Answer:

Explanation:
¡Hola!
En este caso, teniendo en cuenta la información dada por el problema, inferimos que primero se debe usar la ecuación del gas ideal con el fin de calcular las moles de gas que se encuentran al inicio del experimento:

Seguidamente, usamos la ley de Avogadro para calcular las moles finales, teniendo el cuenta que el volumen final es el doble del inicial (8.50 L):

Quiere decir que las moles de N2(g) que se agregaron son:

¡Saludos!
Answer:
1. DNA copying mistakes made during cell division, exposure to ionizing radiation, exposure to chemicals called mutagens, or infection by viruses.
2. both
Explanation: Since errors can occur when DNA replicates itself, mutations can occur in both asexually and sexually reproducing organisms.
Answer: The final concentration of aluminum cation is 0.335 M.
Explanation:
Given:
= 47.8 mL (1 mL = 0.001 L) = 0.0478 L
= 0.321 M,
= 21.8 mL = 0.0218 L,
= 0.366 M
As concentration of a substance is the moles of solute divided by volume of solution.
Hence, concentration of aluminum cation is calculated as follows.
![[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}](https://tex.z-dn.net/?f=%5BAl%5E%7B3%2B%7D%5D%20%3D%20%5Cfrac%7BM_%7B1%7DV_%7B1%7D%20%2B%20M_%7B2%7DV_%7B2%7D%7D%7BV_%7B1%7D%20%2B%20V_%7B2%7D%7D)
Substitute the values into above formula as follows.
![[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}\\= \frac{0.321 M \times 0.0478 L + 0.366 M \times 0.0218 L}{0.0478 L + 0.0218 L}\\= \frac{0.0153438 + 0.0079788}{0.0696}\\= 0.335 M](https://tex.z-dn.net/?f=%5BAl%5E%7B3%2B%7D%5D%20%3D%20%5Cfrac%7BM_%7B1%7DV_%7B1%7D%20%2B%20M_%7B2%7DV_%7B2%7D%7D%7BV_%7B1%7D%20%2B%20V_%7B2%7D%7D%5C%5C%3D%20%5Cfrac%7B0.321%20M%20%5Ctimes%200.0478%20L%20%2B%200.366%20M%20%5Ctimes%200.0218%20L%7D%7B0.0478%20L%20%2B%200.0218%20L%7D%5C%5C%3D%20%5Cfrac%7B0.0153438%20%2B%200.0079788%7D%7B0.0696%7D%5C%5C%3D%200.335%20M)
Thus, we can conclude that the final concentration of aluminum cation is 0.335 M.
Answer:
I think it is a covalent bon
She needs to know the distance from LA to SF.
Call that distance x. Then you can calculate the cost from:
# of gallons consumed: x miles / 38 miles/ gallon = (x/38) gallons
After that,
cost = # gallons * cost per gallon = (x/38) gallons * 4 $/gallon = 4x/38 $