Mg reaction with O₂ gas will produce MgO so the equation will be
2Mg+O₂⇒2MgO. (You have to find the equation in order two figure out the number of moles of O₂ that will react with 1 mole of MgO).
The first step is to find the number of moles of Mg in 4.03g of Mg. You can do this by dividing 4.03g Mg by its molar mass (which is 24.3g/mol) to get 0.1658mol Mg. Then you have to find the number of moles of O₂ that will react with 0.1658mol Mg. To do this you need to use the fact that 1mol O₂ will react with 2mol Mg (this reatio is from the chemical equation) so you have to multiply 0.1658mol Mg by (1mol O₂)/(2mol Mg) to get 0.0829mol O₂. From here you would usually use PV=nRT and solve for V However, the question tells us that we are at STP, that means you can use the fact that 22.4L of gas is 1 mol of gas at STP. Using that information we can find the volume of O₂ gas by mulitlying 0.0829mol O₂ by 22.4L/mol to get 1.857L which equals 1857mL.
therefore, 1857mL of O₂ gas will react with 4.03g of Mg.
I hope this helps. Let me know in the comments if anything is unclear.
D. Drop in barometric pressure, warm ocean water, humid air. The low pressure brings in a cool air mass causing collision of two different masses.
My father rode out a typhoon near Okinawa WWII, onboard the battleship USS Missouri BB-63.
Violent pitching, alarms going off for approaching capsize pitch. The captain came on loudspeaker “ don’t worry men, land is near... about a mile straight down”.
Sand only bc the salt dissolved and sand doesn't
Answer:
B . Changing the material that the fluids container is made of
Explanation:
Changing the material of the container does not affect the pressure in a container whereas increasing the volume, changing the weight of the fluid, and heating/cooling the fluid will all change the pressure.
