Answer:
c. 20.0332 g to 20,0 g
Explanation:
A significant figure is each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first non-zero digit, with the exception of the trailing zeros.
<em>Which of the following examples illustrates a number that is correctly rounded to three significant figures?
</em>
a. 109 526 g to 109 500 g. NO. The rounded number has 4 significant figures: 109 500.
b. 0.03954 g to 0.040 g. NO. The rounded number has 2 significant figures: 0.040.
c. 20.0332 g to 20.0 g. YES. The rounded number has 3 significant figures: 20.0.
d. 04.05438 g to 4.054 g. NO. The rounded number has 4 significant figures: 4.054.
e. 103.692 g to 103.7g. NO. The rounded number has 4 significant figures: 103.7.
Answer:
a. 1.21M
b. 0.119M
c. 0.00496M
Explanation:
Molarity, M, is an unit of concentration defined as the ratio between moles of solute and liters of solution:
a. 4.35 mol LiCl / 3.60L = 1.21M
b. 29.43gC6H12O6 * (1mol / 180.16g) = 0.1634moles / 1.37L = 0.119M
<em>Molar mass C6H12O6: 180.16g/mol</em>
c. 34.5mg NaCl = 0.0345g * (1mol / 58.44g) = 5.9x10⁻⁴moles / 0.1191L = 0.00496M
Answer:
s = 4.41 g/L.
Explanation:
¡Hola!
En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:
Lo cual hace que la expresión de equilibrio se calcule como:
![Ksp=[Pb^{2+}][Cl^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BPb%5E%7B2%2B%7D%5D%5BCl%5E-%5D%5E2)
Y que en términos de la solubilidad molar, s, se resuelve como:
![1.6x10^{-5}=s(2s)^2\\\\1.6x10^{-5}=4s^3\\\\s=\sqrt[3]{\frac{1.6x10^{-5}}{4} } \\\\s=0.0159molPbCl_2/L](https://tex.z-dn.net/?f=1.6x10%5E%7B-5%7D%3Ds%282s%29%5E2%5C%5C%5C%5C1.6x10%5E%7B-5%7D%3D4s%5E3%5C%5C%5C%5Cs%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B1.6x10%5E%7B-5%7D%7D%7B4%7D%20%7D%20%5C%5C%5C%5Cs%3D0.0159molPbCl_2%2FL)
Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):
¡Saludos!
<h3>
Answer:</h3>
200 mL
<h3>
Explanation:</h3>
Concept tested: Dilution formula
We are given;
- Concentration of stock solution as 1.00 M
- Volume of the stock solution as 50 mL
- Molarity of the dilute solution as 0.25 M
We are required to calculate the volume of diluted solution;
- The stock solution is the original solution before dilution while diluted solution is the solution after dilution.
- Using the dilution formula we can determine the volume of diluted solution;
M1V1 = M2V2
Rearranging the formula;
V2 = M1V1 ÷ M2
= (1.00 M × 50 mL) ÷ 0.25 M
= 200 mL
Therefore, a volume of 200mL of 0.25 M solution could be made from the stock solution.
