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erica [24]
3 years ago
12

we know if we apply the force along the left, friction acts along the right and vice versa.But what about an object which is not

moving but at rest on a surface?​
Physics
1 answer:
myrzilka [38]3 years ago
5 0

Answer:

<em>Explanation below</em>

Explanation:

<u>Friction Force</u>

When two objects are in contact, the surface between them exerts a frictional force as long as the objects are tried to move with respect to each other.

If a given external force is applied to the left, then a friction force appears to the right and opposes the intent to move them. A similar situation occurs when the force is applied to the right side.

If the object is at rest and no force is applied, no friction force appears.

It's important to recall that two different friction forces are defined: the kinetic friction force and the static friction force, usually the latter is greater than the first.

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3 years ago
calculate the mass of potassium chlorate (kcio3) required to obtain 10g of oxygen in the following reaction:kclO3-kcl+O2​
igor_vitrenko [27]

First, balance the reaction:

_ KClO₃   ==>   _ KCl + _ O₂

As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :

2 KClO₃   ==>   2 KCl + 3 O₂

Since we start with a known quantity of O₂, let's divide each coefficient by 3.

2/3 KClO₃   ==>   2/3 KCl + O₂

Next, look up the molar masses of each element involved:

• K: 39.0983 g/mol

• Cl: 35.453 g/mol

• O: 15.999 g/mol

Convert 10 g of O₂ to moles:

(10 g) / (31.998 g/mol) ≈ 0.31252 mol

The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need

(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃

KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of

(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g

of KClO₃.

7 0
3 years ago
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