C. 343K
70 degrees C + 273 = 343
Answer:
r = 3.61x
M/s
Explanation:
The rate of disappearance (r) is given by the multiplication of the concentrations of the reagents, each one raised of the coefficient of the reaction.
r = k.![[S2O2^{-8} ]^{x} x [I^{-} ]^{y}](https://tex.z-dn.net/?f=%5BS2O2%5E%7B-8%7D%20%5D%5E%7Bx%7D%20x%20%5BI%5E%7B-%7D%20%5D%5E%7By%7D)
K is the constant of the reaction, and doesn't depends on the concentrations. First, let's find the coefficients x and y. Let's use the first and the second experiments, and lets divide 1º by 2º :
x = 1
Now, to find the coefficient y let's do the same for the experiments 1 and 3:
y = 1
Now, we need to calculate the constant k in whatever experiment. Using the first :
k = 4.01x10^{-3} M^{-1}s^{-1}[/tex]
Using the data given,
r = 
r = 3.61x M/s
The geosphere is about 99.94% of Earth's mass, so C is the answer.
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.
If you don't practice enough it's obviously going to be hard but if you practice enough it's going to be a piece of cake so don't think if it's going to be hard or not just think it's going to be worth the try at the very end
