Answer:
120.0 mL.
Explanation:
- As it is known that the no. of millimoles of a solution before dilution is equal to the no. of millimoles after dilution.
We suppose that the initial W% of methanol is 100.0 %
<em>∴ (W%V) before dilution = (W%V) after dilution.</em>
W% before dilution = 100.0 %, V before dilution = 18.0 mL.
W% after dilution = 15.0 %, V after dilution = ??? mL.
<em>∴ V after dilution = (W%V) before dilution/W% after dilution = </em>(100.0 %)(18.0 mL)/(15.0%)<em> = 120.0 mL.</em>
1. U₂₃₈→α→Th₂₃₄(UX₁)
<span>2. Th₂₂₈→α→Ra₂₂₆(MsTh₁) </span>
<span>α = Alpha decay (release of He Nucleus) </span>
<span>The decay products are meso states that undergo further (β) decay</span>
Answer:
670.68°C
Explanation:
Given that:
volume of water = 50 ml but 1 g = 1 ml. Therefore the mass of water (m) = 50 ml × 1 g / ml = 50 g
specific heat (C) = 4.184 J/g˚C
Initial temperature = 20°C, final temperature = 22°C. Therefore the temperature change ΔT = final temperature - initial temperature = 22 - 20 = 2°C
The quantity of heat (Q) used to raise the temperature of a body is given by the equation:
Q = mCΔT
Substituting values:
Q = 50 g × 4.184 J/g˚C × 2°C = 418.4 J
Since the mass of lead = 5 g and specific heat = 0.129 J/g˚C. The heat used to raise the temperature of water is the same heat used to raise the temperature of lead.
-Q = mCΔT
-418.4 J = 5 g × 0.129 J/g˚C × ΔT
ΔT = -418.4 J / ( 5 g × 0.129 J/g˚C) = -648 .68°C
temperature change ΔT = final temperature - initial temperature
- 648 .68°C = 22°C - Initial Temperature
Initial Temperature = 22 + 648.68 = 670.68°C
A solid.
As you slowly go from the solid state to the gas, the molecules within spread farther and farther away.
