Answer:
1. Its formation requires very strong updrafts = a. Hail
2. Its formation requires falling through a layer of above-freezing air = d. Freezing Rain
3. Precipitation from cumuliform clouds is typically of this nature = c. Shower
4. Precipitation from stratus clouds is typically of this nature = Drizzle
Explanation:
Hail formation requires very strong updrafts, these updrafts are the upward moving air created in a thunderstorm. This period of noticeable thunderstorms creates hails.
Freezing rain requires the presence of warm air, it requires falling through a layer of above-freezing air to the colder air below to produce an ice coating on anything it drops on.
Showers are produced by cumuliform clouds which look like cotton balls. Since cumuliform clouds precipitate too, these clouds can have fluctuating rain in a day in the form of showers.
Drizzle which raises low visibility is considered a type of liquid precipitation since it also falls from a cloud. Drizzle which is obviously smaller in diameter when compared to that of raindrops, however, is common with stratus clouds.
Which equation obeys the law of conservation of
mass?
Answer:2C4H10+2C12+12O2 4CO2+CC14+H20
En la fermentación del alcohol, la levadura convierte la glucosa en etanol y dióxido de carbono:
C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)
Si reaccionan 5.97 g de glucosa y se recolectan 1.44 L de CO2 gaseoso, a 293 K y 0.984 atm, ¿cuál
es el rendimiento porcentual de la reacción
Answer:
88.9%
Explanation:
Primero convertimos 5.97 g de glucosa a moles, usando su masa molar:
5.97 g ÷ 180 g/mol = 0.0332 molDespués calculamos la cantidad máxima de moles de CO₂ que se hubieran podido producir:
0.0332 mol C₆H₁₂O₆ * [tex]\frac{2molCO_2}{1molC_6H_{12}O_6}[/tex] = 0.0664 mol CO₂Ahora calculamos los moles de CO₂ producidos, usando los datos de recolección dados y la ecuación PV=nRT:
0.984 atm * 1.44 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 Kn = 0.0590 molFinalmente calculamos el rendimiento porcentual:
0.0590 mol / 0.0664 mol * 100% = 88.9%Que es la actividad física y en qué mejora
If 0.250 L of a 5.90 M HNO₃ solution is diluted to 2.00 L, what is the molarity of the new solution?
Answer:
0.74 M
Explanation:
From the question given above, the following data were obtained:
Molarity of stock solution (M₁) = 5.90 M
Volume of stock solution (V₁) = 0.250 L
Volume of diluted solution (V₂) = 2 L
Molarity of diluted solution (M₂) =?
The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
5.90 × 0.250 = M₂ × 2
1.475 = M₂ × 2
Divide both side by 2
M₂ = 1.475 / 2
M₂ = 0.74 M
Thus, the molarity of the diluted solution is 0.74 M
