person, there is less force on the left side, the newtons gets subtracted having only 20N. Since you are pulling the rope, the friction is opposite (<---). Since you are pulling the rope, you are using Kinetic force and the rope stays in potential force since it stays constant.

Hope this helps, thank you :) and I am not sure about magnitude I think you can that since there is greater force on the right side, there is more magnitude there.

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What mass of sucrose (C12H22O11) should be combined with 546 g of water to make a solution with an osmotic pressure of 8.80 atm
lesya [120]

<u>Answer:</u> The mass of sucrose required is 69.08 g

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = 8.80 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (sucrose) = ?

Molar mass of sucrose = 342.3 g/mol

Volume of solution = 564 mL    (Density of water = 1 g/mL)

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

T = Temperature of the solution = 290 K

Putting values in above equation, we get:

8.80atm=1\times \frac{\text{Mass of sucrose}\times 1000}{342.3\times 546}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 290K\\\\\text{Mass of sucrose}=\frac{8.80\times 342.3\times 546}{1\times 1000\times 0.0821\times 290}=69.08g

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5 0
3 years ago
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Digiron [165]

Answer:

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