1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N. Details about number of atoms can be found below.
How to calculate number of atoms?
The number of atoms of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.
However, the number of moles of oxygen in glycine can be calculated using the following expression:
Molar mass of C₂H5O2N = 75.07g/mol
Mass of oxygen in glycine = 32g/mol
Hence; 32/75.07 × 7.51 = 3.2grams of oxygen in glycine
Moles of oxygen = 3.2g ÷ 16g/mol = 0.2moles
Number of atoms of oxygen = 0.2 × 6.02 × 10²³ = 1.205 × 10²³ atoms
Therefore, 1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N.
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Answer:
Magnetic Quantum Number (ml): ml = -l, ..., 0, ..., +l. Specifies the orientation in space of an orbital of a given energy (n) and shape (l). This number divides the subshell into individual orbitals which hold the electrons; there are 2l+1 orbitals in each subshell.
Answer:
A) The reaction mixture contains mostly products at equilibrium
Explanation:
For the general reaction
Reactants ⇌ Products
The equilibrium constant is
![K_{\text{eq}} = \dfrac{\text{[Products]}}{\text{[Reactants]}}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Beq%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BProducts%5D%7D%7D%7B%5Ctext%7B%5BReactants%5D%7D%7D)
Thus, if K is large, the concentration of products is greater than that of reactants.
The reaction mixture will contain mostly products at equilibrium.
C) and E) are wrong. The equilibrium constant gives us information only on the position of equilibrium, not on how fast it is achieved.
Answer:
1.57 mol NaN₃
Explanation:
- 2 NaN₃ (s) → 2 Na (s) + 3 N₂ (g)
First we <u>use PV=nRT to calculate the number of N₂ moles that need to be produced</u>:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 23.7 °C ⇒ 23.7 + 273.16 = 296.86 K
<u>Inputing the data</u>:
- 1.07 atm * 53.4 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.86
And <u>solving for n</u>:
Finally we <u>convert N₂ moles into NaN₃ moles</u>, using <em>the stoichiometric coefficients of the balanced reaction</em>:
- 2.35 mol N₂ *
= 1.57 mol NaN₃