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trapecia [35]
3 years ago
13

A 5.0-kg box is pushed across the floor with a force of 32 N. What is the frictional force on the box if the box's acceleration

is 3.6 m/s2?
Chemistry
2 answers:
brilliants [131]3 years ago
4 0
In order to accelerate 5.0 kg of mass at the rate of 3.6 m/s², the force required is

F = m a = (5) (3.6) = 18 newtons.

The force applied is 32 newtons.  So 18 of them are going to accelerate the box,
and friction is eating up the other 14 newtons by acting in the opposite direction.
Slav-nsk [51]3 years ago
3 0
From\ 2nd\ Newton\ Law\ resultuant\ force: \\\\\ F_R=m*a=5kg*3,6\frac{m}{s^2}=18N\\\\
F=32N\\\\
F_f- friction\ force\\\\
F_R=F-F_f\\\\
F_f=F-F_R=32N-18N=14N\\\\Friction\ force\ is\ equal\ to\ 14\ N.
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If 500.0 ml of 0.10 m ca2+ is mixed with 500.0 ml of 0.10 m so42−, what mass of calcium sulfate will precipitate? ksp for caso4
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Answer : The mass of calcium sulfate precipitate will be, 6.12 grams

Solution :

First we have to calculate the moles of Ca^{2+} and SO_4^{2-}.

\text{Moles of }Ca^{2+}=\text{Molarity of }Ca^{2+}\times \text{Volume of }Ca^{2+}=0.10mole/L\times 0.5L=0.05\text{ moles}

\text{Moles of }SO_4^{2-}=\text{Molarity of }SO_4^{2-}\times \text{Volume of }SO_4^{2-}=0.10mole/L\times 0.5L=0.05\text{ moles}

As, 0.05 moles of Ca^{2+} is mixed with 0.05 moles of  SO_4^{2-}, it gives 0.05 moles of calcium sulfate.

Now we have to calculate the solubility of calcium sulfate.

The balanced equilibrium reaction will be,

CaSO_4\rightleftharpoons Ca^{2+}+SO_4^{2-}

The expression for solubility constant for this reaction will be,

K_{sp}=(s)\times (s)

K_{sp}=(s)^2

Now put the value of K_{sp} in this expression, we get the solubility of calcium sulfate.

2.40\times 10^{-5}=(s)^2

s=4.89\times 10^{-3}M

Now we have to calculate the moles of dissolved calcium sulfate in one liter solution.

\text{Moles of }CaSO_4=\text{Molarity of }CaSO_4\times \text{Volume of }CaSO_4=4.89\times 10^{-3}mole/L\times 1L=4.89\times 10^{-3}\text{ moles}

Now we have to calculate the moles of calcium sulfate that precipitated.

\text{Moles of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ present}-\text{Moles of }CaSO_4\text{ dissolved}

\text{Moles of }CaSO_4\text{ precipitated}=0.05-4.89\times 10^{-3}=0.045\text{ moles}

Now we have to calculate the mass of calcium sulfate that precipitated.

\text{Mass of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ precipitated}\times \text{Molar mass of }CaSO_4

\text{Mass of }CaSO_4\text{ precipitated}=0.045moles\times 136g/mole=6.12g

Therefore, the mass of calcium sulfate precipitate will be, 6.12 grams

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