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Vitek1552 [10]
4 years ago
11

jake pushes a heavy box across a carpeted floor.Name three forces that act on the box, and explain how you know these are unbale

nced
Chemistry
1 answer:
Fantom [35]4 years ago
6 0
I actually learned about this stuff about 3 years ago and this is what I can tell you what I remember and hopefully it helps in a way...
[Newton’s law of motion]
There’s unbalanced forces, friction, and balanced forces.
Unbalanced forces is where the object (in this case the box) is moving in one direction because one of the forces is applying more pressure than the other
Friction is where the box would be harder to push but if it’s still an unbalanced force, it will still move in one direction. The surfaces of the ground and or object isn’t smooth. Therefore creating difficulty when moving it.
A Balanced force will not move the object in any direction. This is because there is an equal amount of force being applied to all opposing sides. The net force will always equal 0 when there’s a balanced force!

-HOPE THAT HELPED IN A WAY! <3
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What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 (formula weight = 164 g/mol) in 115 g of
Citrus2011 [14]

Answer:

freezing point   (°C) of the solution =  - 3.34° C

Explanation:

From the given information:

The freezing point (°C) of a solution can be prepared by using the formula:

\Delta T = iK_fm

where;

i = vant Hoff factor

the vant Hoff factor is the totality of the number of ions in the solution

Since there are 1 calcium ion and 2 nitrate ions present in Ca(NO3)2, the vant Hoff factor = 3  

K_f = 1.86 °C/m

m = molality of the solution and it can be determined by using the formula

molality = \dfrac{mole \ of \ solute }{kg \ of \ solvent }

which can now be re-written as :

molality = \dfrac{mole \ of \ Ca(NO_3)_2}{kg \ of \  water}

molality = \dfrac{\dfrac{mass \ of \  \ Ca(NO_3)_2}{molar \  mass of \ Ca(NO_3)_2} }{kg \ of \  water}

molality = \dfrac{\dfrac{11.3 \ g }{164 \ g/mol} }{0.115 \ kg }

molality = 0.599 m

∴

The freezing point (°C) of a solution can be prepared by using the formula:

\Delta T = iK_fm

\Delta T =3 \times (1.86 \ ^0C/m) \times (0.599 \ m)

\Delta T =3.34^0 \ C

\Delta T = the freezing point of water - freezing point of the solution

3.34° C = 0° C -  freezing point of the solution

freezing point  (°C) of the solution =  0° C - 3.34° C

freezing point   (°C) of the solution =  - 3.34° C

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