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3 years ago

Which part of an atom is gained, lost, or shared during a chemical reaction? Electron

Physics

2 answers:

slamgirl [31]3 years ago

8 0

Answer:

Electrons

Explanation:

when they are trying to balanced each other electrons is what change

siniylev [52]3 years ago

5 0

Answer:

The gain,loss, or sharing of electrons occurs in every chemical bond.

Explanation:

Should electrons be lost by one particle, and gained by another particle, the now charged particles can be chemically bound by an ionic bond.

Should a pair of electrons form a region of high electron density between two positively charged atomic nuclei, a bond can occur in that mutual attraction of the positively charged nuclei towards the region of high electron density operates. This is a long-winded description of a covalent bond.

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A 5,000 kg truck moving at 8 m/s has the same momentum as a 2,500 kg car. What is the velocity of

a_sh-v [17]

Answer:

16m/s

Explanation:

Mv=mv

5000x8=2500x v

V=5000x8/2500

V=40000/2500

= 16m/s

8 0

3 years ago

When the daughter nucleus produced in a radioactive decay is itself unstable, it will eventually decay and form its own daughter

Dmitrij [34]

Answer:

4.981 MeV

Explanation:

The quantity of energy Q can be calculated using the formula

Q = (mass before - mass after) × c²

Atomic Mass of thorium = 232.038054 u, atomic of Radium = 228.0301069 u and mass of Helium = 4.00260. The difference of atomic number and atomic mass between the thorium and radium ( 232 - 228) and ( 90 - 88) show α particle was emitted.

1 u = 931.494 Mev/c²

Q = (mass before - mass after) × c²

Q = ( mass of thorium - ( mass of Radium + mass of Helium ) )× c²

Q = 232.038054 u - ( 228.0301069 + 4.00260) × c²

Q = 0.0053471 u × c²

replace 1 u = 931.494 MeV/ c²

Q = 0.0053471 × c² × (931.494 MeV / c²)

cancel c² from the equation

Q = 0.0053471 × 931.494 MeV = 4.981 MeV

3 0

3 years ago

Is the SI unit of work newton?

Naya [18.7K]

Answer:

Newton is the SI unit for force . Newton is kg m2

7 0

4 years ago

Read 2 more answers

A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attached to the ceiling of an elevator.(a) What does t

tresset_1 [31]

Answer:

Part a)

$Reading = 2.00 kg$

Part b)

$Reading = 2.00 kg$

Part c)

$Reading = 4.04 kg$

Part d)

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

Explanation:

Part a)

When elevator is ascending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part b)

When elevator is decending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

$F_{net} = ma$

$T - mg = ma$

$T = mg + ma$

Reading is given as

$Reading = \frac{mg + ma}{g}$

$Reading = 2.00\frac{9.81 + 10}{9.81}$

$Reading = 4.04 kg$

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

3 0

3 years ago

Mass m moves to the right with speed =v along a frictionless horizontal surface and crashes into an equal mass m initially at re

Amiraneli [1.4K]

After the collision the magnitude of the momentum of the system is Mv

Given:

mass of 1st object = M

speed of 1st object = v

mass of 2nd object = M

speed of 2nd object = 0

To Find:

magnitude of the momentum after collision

Solution: Product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.

Applying conservation of linear momentum

Mv + M(0) = 2MV

Mv = 2MV

V = v/2

So, after collision momentum is

p = 2MV = 2xMxv/2 = Mv

So, after collision momentum is Mv

Learn more about Momentum here:

brainly.com/question/1042017

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4 0

2 years ago