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:
2AlCl3 + Ca3N2 - 2AlN+ 3CaCl2
Answer:
Here:
Explanation:
To familiarize students with experimental apparatus, the scientific method, and methods of data analysis so that they will have some idea of the inductive process by which the ideas were originated. To teach how to make careful experimental observations and how to think about and draw conclusions from such data.
The total pressure when the new equilibrium is stabilized is half of the initial pressure of the system.
The given chemical reaction at a stable equilibrium is,
2H₂O(g)+O₂(g) = 2H₂O₂(g)
According to the ideal gas equation,
PV = nRT
P is pressure,
V is volume,
n is moles
R is gas constant,
T is temperature.
Assuming the temperature is constant.
If the volume of the system is twice the initial volume then the total pressure at the new equilibrium can be found out as,
P₁V₁ = P₂V₂
Where, P₁ and V₁ are initial volume and pressure while P₂ and V₂ are final pressure and volume.
If V₂ = 2V₁,
P₂ = P₁/2
So, the final total pressure will be half of the initial pressure.
To know more about equilibrium, visit,
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Answer:
First, balance the half-reactions
Second, equalize the electrons
Third,add two reaction equations to get final answer
Explanation:
For example
H₂C₂0₄ + MnO⁻₄ ---------->CO₂+Mn²⁺
(i) Balancing the half reactions
H₂C₂O₄-------->2CO₂+2H⁺+2e⁻
5e⁻ +8H⁺+MnO₄⁻----------->Mn²⁺+4H₂O
(ii)
Equalizing the electrons
5H₂C₂O₄--------->10CO₂+10H⁺+10e⁻ ---here there is a factor of 5
10e⁻+16H⁺+2MnO₄⁻--------->2Mn²⁺+8H₂O -----here there is a factor of 2
(iii)
Add the two where electrons and some Hydrogen ions will cancel out
5H₂C₂O₄+6H⁺+2MnO₄⁻---->10CO₂+2Mn²⁺+8H₂O
