Ethylene glycol is termed as the primary ingredients in antifreeze.
The ethylene glycol molecular formula is C₂H₆O₂.
Molar mass of C₂H₆O₂ is = (2×12) +(6×1) + (216) = 62g/mol
Now that antifreeze by mass is 50%, then there is 1kg of ethylene glycol which is present in 1kg of water.
ΔTf = Kf×m
ΔTf = depression in the freezing point.
= freezing point of water freezing point of the solution
= O°c - Tf
= -Tf
Kf = depression in freezing constant of water = 1.86°C/m
M is the molarity of the solution.
=(mass/molar mass) mass of solvent in kg
=1000g/62 (g/mol) /1kg
=16.13m
If we plug the value we get
-Tf = 1.86 × 16.13 = 30
Tf = -30°c
The answer is going to be B
Answer:
Physical Properties of Sodium
Atomic number 11
Melting point 97.82°C (208.1°F)
Boiling point 881.4°C (1618°F)
Volume increase on melting 2.70%
Latent heat of fusion 27.0 cal/g
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Home Periodic table Elements Sodium
Sodium - Na
Chemical properties of sodium - Health effects of sodium - Environmental effects of sodium
Atomic number
11
Atomic mass
22.98977 g.mol -1
Electronegativity according to Pauling
0.9
Density
0.97 g.cm -3 at 20 °C
Melting point
97.5 °C
Boiling point
883 °C
Vanderwaals radius
0.196 nm
Ionic radius
0.095 (+1) nm
Isotopes
3
Electronic shell
[Ne] 3s1
Energy of first ionisation
495.7 kJ.mol -1
Answer:
T₂ = 95.56°C
Explanation:
The final resistance of a material after being heated is given by the relation:
R' = R(1 + αΔT)
where,
R' = Final Resistance = 207.4 Ω
R = Initial Resistance = 154.9 Ω
α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹
ΔT = Change in Temperature = ?
Therefore,
207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]
207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT
1.34 - 1 = (0.0045°C⁻¹)ΔT
ΔT = 0.34/0.0045°C⁻¹
ΔT = 75.56°C
but,
ΔT = Final Temperature - Initial Temperature
ΔT = T₂ - T₁ = T₂ - 20°C
T₂ - 20°C = 75.56°C
T₂ = 75.56°C + 20°C
<u>T₂ = 95.56°C</u>
Answer:
c.) 25 N
Explanation:
We find the volume of the brick, knowing that the volume of a cube is given by the formula:
being l the side of the cube, which in this case is 10 cm or 0,1 m. Now we find the mass of the object, knowing the density and the Volume of the cube:
We find the weight by multiplying the mass of the object with the gravity constant.
