Answer:
Protons: 2.
Electrons: 2.
Neutrons: 2.
Explanation:
Hello,
In this case, since an atom's atomic number is equal to the number of electrons, considering the electron configurations, taking into account that helium-4 is neither positively nor negatively charged, we can infer that the number of electrons equal the number of protons, which in this case are 2, due to the fact that is atomic number is 2.
Moreover, as helium-4's atomic number is 4 as a whole number, we compute the number of neutrons by using the shown below equation:

Regards.
Answer:
76.5g KCl/74.55 grams per mole Kcl = x
molality= x/.085 kg H2O
Explanation:
well remember molality is moles of solute/kilograms of solvent. So it's the moles of KCl over 85 g of h20 converted into kg. if this makes sense.
<u>Answer:</u> The specific heat of metal is 0.821 J/g°C
<u>Explanation:</u>
When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

The equation used to calculate heat released or absorbed follows:

......(1)
where,
q = heat absorbed or released
= mass of metal = 30 g
= mass of water = 100 g
= final temperature = 25°C
= initial temperature of metal = 110°C
= initial temperature of water = 20.0°C
= specific heat of metal = ?
= specific heat of water = 4.186 J/g°C
Putting values in equation 1, we get:
![30\times c_1\times (25-110)=-[100\times 4.186\times (25-20)]](https://tex.z-dn.net/?f=30%5Ctimes%20c_1%5Ctimes%20%2825-110%29%3D-%5B100%5Ctimes%204.186%5Ctimes%20%2825-20%29%5D)

Hence, the specific heat of metal is 0.821 J/g°C
Answer:
3.064 $
Solution:
First of all we will find the number of pounds contained by one Kg. Therefore, it is found that,
1 Kilogram = 2.20462 Pounds
Hence, as given,
1 Pound of Margarine costs = 1.39 Dollars
So,
2.20462 Pounds Margarine will cost = X Dollars
Solving for X,
X = (2.20462 Pounds × 1.39 Dollars) ÷ 1 Pound
X = 3.064 Dollars
A sodium chloride is like most of the ionic compounds
existing here on earth in which they are composed of having a high melting
point and by this, if found in underground rock deposits, they are usually in a
form of solid.