Answer : 17.12 g
Explanation:
= elevation in boiling point
= boiling point elevation constant
m= molality
given 
Molar mass of solute = 46.0 
Weight of the solvent = 150.0 g = 0.15 kg
Putting in the values
x = 17.12 g
Answer:
None of the alpha particles fired at the foil are being repelled back, like they were in the Rutherford atom simulation.I hope this is correct.
MAl₂(SO₄)₃·xH₂O:
(mAl×2) + (mS×3) + (mO×12) + (mH₂O×x)
(27×2)+(32×3)+(16×12)+(x×18) = 342 + 18x [g]
mAl₂: 27×2 = 54 [g]
54g ---------- 13,63%
342+18x ---- 100%
0,1363(342+18x) = 54
46,6146 + 2,4534x = 54
2,4534x = 7,3854
x ≈ 3
>>> Al₂(SO₄)₃·3H₂O <<<<
:)
When the specific heat capacity of the water is 4.18 J/g.°C so, we are going to use this formula to get the heat for cooling three phases changes from steam to liquid and from liquid to ice (solid) :
when Q = M*C*ΔT
Q is the heat in J
and M is the mass in gram = 1 mol H2O * 18 g/mol(molar mass) = 18 g
C is the specific heat J/g.°C
ΔT is the change in temperature
Q = Mw *[ ( Csteam * ΔTsteam)+(Cw*ΔTw) + (Cice * ΔT ice)]
= 18 g * [(2.01 * (155-100°C)) + (4.18 * (100-0°C)) + (2.09 * (0 - 55 °C))]
∴Q = 7444.8 J
and when we know that the heat of fusion for water = 334J/g
and heat of vaporization for water = 2260J/g
∴Q for the two phases changes = M * (2260+334)
= 18 * (2260+334)
= 46692 J
∴ Q total = 7444.8 + 46692 = 54136.8 J
Answer:
Single replacement reaction
Explanation:
