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.
A: making s sandcastle. This is because water and sand is only a mixture, so they do not react with each other. All the rest include chemical reactions!
Charles law gives the relationship between volume and temperature of gas.
It states that at constant pressure volume is directly proportional to temperature
Therefore
V/ T = k
Where V - volume T - temperature in kelvin and k - constant
V1/T1 = V2/T2
Parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
Substituting the values in the equation
267 L/ 480 K = V / 750 K
V = 417 L
Final volume is 417 L
Saturated hydrocarbons consists of C-C single bond whereas Unsaturated hydrocarbons consists C-C double/triple bond.
Answer:
The volume of a sample of gas (2.49 g) was 752 mL at 1.98 atm and 62∘C 62 ∘ C .
