Answer:
178.67K
Explanation:
PV=nRT
T=PV/nR
= 1.072atm*20L/1.485mol*0.0821LatmK^-1
=178.67K
V(NaOH)=15 mL =0.015 L
C(NaOH)=0.1 mol/L
C(H₂SO₄)=0.05 mol/L
2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O
n(NaOH)=V(NaOH)C(NaOH)=2n(H₂SO₄)
n(H₂SO₄)=V(H₂SO₄)C(H₂SO₄)
V(NaOH)C(NaOH)=2V(H₂SO₄)C(H₂SO₄)
V(H₂SO₄)=V(NaOH)C(NaOH)/{2C(H₂SO₄)}
V(H₂SO₄)=0.015*0.1/{2*0.05}=0.015 L = 15 mL
40% solution of glucose is where the solution contains, by weight, 40% glucose and 60% water.
Therefore, if the total weight of the solution is 250 g,
mass of the glucose (C6H12O6) = 250 g * 40% = 100 g
mass of water (H2O) = 250 g * 60% = 150 g
Mass of water can also be calculated by subtracting the weight of glucose from the total weight of the solution:
mass of water = 250g-100g = 150g.
Answer:
litre.50000665432158900643212lo
When heat energy is supplied to a material it can raise the temperature of mass of the material.
Specific heat is the amount of energy required by 1 g of material to raise the temperature by 1 °C.
equation is
H = mcΔt
H - heat energy
m - mass of material
c - specific heat of the material
Δt - change in temperature
substituting the values in the equation
120 J = 10 g x c x 5 °C
c = 2.4 Jg⁻¹°C⁻¹
