Answer:
Volume of container = 0.0012 m³ or 1.2 L or 1200 ml
Explanation:
Volume of butane = 5.0 ml
density = 0.60 g/ml
Room temperature (T) = 293.15 K
Normal pressure (P) = 1 atm = 101,325 pa
Ideal gas constant (R) = 8.3145 J/mole.K)
volume of container V = ?
Solution
To find out the volume of container we use ideal gas equation
PV = nRT
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
First we find out number of moles
<em>As Mass = density × volume</em>
mass of butane = 0.60 g/ml ×5.0 ml
mass of butane = 3 g
now find out number of moles (n)
n = mass / molar mass
n = 3 g / 58.12 g/mol
n = 0.05 mol
Now put all values in ideal gas equation
<em>PV = nRt</em>
<em>V = nRT/P</em>
V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa
V = 121.87 ÷ 101,325 pa
V = 0.0012 m³ OR 1.2 L OR 1200 ml
Answer:
The atomic mass of phosphorus is 29.864 amu.
Explanation:
Given data:
Atomic mass of phosphorus = ?
Percent abundance of P-29 = 35.5%
percent abundance of P-30 = 42.6%
Percent abundance of P-31 = 21.9%
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass / 100
Average atomic mass = (29×35.5)+(30×42.6) + (31×21.9) /100
Average atomic mass = 1029.5 + 1278 + 678.9/ 100
Average atomic mass = 2986.4 / 100
Average atomic mass = 29.864 amu.
The atomic mass of phosphorus is 29.864 amu.
Answer:
b
. Irradiated food is shown to not be radioactive.
Explanation:
If it can be proven that irradiated food is not radioactive, then it will effective dispute the idea that irradiated food are less safe to eat.
- An irradiated food is one in which ionizing radiations have been employed to improve food quality.
- Thus, bacteria and other food spoilers can be exterminated from the food.
- Most irradiated food do not contain radiation and are fit for consumption.
If it can be proven, that this is true, then it will challenge the idea that irradiated foods are not safe.
This equation relates all three variables, so just plug in all of the given values. Don’t forget to convert deltaS into kJ/K because Gibbs free energy is measured in kJ. The K used in temperature cancels out the K in kJ/K, leaving only kJ as your units. Don’t forget sig figs!
Answer:
Because both compounds (CaCl2 and CaBr2) contain elements (bromine and chlorine) from the same group (group 7)
Explanation:
Elements are organized into groups on the periodic table based on the number of valence electrons contained in their outermost shell. These elements in the same group i.e. same number of valence electrons, will behave in a similar manner chemically.
CaCl2 and CaBr2 are two compounds that contains elements (Chlorine and Bromine) from the same group, which is group 7. Elements in group 7 are called HALOGENS and have 7 valence electrons in their outermost shell.
Bromine (Br) and Chorine (Cl) are responsible for the similarity in the properties of CaCl2 and CaBr2 because the Chlorine and Bromine contained in them will cause them to react similarly and behave similarly in the presence of other compounds.
