Answer:
See Explanation
Explanation:
Given
(a) to (d)
Required
Determine whether the given parameters can calculate the required parameter
To calculate either Density, Mass or Volume, we have



(a) 432 g of table salt occupies 20.0 cm^3 of space
Here, we have:


The above can be used to calculate Density as follows;



(b) 5.00 g of balsa wood, density of balsa wood : 0.16 g/cm^3
Here, we have:


This can be used to solve for Volume as follows:



(c) 32 cm^3 sample of gold density of 19.3 g/cm^3
Here, we have:


This can be used to calculate Mass as follows:



(d) 150 g of iron, density of Iron = 79.0 g/cm^3
Here, we have


This can be used to calculate volume as follows:


<em>Approximated</em>
Answer:
2.00 M
Explanation:
The concentration of a solution is given by

where
m is the mass of solute
V is the volume of the solution
At the beginning, the solution has:
M = 6.00 M is the concentration
V = 100 mL = 0.1 L is the volume
So the mass of solute (HCl) is

Then, the HCL is diluted into a solution with volume of
V = 300 mL = 0.3 L
Therefore, the final concentration is:

Data:
Molar Mass of HNO2
H = 1*1 = 1 amu
N = 1*14 = 14 amu
O = 3*16 = 48 amu
------------------------
Molar Mass of HNO2 = 1 + 14 + 48 = 63 g/mol
M (molarity) = 0.010 M (Mol/L)
Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:
M (molarity) = 0.010 M (Mol/L)
Use: Ka (ionization constant) =









Now, we will calculate the amount of Hydronium [H3O+] in nitrous acid (HNO2), multiply the acid molarity by the degree of ionization, we will have:
![[ H_{3} O^+] = M* \alpha](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%3D%20M%2A%20%5Calpha%20)
![[ H_{3} O^+] = 0.010* 2.23*10^{-3}](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%3D%200.010%2A%202.23%2A10%5E%7B-3%7D)
![[ H_{3} O^+] \approx 0.0223*10^{-3}](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%5Capprox%200.0223%2A10%5E%7B-3%7D)
![[ H_{3} O^+] \approx 2.23*10^{-5} \:mol/L](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%5Capprox%202.23%2A10%5E%7B-5%7D%20%5C%3Amol%2FL)
And finally, we will use the data found and put in the logarithmic equation of the PH, thus:
Data:
log10(2.23) ≈ 0.34
pH = ?
![[ H_{3} O^+] = 2.23*10^{-5}](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%3D%202.23%2A10%5E%7B-5%7D)
Formula:
![pH = - log[H_{3} O^+]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%5BH_%7B3%7D%20O%5E%2B%5D)
Solving:
![pH = - log[H_{3} O^+]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%5BH_%7B3%7D%20O%5E%2B%5D)




Note:. The pH <7, then we have an acidic solution.
Answer:
.
Explanation:
Eardrum or tympanic membrane changes sound to vibration and pass through malleus, incus and stapes.
Answer:
32. Opcion E
Explanation:
ZO₂ podemos entenderlo como dióxido de Z
1 mol de ZO₂ contiene 1 mol de Z y 2 moles de O₂ por lo que si la masa molar es de 64 g/mol, podemos plantear lo siguiente
Masa de Z + 2 masa de O = 64 g/mol
Sabemos que la masa del oxígeno es 16 g/mol
x + 2. 16 g/mol = 64 g/mol
x = 64 - 32 → 32 g/mol
La masa atómica coincide con la masa molar, por lo tanto la masa atomica de Z es 32.