Answer:
The product is significantly impure
Explanation:
In order to test for the purity of a specific sample that was synthesized, the melting point of a compound is measured. Basically speaking, the melting point identifies how pure a compound is. There are several cases that are worth noting:
- if the measured melting point is significantly lower than theoretical, e. g., lower by 3 or more degrees, we conclude that our compound contains a substantial amount of impurities;
- wide range in the melting point indicates impurities, unless it agrees with the theoretical range.
Since our compound is even 10 degrees Celsius lower than expected, it indicates that the compound is significantly impure.
By using flame test we can identify the elements because colors which are given by elements with flame test are unique.
During the flame test, the electrons of the atom which are in ground state absorb energy and go to upper level. This is called electron excitation. Excited electrons are unstable. Hence, they come back to the ground state by emitting the energy as photons. If that released energy has a frequency which belongs to visual light, then that wave gives a color.
The standard addition equation is as followsI_(S+X) (V/V_O )=I_X+I_X/[X]_i [S]_4 (V_S/V_0 ) Here, [X]_i is the initial concentration of analyte, [S]_i is the initial concentration of standard, I_X is signal for analyte, I_(S+X) is signal for standard and analyte, V_0 is the initial volume, V_S is the added standard volume, and V is the total volume.Added volume of standard V_S is-23.3 mL. Initial volume of the sample V_0 is 10.00 mL. Initial concentration of standard ([S]_i) is 0.156 ng/mL.[X]_i= -[S]_i (V_S/V_0 )〖[X]〗_(i )= -(0.156 ng/mL)((-23.3 mL)/(10.00 mL))=0.363 ng/mL
Concentration of U(III) in ground sample is 0.363 ng/mL
Answer:
The new volume of the balloon when the pressure equalised with the pressure of the atmosphere = 494 L.
The balloon expands by am additional 475 L.
Explanation:
Assuming Helium behaves like an ideal gas and temperature is constant.
According to Boyle's law for ideal gases, at constant temperature,
P₁V₁ = P₂V₂
P₁ = 26 atm
V₁ = 19.0 L
P₂ = 1 atm (the balloon is said to expand till the pressure matches the pressure of the atmpsphere; and the pressure of the atmosphere is 1 atm)
V₂ = ?
P₁V₁ = P₂V₂
(26 × 19) = 1 × V₂
V₂ = 494 L (it is assumed the balloon never bursts)
The new volume of the balloon when the pressure equalised with the pressure of the atmosphere = 494 L.
The balloon expands by am additional 475 L.
Hope this Helps!!!
