Answer:
CS₂
Explanation:
To find the empirical formula we need to determine first the percentage of each atom in the molecule. Then, we need to find the moles and, as empirical formula is the simplest whole-number ratio of atoms we can solve the empirical formula:
<em>%C:</em>
0.33gC / 2.12g * 100 = 15.6%
<em>%S:</em>
1.53g S / 1.82g * 100 = 84.1%
In a basis of 100, the moles of each atom are:
<em>C:</em>
15.6g C * (1mol / 12.01g) = 1.30 moles
<em>S:</em>
84.1g S * (1mol / 32.065g) = 2.62 moles
The ratio of Sulphur-Carbon is:
2.62mol / 1.30mol = 2
That means empirical formula is:
<h3>CS₂</h3>
The effective nuclear charge is the net positive charge experienced by valence electrons
Note: The question is incomplete. The complete question is given below :
Suppose a substance has a heat of fusion equal to 45 cal/g and a specific heat of 0.75 cal/g°C in the liquid state. If 5.0 kcal of heat are applied to a 50 g sample of the substance at a temperature of 24°C, what will its new temperate be? What state will the sample be in? (melting point of the substance = 27°C; specific heat of the solid =0.48 cal/g°C; boiling point of the substance = 700°C)
Explanation:
1.a) Heat energy required to raise the temperature of the substance to its melting point, H = mcΔT
Mass of solid sample = 50 g; specific heat of solid = 0.75 cal/g; ΔT = 27 - 24 = 3 °C
H = 50 × 0.75 × 3 = 112.5 calories
b) Heat energy required to convert the solid to liquid at its melting point at 27°C, H = m×l, where l = 45 cal/g
H = 50 × 45 = 2250 cal
c) Total energy used so far = 112.5 cal + 2250 cal = 2362.5 calories.
Amount of energy left = 5000 - 2362.5 = 2637.5 cal
The remaining energy is used to heat the liquid
H = mcΔT
Where specific heat of the liquid, c = 0.75 cal/g/°C, H = 2637.5 cal, ΔT = temperature change
2637.5 = 50 × 0.75 x ΔT
ΔT = 2637.5 / ( 50*0.75)
ΔT = 70.3 °C
Final temperature of sample = (70.3 + 27) °C = 97.3 °C
The substance will be in liquid state at a temperature of 97.3 °C
i hope that this eg gonna help u
The answer is C. <span>The density of the fluid is greater than the density of the object.</span>
