When a conductor is passed through a magnetic field, a current is induced in the conductor. This phenomenon is called electromagnetic induction. A changing magnetic field creates an electric field or current. This happens because of the movement of electrons in conductor as affected by the presence of the changing magnetic field.
The moles of each element found in a 5.00 g nickel coin is calculated as below
moles =mass/molar mass
calculate the mass of each element =% composition of element/100 x total mass of nickel
Mn = 9/100 x5 = 0.45g
Cu=56/100 x5= 2.8 g
Ag= 35/100x5= 1.75 g
moles of each element is therefore=
Mn = 0.45g/54.94 = 8.19 x10^-3 moles
Ag=1.75g/107.87 g/mol = 0.0162 moles
Cu = 2.8 g/63.5 g/mol=0.0441 moles
Answer:
this reaction is a double displacement reaction in which, Aluminium nitrate reacts with hydrogen sulphide to form nitric acid and Aluminium sulphide.
here is the balanced chemical equation for the above reaction :
2 Al (NO3)3 + 3 H2S -> 6 HNO3 + Al2S3
Answer:
1. ![K_eq = [Ca^{2+][OH^-]^2 = K_{sp}](https://tex.z-dn.net/?f=K_eq%20%3D%20%5BCa%5E%7B2%2B%5D%5BOH%5E-%5D%5E2%20%3D%20K_%7Bsp%7D)
2. a. No effect;
b. Products;
c. Products;
d. Reactants
Explanation:
1. Equilibrium constant might be written using standard guidelines:
- only aqueous species and gases are included in the equilibrium constant excluding solids and liquids;
- the constant involves two parts: in the numerator of a fraction we include the product of the concentrations of products;
- the denominator includes the product of the concentrations of reactants;
- the concentrations are raised to the power of the coefficients in the balanced chemical equation.
Based on the guidelines, we have two ions on the product side, a solid on the left side. Thus, the equilibrium constant has the following expression:
![K_eq = [Ca^{2+][OH^-]^2 = K_{sp}](https://tex.z-dn.net/?f=K_eq%20%3D%20%5BCa%5E%7B2%2B%5D%5BOH%5E-%5D%5E2%20%3D%20K_%7Bsp%7D)
2. a. In the following problems, we'll be considering the common ion effect. According to the principle of Le Chatelier, an increase in concentration of any of the ions would shift the equilibrium towards the formation of our precipitate.
In this problem, we're adding calcium carbonate. It is insoluble, so it wouldn't have any effect on the equilibrium.
b. Sodium carbonate is completely soluble, it would release carbonate ions. The carbonate ions would combine with calcium cations and more precipitate would dissolve. This would shift the equilibrium towards formation of the products to reproduce the amount of calcium cations.
c. HCl would neutralize calcium hydroxide to produce calcium chloride and water, so the amount of calcium ions would increase, therefore, the products are favored.
d. NaOH contains hydroxide anions, so we'd have a common ion. An increase in hydroxide would produce more precipitate, so our reactants are favored.
Answer:
the quantity required can go from 117 ml (for maximum concentration) up to 2900 ml ( if the concentrated solution has molarity =0.420 M)
Explanation:
the amount of water required to dilute a solution V₁ liters of Molarity M₁ to V₂ liters of M₂
moles of hydrochloric acid = M₁ * V₁= M₂ * V₂
V₁ = V₂ * M₂/M₁
where
M₂ = 0.420 M
V₂ =2.90 L
Since the hydrochloric acid can be concentrated up to 38% p/V ( higher concentrations are possible but the evaporation rate is so high that handling and storage require extra precautions, like cooling and pressurisation)
maximum M₁ =38% p/V = 38 gr/ 0.1 L / 36.5 gr/mol = 10.41 M
then
min V₁ = V₂ * M₂/ max M₁ = 2.90 L* 0.420 M/ 10.41 M= 0.117 L = 117 ml
then the quantity required can go from 117 ml up to 2900 ml ( if M₁ = M₂)