Se ha quemado magnesio (reacción con el oxígeno) y se obtuvieron 12g de óxido de magnesio (II). ¿Cuánto magnesio se quemó? ¿Qué volumen de oxígeno medido en condiciones normales se quemó?

The first answer is 7,24g Mg

The second answer is 3,36L O2

Explanation:

Atomic masses: Mg= 24,3 ; O=16

Solution: MMgO = 40,3 g/mol

2 Mg + O2 => 2 MgO

1) 12 g MgO x (2x24,3 g Mg / 2x 40,3 MgO) = 7,24 g Mg

2)

At  first calculate moles from Oxygen:

12 g MgO x ( 1 mol from Oxygen/ 2 x 40,3 g MgO ) = 0,15 moles from Oxygen

then

0,15 moles from Oxygen x (22,4 lts / 1 mol from Oxygen ) = 3,36 lts

22.4 liters is the constant that we use in this equation, it is a value that is repeated since we are in the presence of a gas with ideal behavior, so it is considered that the normal molar volume of ideal gaseous substances is always 22.4 liters, estimated with a temperature of 0º C and a pressure of 1 atmosphere. (constant temperature and pressure)

15.3 cos 65 degrees trig

can u provide the graph and if its not possible

then i can tell u the segment where the

average time interval starts degrading is the segment where the car is slowing down