The answer is (4). You may recall the term "radiometric dating," which refers to the dating of old artifacts by measuring proportions of certain radioactive isotopes they contain and making calculations based on their estimated half-lives. Geological formations are dated in this way.
The first element discovered through synthesis was technetium<span>—its discovery being definitely confirmed in 1936. Hope that helps.</span>
Answer: ![Bi(OH)_3+3HNO_3\rightarrow 3H_2O+Bi(NO_3)_3](https://tex.z-dn.net/?f=Bi%28OH%29_3%2B3HNO_3%5Crightarrow%203H_2O%2BBi%28NO_3%29_3)
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
The balanced equation will be:
![Bi(OH)_3+3HNO_3\rightarrow 3H_2O+Bi(NO_3)_3](https://tex.z-dn.net/?f=Bi%28OH%29_3%2B3HNO_3%5Crightarrow%203H_2O%2BBi%28NO_3%29_3)
Answer:
Following are the response to the given question:
Explanation:
In the given scenario, When the balance has never been tainted before the KHP is weighted, which can affect the molar concentration of NaOH because its molarity is directly proportional to the weight including its substance. In this question it is the mistake is systemic because it may be corrected by modifying balancing parameters.
Answer:
5.521 × 10⁻² mol/m².s
Explanation:
Given:
Pressure of the Methane and Helium gas = 101.32 kPa
Temperature of the Methane and Helium gas = 298 K
Partial pressure of Methane, pA₁ = 60.79 kPa
Partial pressure of Methane at point 0.02 m away, pA₂ = 20.26 kPa
Now,
Molar flux is given as:
![J_A^* = -D_{AB} \times\frac{pA_2 - pA_1}{RT(z_2 - z_1)}](https://tex.z-dn.net/?f=J_A%5E%2A%20%3D%20-D_%7BAB%7D%20%5Ctimes%5Cfrac%7BpA_2%20-%20pA_1%7D%7BRT%28z_2%20-%20z_1%29%7D)
Here,
= 0.675 × 10⁻⁴ m²/s (for He-CH4 at 101.32 kPa and 298 K)
Z₂ - Z₁ = 0.02 m
R is the ideal gas constant = 8.314 J/mol.K
T is the temperature = 298 K
On substituting the respective values, we get
![J_A^* = -0.675\times10^{-4} \times\frac{20.26 - 60.79}{8.314\times298(0.02)}](https://tex.z-dn.net/?f=J_A%5E%2A%20%3D%20-0.675%5Ctimes10%5E%7B-4%7D%20%5Ctimes%5Cfrac%7B20.26%20-%2060.79%7D%7B8.314%5Ctimes298%280.02%29%7D)
or
= 5.521 × 10⁻² mol/m².s