Assuming we have a 100g sample, the mass of each element is as follows:
C: 74 g
H: 7.4 g
N: 8.6 g
O: 10 g
Next, we calculate the moles of each by dividing the mass of each element by its molar mass:
C: (74 / 12) = 6.17
H: (7.4 / 1) = 7.4
N: (8.6 / 14) = 0.61
O: (10 / 16) = 0.625
Now, we take the smallest value to determine the ratio:
C: 10
H: 12
N: 1
O: 1
Thus, the empirical formula can be expressed as
C10H12NO
Answer:
Protons: 19
Neutrons: 25
Electrons: 19
Explanation:
Protons:
The atomic number determines the number of protons in an atom. Consequently, with Potassium's atomic number being 19, it contains 19 protons.
Neutrons:
The formula to find neutrons is:
# of Neutrons = Atomic Mass - # of Protons
Given:
Atomic Mass = 43
# of Protons = 19
Thus,
# of Neutrons = 43 - 19
# of Neutrons = 24
Electrons:
In a neutral atom, the quantity of electrons matches that of protons. Therefore, a neutral Potassium atom with 19 protons must equally have 19 electrons.
Answer:
Explanation:
Given data:
Initial temperature T₁ = 25.2°C = 298.2K
Initial pressure P₁ = 0.6atm
Final temperature = 72.4°C = 345.4K
What we need to find:
Final pressure = ?
To determine this, we apply a modified version of the combined gas law with constant volume. This simplifies our calculations to:

Here, P and T signify pressure and temperatures, 1 refers to initial and 2 to final temperatures.
Now we can substitute the known variables:

P₂ = 0.7atm
The formula for a monoprotic acid can be represented as HA, and its reaction with a base is shown as follows: HA + NaOH ---> NaA + H₂O. The stoichiometry between the acid and the base is 1:1. At the point of neutralization, the moles of HA equals the moles of the base. The moles of NaOH that reacted can be calculated as 0.100M / 1000 mL/L x 30.0 mL = 0.003 mol. Consequently, the moles of HA that reacted equal 0.003 mol. The mass of the acid is 0.384 g, yielding a molar mass of 0.384 g / 0.003 mol = 128 g/mol.
The enthalpy change associated with the precipitation reaction is 84 kJ/mole
Why?
The chemical equation for the reaction can be written as
AgNO₃(aq) + NaCl (aq) → AgCl(s) + NaNO₃(aq)
To determine the enthalpy change, the following equation applies

To calculate the heat (Q):

Next, we need to calculate the number of moles involved in the reaction (n):
![n=[AgNO_3]*v(L)=(0.1M)*(0.05L)=0.005moles](https://tex.z-dn.net/?f=n%3D%5BAgNO_3%5D%2Av%28L%29%3D%280.1M%29%2A%280.05L%29%3D0.005moles)
With these two values, we can substitute them into the first equation:

Have a great day!