Answer:
A
Explanation:
An antifreeze is a substance which when added to a liquid reduces the freezing point of the liquid. Hence the solution can only freeze at a lower temperature compared to the solvent. Examples of antifreeze includes salt and propylene glycol. Antifreeze are used in low temperature environments to prevent water from freezing quicky. Hence the answer is A.
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Answer: 585 J
Explanation:
We can calculate the work done during segment A by using the work-energy theorem, which states that the work done is equal to the gain in kinetic energy of the object:

where Kf is the final kinetic energy and Ki the initial kinetic energy. The initial kinetic energy is zero (because the initial velocity is 0), while the final kinetic energy is

The mass is m=1.3 kg, while the final velocity is v=30 m/s, so the work done is:

Answer:
2 electrons will be needed by unbound oxygen in order to fill its 2nd shell.
Explanation:
The chemical reaction between magnesium and oxygen gives magnesium oxide as a product.The reaction is chemically represented as:

Magnesium is a metal of group-2 with 2 valence electrons.It has atomic number of 12.
![[Mg]=1s^22s^22p^63s^2](https://tex.z-dn.net/?f=%5BMg%5D%3D1s%5E22s%5E22p%5E63s%5E2)
In order to attain noble gas configuration it will loose two electrons.
![[Mg]^{2+}=1s^22s^22p^6](https://tex.z-dn.net/?f=%5BMg%5D%5E%7B2%2B%7D%3D1s%5E22s%5E22p%5E6)
...[1]
Oxygen is a non metal of group-16 with 6 valence electrons..It has atomic number of 8.
![[O]=1s^22s^22p^4](https://tex.z-dn.net/?f=%5BO%5D%3D1s%5E22s%5E22p%5E4)
In order to attain noble gas configuration it will gain two electrons.
![[O]^{2-}=1s^22s^22p^6](https://tex.z-dn.net/?f=%5BO%5D%5E%7B2-%7D%3D1s%5E22s%5E22p%5E6)
..[2]
2 electrons will be needed by unbound oxygen in order to fill its 2nd shell.
The electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.
To determine σ:
σ = Q/A
Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:
σ = Q/d²
Make this substitution in the equation for E:
E = Q/(2ε₀d²)
We see that E is inversely proportional to the square of d:
E ∝ 1/d²
The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude:
