Answer:
300 gr.
Explanation:
if 204gr/100gr=solubility, then according to the condition solubility=612/x.
It means, 204/100=612/x, where x=612*100/204=300 gr.
There are two oxygen atoms in an oxygen molecule. Denoted by the 2 following O
Answer:
Option D
Explanation:
Rutherford deduced that the atomic nucleus was positively charged because the alpha particles that he fired at the metal foils were positively charged, and like charges repel. Alpha particles consist of two protons and two neutrons, so they are positively charged. In Rutherford's experiments most of the alpha particles passed straight through the foil without being deflected. However, occasionally the alpha particles were deflected in their paths, and rarely the alpha particles were deflected backward at a 180 degree angle.
Since like charges repel, Rutherford concluded that the cause of the deflections of the positively charged alpha particles had to be something within the atom that was also positively charged. Rutherford concluded from his metal foil experiments that most of an atom is empty space with a tiny, dense, positively charged nucleus at the center that contains most of the mass of the atom.
Answer : The boiling point of water increases, 
Solution : Given,
Moles of solute (sugar) = 4 moles
Mass of solvent (water) = 1 Kg
i = 1 for sugar
Formula used :
Where,
= elevation in boiling point
= elevation constant
m = molality
= moles of solute (sugar)
= mass of solvent (water)
i = van't Hoff factor
Now put all the given values in this formula, we get the elevation in boiling point of water.
Therefore, the elevation in boiling point of water is
Answer:
241.1 mL.
Explanation:
- From the general law of ideal gases:
<em>PV = nRT.</em>
where, P is the pressure of the gas.
V is the volume of the container.
n is the no. of moles of the gas.
R is the general gas constant.
T is the temperature of the gas (K).
- For the same no. of moles of the gas at two different (P, V, and T):
<em>P₁V₁/T₁ = P₂V₂/T₂.</em>
<em></em>
- P₁ = 101.3 kPa = 1.0 atm, V₁ = 260.0 mL, T₁ = 21°C + 273 = 294.0 K.
- P₂ = 1.0 atm (standard P), V₂ = ??? mL, T₂ = 0.0°C + 273 = 273.0 K (standard T).
<em>∴ V₂ = (P₁V₁T₂)/(T₁P₂)</em> = (1.0 atm)(260.0 mL)(273.0 K)/(294.0 K)(1.0 atm) = <em>241.1 mL.</em>
