Answer:
The empirical formula of the compound C₃H₆N₂ is C₃H₆N₂
Explanation:
The empirical formula of a compound is the formula of the compound given in the (smallest) whole number ratio of the elements of the compound
The empirical formula of S₂O₂ is SO
The empirical formula of C₃H₆O₃ is CH₂O
The given compound's molecular formula is C₃H₆N₂
The smallest whole number ratio of of the elements of the compound is 3:6:2, therefore, the empirical formula of the compound C₃H₆N₂ is C₃H₆N₂.
Answer:
The electron configuration for a
Mn3+ ion is [Ar]3d4
Explanation:
3.52g BiCl3 × 1 mol BiCl3/ 315.34g BiCl3 × 3 mol Cl/ 2 mol BiCl3 × 70.906g Cl/ 1 mol Cl= 1.187 g Cl
Answer:
Explanation:
Name: Tantalum
Symbol: Ta
Atomic Number: 73
Atomic Mass: 180.9479 amu
Melting Point: 2996.0 °C (3269.15 K, 5424.8 °F)
Boiling Point: 5425.0 °C (5698.15 K, 9797.0 °F)
Number of Protons/Electrons: 73
Number of Neutrons: 108
Classification: Transition Metal
Crystal Structure: Cubic
Density @ 293 K: 16.654 g/cm3
Color: gray
Answer:
% = 76.75%
Explanation:
To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:
A = A₀ e^(-kt) (1)
Where:
A and A₀: concentrations or mass of the compounds, (final and initial)
k: constant decay of the compound
t: given time
Now to get the value of k, we should use the following expression:
k = ln2 / t₁/₂ (2)
You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.
Now, let's calculate k:
k = ln2 / 956.3
k = 7.25x10⁻⁴ d⁻¹
With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:
A = 250 e^(-7.25x10⁻⁴ * 365)
A = 250 e^(-0.7675)
A = 191.87 g
However, the question is the percentage left after 1 year so:
% = (191.87 / 250) * 100
<h2>
% = 76.75%</h2><h2>
And this is the % of isotope after 1 year</h2>
