Answer:
d and e - Sodium and antimony
Explanation:
The atomic numbers remain the same, while the mass numbers change (because neutrons are being added or taken away).
sodium has an atomic number of 19 and a mass number of 39 - in d, it has an atomic number of 19 but a mass number of 40. therefore, it is an isotope
antimony has an atomic number of 51 and a mass number of 121.60 - in e, it has an atomic number of 51, but a mass number of 123. therefore, it is an isotope
There's 6.022×10^23 particles in 1 mole of anything
like there is 1000 grams in 1 kilogram of anything
One way of expressing concentration is by percent. It may be on the basis of mass, mole or volume. Percent is expressed as the amount of solute per amount of the solution. For this case, we are given the percent by mass. In order to solve the amount of solute, we multiply the percent with the amount of the solution.
Mass of solute = percent by mass x mass solution
Mass of solute = 0.0350 x 2.50 x10^2 = 8.75 grams of solute
Answer: concentration
Explanation:
Concentration refers to the amount of a substance present in a sample. The more molecules of a substance present in a sample, the greater its concentration. The less molecules of a substance in a sample, the lesser the concentration. We are often concerned about analytically determining the concentration of a substance using diverse analytical methods in chemistry.
Given:
175 kilograms of Methane (CH4) to be synthesized into Hydrogen Cyanide (HCN)
The balanced chemical equation is shown below:
2 CH4<span> + 2 NH</span>3<span> + 3 O</span>2<span> → 2 HCN + 6 H</span>2<span>O
</span>
To calculate for the masses of ammonia and oxygen needed, our basis will be 175 kg CH4.
Molar mass:
CH4 = 16 kg/kmol
NH3 = 17 kg/kmol
O2 = 32 kg/kmol
mass of NH3 = 175 kg CH4 / 16 kg/kmol * (2/2) * 17 kg/kmol
mass of NH3 = 185.94 kg NH3 needed
mass of O2 = 175 kg CH4 / 16 kg/kmol * (3/2) * 32 kg/kmol
mass of O2 = 525 kg
mass of O = 525 kg / 32 kg/kmol * (1/2) * 16 kg/kmol
mass of O = 131.25 kg O
