Answer: B. Unsaturated
Explanation: in a saturated/super saturated solution, more solute will not be able to dissolve.
The number of atoms in one mole of any substance is measured by Avogadro's number. The value of Avogadro's number is 6.023 x 10 ^23. It is named after scientist Avogadro who proposed this number. 12 grams of carbon-12 represents 1 mole of carbon-12. For this reason, the number of atoms present in 1 mole of any substance is 6.023 x 10 ^23. Therefore, the number of atoms present in 1 mole carbon-12 is 6.023 x 10^23.
(Answer) This unit is the number of atoms in 12 grams of carbon-12 and known as Avogadro's number.
It allows electrons to flow from the anode to the cathode.
We are given with the initial volume of the substance and the molarity. The first thing that needs to be done is to multiply the equation in order to obtain the number of moles such as shown below.
number of moles = (40 mL) x (1 L / 1000 mL) x (0.3433 moles / L)
number of moles = 0.013732 moles
To get the value of the molarity of the diluted solution, we divide the number of moles by the total volume.
molarity = (0.013732 moles) / (750 mL / 1000 mL/L) = 0.0183 M
Similarly, we can solve for the molarity by using the equation,
M₁V₁ = M₂V₂
Substituting the known values in the equation,
(0.3433 M)(40 mL) = M₂(750 mL)
M₂ = 0.0183 M
Answer: The empirical formula for the given compound is ![CH_5N](https://tex.z-dn.net/?f=CH_5N)
Explanation : Given,
Percentage of C = 38.8 %
Percentage of H = 16.2 %
Percentage of N = 45.1 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 38.8 g
Mass of H = 16.2 g
Mass of N = 45.4 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =![\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{38.8g}{12g/mole}=3.23moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BGiven%20mass%20of%20Carbon%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20Carbon%7D%7D%3D%5Cfrac%7B38.8g%7D%7B12g%2Fmole%7D%3D3.23moles)
Moles of Hydrogen = ![\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{16.2g}{1g/mole}=16.2moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BGiven%20mass%20of%20Hydrogen%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20Hydrogen%7D%7D%3D%5Cfrac%7B16.2g%7D%7B1g%2Fmole%7D%3D16.2moles)
Moles of Nitrogen = ![\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{45.4g}{14g/mole}=3.24moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BGiven%20mass%20of%20nitrogen%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20nitrogen%7D%7D%3D%5Cfrac%7B45.4g%7D%7B14g%2Fmole%7D%3D3.24moles)
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.23 moles.
For Carbon = ![\frac{3.23}{3.23}=1](https://tex.z-dn.net/?f=%5Cfrac%7B3.23%7D%7B3.23%7D%3D1)
For Hydrogen = ![\frac{16.2}{3.23}=5.01\approx 5](https://tex.z-dn.net/?f=%5Cfrac%7B16.2%7D%7B3.23%7D%3D5.01%5Capprox%205)
For Oxygen = ![\frac{3.24}{3.23}=1.00\approx 1](https://tex.z-dn.net/?f=%5Cfrac%7B3.24%7D%7B3.23%7D%3D1.00%5Capprox%201)
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : N = 1 : 5 : 1
Hence, the empirical formula for the given compound is ![C_1H_5N_1=CH_5N](https://tex.z-dn.net/?f=C_1H_5N_1%3DCH_5N)