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melamori03 [73]
4 years ago
10

Calculate the net force of an object that is being moved 10 Newtons to the right and 18 Newtons to the left. Explain how the unb

alanced force will change the object’s speed and direction.
Chemistry
2 answers:
Pachacha [2.7K]4 years ago
5 0
If an object<span> has a net </span>force<span> acting on it, it </span>will<span> accelerate. The </span>object will speed<span> up, slow down or </span>change direction<span>. An </span>unbalanced force<span> (net </span>force) acting on anobject<span> changes its </span>speed<span> and/or </span>direction<span> of motion. An </span>unbalanced force<span> is an unopposed </span>force<span> that causes a </span>change<span> in motion.</span>
makvit [3.9K]4 years ago
5 0

If an object has a net force acting on it, it will accelerate. The object will speed up, slow down or change direction. An unbalanced force (net force) acting on an object changes its speed and/or direction of motion. An unbalanced force is an unopposed force that causes a change in motion.

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Phosphorus-32 has a half-life of 14.0 days. Starting with 6.00 g of 32P, how many grams will remain after 42.0 days ?
aleksandrvk [35]
6g...........3g..........1,5g........0,75g
0d..........14d.........28d.........42d

After 42 days 0,75g of Phoshorus-32 will remain.
4 0
3 years ago
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At constant pressure increasing the temperature of a fixed amount of gas will do what
Naddika [18.5K]

Answer:

For a fixed mass of gas at constant pressure, the volume is directly proportional to the kelvin temperature. That means, for example, that if you double the kelvin temperature from, say to 300 K to 600 K, at constant pressure, the volume of a fixed mass of the gas will double as well.

5 0
3 years ago
Is each of these statements true? If not, explain why.(p) A bimolecular reaction is generally twice as fast as a unimolecular re
Ede4ka [16]

A bimolecular reaction typically takes twice as long as a unimolecular reaction. The answer is true

<h3>How is activation energy related to temperature?</h3>
  • Consider what must occur for ClNO2 to react with NO to understand why reactions have an activation energy.
  • First and foremost, these two molecules must clash in order to organise the system.
  • Not only must they be brought together, but they must also be kept in precisely the appropriate orientation relative to one other in order for the reaction to occur.
  • Some energy must also be used in order to break the Cl-NO2 bond and allow the Cl-NO bond to form.

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5 0
2 years ago
The formation of CsCl from Cs(s) and Cl2 (g) involves the following steps:
scoray [572]

Explanation:

1) Cs(s)\rightarrow Cs(g)

Energy will be absorbed by the Cs(s) to get converted into Cs(g). In order to break the strong association of particles present in solid to convert it into gaseous state energy will be needed. So, energy will be absorbed by the Cs solid in this reaction.

2) \frac{1}{2}Cl_2(g)\rightarrow Cl(g)

Energy will be absorbed by Cl_2 gas in order to break bond between two Cl atoms to form isolated(alone) single chlorine atom.So, the energy will be absorbed by the chlorine gas molecule in this reaction.

3) Cs(g)\rightarrow Cs^+(g)+e^-

Energy will be absorbed by the Cs(g) to get converted into Cs^+(g) cation. In order to remove an electron from the outer most shell of Cs atom energy will be required by Cs atom, So, energy will be absorbed in this reaction.

4) Cl(g)+e^-\rightarrow Cl^-(g)

Energy will get release in formation of Cl^- anion from Cl atom in gaseous state. Chlorine atom need one electron to attain noble gas configuration. So, when an electron is added to the outer most shell of chlorine it attains stability of fully filled outermost shell by which it releases energy on addition of an electron.

5) Cs^+(g)+Cl^-(g)\rightarrow CsCl(s)

Energy will get release on formation of CsCl(s) fromCs^+ cation and Cl^- anion. since both are oppositely charged ion and due to strong electrostatic interaction will get converted into stable molecule of CsCl (s) with release in an energy.

6 0
3 years ago
Draw 2-methyl, 3- methyl, hex-1-ene
Dennis_Churaev [7]

Answer: I found the answer it is.

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