The specific gravity of a sample is the ratio of the density of the sample with respect to one standard sample. The standard sample used in specific gravity calculation is water whose density is 1 g/mL. The solution having specific gravity 1.30 is the density of the sample that is 1.30 g/mL. Thus the weight of the 30 mL sample is (30×1.30) = 39 g.
Now the mass of the 10 mL of water is 10 g as density of water is 10 g/mL. Thus after addition the total mass of the solution is (39 + 10) = 49g and the volume is (30 + 10) = 40 mL. Thus the density of the mixture will be 
g/mL. Thus the specific gravity of the mixed sample will be 1.225 g/mL.
Answer:
The pH of the solution is 8, 40-
Explanation:
The pH indicates the acidity or basicity of a substance. PH values between 0 and less than 7 indicate acidic solutions, 7 neutral and greater than 7 to 14 basic. It is calculated as
pH = -log (H30+)
pH= -log (4x10-9M)
<em>pH=8,40</em>
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
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