Answer:
4.8× 10²³ atoms
Explanation:
Given data:
Number of moles of San element = 0.796 mol
Number of atoms present = ?
Solution:
Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms
0.796 mol × 6.022 × 10²³ atoms / 1mol
4.8× 10²³ atoms
Answer:
<h2>Actin and myosin.</h2>
Explanation:
The cells that allow your bones to move, the movement of thick (myosin) and thin (actin) filaments during contraction
.
During a contraction thick and thin filaments do not shorten but increase their overlap of each other.
Thin filaments slide past thick filaments extending more deeply into the A band.
The I bands and H bands decrease in lenght as Z discs are come closer together
.
Sarcomere represents area between two Z disc, so the sarcomere gets smaller during a contraction
.
Hello!
The H₃O⁺ concentration can be found using the definition of pH and clearing the equation for [H₃O⁺]. The solution has a pH lower than 7, so the Sauvignon Blanc is
acid. The calculation for [H₃O⁺] is shown below:
![pH=-log [H_3O^{+}]](https://tex.z-dn.net/?f=pH%3D-log%20%5BH_3O%5E%7B%2B%7D%5D%20)
![[H_3O^{+}]= 10^{-pH}=10^{-3,24}=0,00058M](https://tex.z-dn.net/?f=%5BH_3O%5E%7B%2B%7D%5D%3D%2010%5E%7B-pH%7D%3D10%5E%7B-3%2C24%7D%3D0%2C00058M%20)
So, the concentration of H₃O⁺ in a Sauvignon Blanc with a pH of 3,24 is
0,00058 MHave a nice day!
This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:

Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.