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:
volume is 7.0 liters
Explanation:
We are given;
- Molarity of the aqueous solution as 2.0 M
- Moles of the solute, K₂S as 14 moles
We are required to determine the volume of the solution;
We need to know that;
Molarity = Moles ÷ volume
Therefore;
Volume = Moles ÷ Molarity
Thus;
Volume of the solution = 14 moles ÷ 2.0 M
= 7.0 L
Hence, the volume of the molar solution is 7.0 L
Answer:
Both are the indicators which helps to identity the acid and base
Answer:
[KOH] = 0.10M in KOH
Explanation:
Molar Concentration [M] = moles solute/volume solution in liters
moles KOH = 0.56g/56g/mole = 0.01mole
Volume of solution = 100cm³ = 100ml = 0.10 liter
[KOH] = 0.01 mole KOH / 0.10 liter solution = 0.10M in KOH
