The empirical formula is N₂O₅.
The empirical formula is the <em>simplest whole-number ratio of atoms</em> in a compound.
The ratio of atoms is the same as the ratio of moles, so our job is to calculate the <em>molar ratio of N:O</em>.
I like to summarize the calculations in a table.
<u>Element</u> <u>Moles</u> <u>Ratio¹ </u> <u> ×2² </u> <u>Integers</u>³
N 1.85 1 2 2
O 4.63 2.503 5.005 5
¹To get the molar ratio, you divide each number of moles by the smallest number (1.85).
²Multiply these values by a number (2) that makes the numbers in the ratio close to integers.
³Round off the number in the ratio to integers (2 and 5).
The empirical formula is N₂O₅.
Answer:
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A carbonated soft drink hs a large amount of water dissolved with ample amount of carbon dioxide. In this case, the solvent is water and carbon dioxide is the solute. In 2, for the reaction <span>CH3COOH = CH3COO + H+, CH3COOH is a Bronsted-Lowry acid because it releases a proton which is H+. </span>
Answer:
e. 3.08 x 10⁻² mol of ions.
Explanation:
- Every 1.0 mole of any compound contains Avogadro's number of molecules (6.022 x 10²³).
- We can get the no. of moles of NiCl₂ using cross multiplication:
1.0 mol NiCl₂ contains → 6.022 x 10²³ molecules.
??? mol NiCl₂ contains → 6.188 x 10²¹ molecules.
∴ The no. of moles of NiCl₂ = (1.0 mol)(6.188 x 10²¹ molecules)/(6.022 x 10²³ molecules) = 1.028 x 10⁻² mol.
- NiCl₂ is ionized according to the equation:
NiCl₂ → Ni²⁺ + 2Cl⁻.
Which means that every 1.0 mol of NiCl₂ is ionized to produce 3.0 moles (1.0 mol of Ni²⁺ and 2 moles of Cl⁻).
<em>∴ The total moles of ions are released</em> = 3 x 1.028 x 10⁻² mol = <em>3.083 x 10⁻² mol of ions.</em>
Answer:
2578.99 years
Explanation:
Given that:
100 g of the wood is emitting 1120 β-particles per minute
Also,
1 g of the wood is emitting 11.20 β-particles per minute
Given, Decay rate = 15.3 % per minute per gram
So,
Concentration left can be calculated as:-
C left = ![[A_t]=\frac{11.20\ per\ minute}{15.3\ per\ minute\ per\ gram}\times [A_0]= 0.7320[A_0]](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5Cfrac%7B11.20%5C%20per%5C%20minute%7D%7B15.3%5C%20per%5C%20minute%5C%20per%5C%20gram%7D%5Ctimes%20%5BA_0%5D%3D%200.7320%5BA_0%5D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Also, Half life of carbon-14 = 5730 years
Where, k is rate constant
So,
The rate constant, k = 0.000120968 year⁻¹
Time =?
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
So,
![\frac {[A_t]}{[A_0]}=e^{-0.000120968\times t}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%3De%5E%7B-0.000120968%5Ctimes%20t%7D)
![\frac {0.7320[A_0]}{[A_0]}=e^{-0.000120968\times t}](https://tex.z-dn.net/?f=%5Cfrac%20%7B0.7320%5BA_0%5D%7D%7B%5BA_0%5D%7D%3De%5E%7B-0.000120968%5Ctimes%20t%7D)
<u>t = 2578.99 years</u>
