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irina1246 [14]
3 years ago
14

How does the electric force vary with distance

Chemistry
1 answer:
Yanka [14]3 years ago
5 0

Answer:

increasing the separation distance between objects decreases the force of attraction

Explanation:

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What gas is used in freezers and fridges? ​
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Answer:

Chloro-Flouro-Carbon

Explanation:

Chloro-Flouro-Carbon (CFC Gas) is used as a refrigerant in fridges and freezers.

7 0
3 years ago
The atomic number of an element:
Aleonysh [2.5K]

Answer:

I think #3 might be the answer .

Because in #1 atomic mass tells mass .

#4 Atomic number doesnot only tell about the electrons atom has

6 0
3 years ago
Which of the following represents a physical change?
julsineya [31]
A. Phase changing. When phase changes nothing chemically changes about the substance, its still the same thing.
8 0
3 years ago
Read 2 more answers
What is hydrogen bonding
yKpoI14uk [10]

Answer:

a weak bond between two molecules resulting from an electrostatic attraction between a proton in one molecule and an electronegative atom in the other.

Explanation:

For example, in water molecules (H2O), hydrogen is covalently bonded to the more electronegative oxygen atom. Therefore, hydrogen bonding arises in water molecules due to the dipole-dipole interactions between the hydrogen atom of one water molecule and the oxygen atom of another H2O molecule.

4 0
3 years ago
The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W
Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹  

Explanation:

Assume the rate law is  

rate = k[A][B]²

If you are comparing two rates,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}

You are cutting each concentration in half, so

\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}

Then,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}

8 0
3 years ago
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