Answer:
History of composition
Years Material Weight (grains)
1944–1946 gilding metal (95% copper, 5% zinc) 48 grains
1947–1962 bronze (95% copper, 5% tin and zinc) 48 grains
1962 – September 1982 gilding metal (95% copper, 5% zinc) 48 grains
October 1982 – present copper-plated zinc (97.5% zinc, 2.5% copper) 38.6 grains
Answer:
1.73 M
Explanation:
Molarity is moles per liter, so we need to divide 7.47 moles by 4.31 liters to get the molarity of the solution.
7.47/4.31 ≈ 1.73 M
Answer:
1 mole of a gas would occupy 22.4 Liters at 273 K and 1 atm
Explanation:
An ideal gas is a set of atoms or molecules that move freely without interactions. The pressure exerted by the gas is due to the collisions of the molecules with the walls of the container. The ideal gas behavior is at low pressures, that is, at the limit of zero density. At high pressures the molecules interact and intermolecular forces cause the gas to deviate from ideality.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P * V = n * R * T
In this case:
- P= 1 atm
- V= 22.4 L
- n= ?
- R= 0.082

- T=273 K
Reemplacing:
1 atm* 22.4 L= n* 0.082
*273 K
Solving:

n= 1 mol
Another way to get the same result is by taking the STP conditions into account.
The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C (or 273 K) are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.
<u><em>1 mole of a gas would occupy 22.4 Liters at 273 K and 1 atm</em></u>
Answer:
![K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BHCO_3%5E-%5D%7D%7B%5BH_2CO_3%5D%7D)
Explanation:
Several rules should be followed to write any equilibrium expression properly. In the context of this problem, we're dealing with an aqueous equilibrium:
- an equilibrium constant is, first of all, a fraction;
- in the numerator of the fraction, we have a product of the concentrations of our products (right-hand side of the equation);
- in the denominator of the fraction, we have a product of the concentrations of our reactants (left-hand side o the equation);
- each concentration should be raised to the power of the coefficient in the balanced chemical equation;
- only aqueous species and gases are included in the equilibrium constant, solids and liquids are omitted.
Following the guidelines, we will omit liquid water and we will include all the other species in the constant. Each coefficient in the balanced equation is '1', so no powers required. Multiply the concentrations of the two products and divide by the concentration of carbonic acid:
![K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BHCO_3%5E-%5D%7D%7B%5BH_2CO_3%5D%7D)