Answer:
-30 °C
Explanation:
First, we have to calculate the molality (m) of the solution. If the solution is 50% C₂H₆O₂ by mass. It means that in 100 g of solution, the are 50 g of solute (C₂H₆O₂) and 50 g of solvent (water).
The molar mass of C₂H₆O₂ is 62.07 g/mol. The moles of solute are:
50 g × (1 mol / 62.07 g) = 0.81 mol
The mass of the solvent is 50 g = 0.050 kg.
The molality is:
m = 0.81 mol / 0.050 kg = 16 m
The freezing-point depression (ΔT) can be calculated using the following expression.
ΔT = Kf × m = (1.86 °C/m) × 16 m = 30 °C
where,
Kf: freezing-point constant
The normal freezing point for water is 0°C. The freezing point of the radiator fluid is:
0°C - 30°C = -30 °C
Answer:
C) three
Explanation:
Let gram of gold required be m . Let temperature change in both be Δ t .
heat absorbed = mass x specific heat x change in temperature
for copper
heat absorbed = 1 x .385 x Δt
for gold
heat absorbed = m x .129 x Δt
So
m x .129 x Δt = 1 x .385 x Δt
m = 2.98
= 3 g approximately .
Answer:
-2.8 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity The S. I unit of acceleration is m/s²
Using the equation of motion,
v² = u² + 2as................... Equation 1
Where v = Final velocity, u = initial velocity, a = acceleration, s = distance,
Given: v = 6.0 m/s, u = 8.0 m/s, s = 5.0 m.
Substituting into equation 1
6² = 8²+2(a)5
36 = 64 + 10a
10a = 36-64
10a = -28
10a/10 = -28/10
a = -2.8 m/s²
Note: a is negative because because the skater decelerate on the rough ice
Hence the magnitude of her acceleration is = -2.8 m/s²
The best and most correct answer among the choices provided by the question is the first choice, larger.
Rankine is Fahrenheit + 460 , while Kelvin is Celsius + 273. We all know that Fahrenheit has larger number compared to kelvin , thus rankine is much larger.
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