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Sergio [31]
3 years ago
12

HELP ASAP

Chemistry
1 answer:
WARRIOR [948]3 years ago
4 0
The snake,raccoon, and preying mantis
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What would ne the acceleration in a body moving wit uniform velocity and why​
meriva

Answer: The derivative of a constant term is always 0. So the acceleration of the body would be zero. Hence, the acceleration of a body moving with uniform velocity will always be zero.

Hope this helps :) :)

3 0
3 years ago
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What was thompson working with when he discovered the cathode rays?
Viefleur [7K]
He was working with electrons
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A particular voltaic cell operates with the reaction: Zn(s)+Cl2(g)--->Zn^2+(aq)+2Cl^-(aq) giving a cell potential of .853 V.
nadya68 [22]

Answer:

The amount of energy liberated will be 49.38 J.

Explanation:

The amount of energy liberated (gibbs free energy) can be calculated using the following equation:

ΔG° = -nFε

n: amount of moles of electrons transfered

F: Faraday's constant

ε: cell potential

20.0 g of Zn is equal to 0.30 mol.

Two electrons are transfered during the reaction.

Therefore, n = 2x0.30 ∴ n = 0.60

ΔG° = - 0.60 x 96.485 x 0.853

ΔG° = 49.38 J

3 0
3 years ago
How many oxygen atoms in 1 mole of CO2
Romashka-Z-Leto [24]
There are two oxygen atoms
8 0
3 years ago
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Use the information below to explain why the atomic radius decreases down a group.
notsponge [240]

Answer:

Detail is given below

Explanation:

Atomic radii trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As the size of atom increases the ionization energy from top to bottom also  decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus.

In A we can see that there is one positive charge and force of attraction is 2.30×10⁻⁸ N and distance is 0.10 nm

In B we can see that negative charge is further away from nucleus because of greater distance thus force of attraction will be less. 0.58×10⁻⁸ N

In C this distance further increases and force also goes in decreasing 0.26×10⁻⁸ N.

3 0
3 years ago
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