How is the air in tube 4 kept out of contact with the nail
1 answer:
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Answer:
The mass of 10 cm³of a 0.4 g/dm³ solution of sodium carbonate is 0.004 grams
Explanation:
The question is with regards to density calculations
The density of the given sodium carbonate solution, ρ = 0.4 g/dm³
The volume of the given solution of sodium carbonate, V = 10 cm³ = 0.01 dm³
Therefore, we have;
The mass, "m", of the sodium carbonate in = ρ×V = 0.4 g/dm³ × 0.01 dm³ = 0.004 g
The mass of 10 cm³ (10 cm³ = 0.01 dm³) of a 0.4 g/dm³ solution of sodium carbonate, m = 0.004 g.
Since you have not included the given reaction, I am going to explain you how to solve these kind of problems.
1) The chemical equilibrium is a dynamic process. It means that in an equilibrim reaction there are two rectaions, the forward reaction and the reverse reaction whose velocities are the same.
2) The general equation of a a chemical reaction in equlibrium is:
aA + bB ⇄ cC + dD
Where A and B are the reactants, C and D are the products, and a, b, c, d, are the coefficientes in the balanced equation.
3) So, the equilibrium law is:
![Keq= \frac{C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=Keq%3D%20%5Cfrac%7BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D%20)
Where Keq is the constant of equilibrium
4) To complete the explanation, I am going to deal with an example:
i) Consider the equlibrium reaction between hydrogen and iodine:
H₂ (g) + I₂(g) ⇄ 2HI(g)
ii) The forward reaction is H₂ (g) + I₂(g) → 2HI(g)
iii) The reverse reaction is 2HI (g) → H₂ (g) + I₂(g)
iv) The law of equilibrium is:
![Keq= \frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=Keq%3D%20%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D%20)
1) The chemical equilibrium is a dynamic process. It means that in an equilibrim reaction there are two rectaions, the forward reaction and the reverse reaction whose velocities are the same.
2) The general equation of a a chemical reaction in equlibrium is:
aA + bB ⇄ cC + dD
Where A and B are the reactants, C and D are the products, and a, b, c, d, are the coefficientes in the balanced equation.
3) So, the equilibrium law is:
Where Keq is the constant of equilibrium
4) To complete the explanation, I am going to deal with an example:
i) Consider the equlibrium reaction between hydrogen and iodine:
H₂ (g) + I₂(g) ⇄ 2HI(g)
ii) The forward reaction is H₂ (g) + I₂(g) → 2HI(g)
iii) The reverse reaction is 2HI (g) → H₂ (g) + I₂(g)
iv) The law of equilibrium is: