<span>12.4 g
First, calculate the molar masses by looking up the atomic weights of all involved elements.
Atomic weight manganese = 54.938044
Atomic weight oxygen = 15.999
Atomic weight aluminium = 26.981539
Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol
Now determine the number of moles of MnO2 we have
30.0 g / 86.936044 g/mol = 0.345081265 mol
Looking at the balanced equation
3MnO2+4Al→3Mn+2Al2O3
it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So
0.345081265 mol / 3 * 4 = 0.460108353 mol
So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum.
0.460108353 mol * 26.981539 g/mol = 12.41443146 g
Finally, round to 3 significant figures, giving 12.4 g</span>
Answer:
0.33 cal⋅g-1°C-1
Explanation:
The amount of heat required is determined from the formula:
q= mcΔT
To see more:
https://api-project-1022638073839.appspot.com/questions/what-is-the-specific-heat-of-a-substance-if-1560-cal-are-required-to-raise-the-t#235434
Number of moles is found by formula n=mass/molar mass, or m/M. the molar mass is found by adding together the atomic masses of Na and Cl (22.99 + 35.45) to give 58.44 g/mol. Since the mass of NaCl is 75.0g, we find the number of moles as follows:
n = 75.0 / 58.44 = 1.28 mol
Answer:
![Kc=[Cl2]2/[HCl]4[O2]](https://tex.z-dn.net/?f=Kc%3D%5BCl2%5D2%2F%5BHCl%5D4%5BO2%5D)
Explanation:
Hello,
In this case, the law of mass action, allows us to study the mathematical expression regarding an equilibrium chemical reaction allowing us to see a relationship between the equilibrium constant and the concentration of both the products and reactants at equilibrium for either gaseous or aqueous substances only. Such relationship is assembled as the quotient between the concentration of products at equilibrium over the concentration of reactants at equilibrium equaling the equilibrium constant. Thus, for the given chemical reaction, such expression will have the concentration of chlorine at the numerator and both the concentrations of hydrogen chloride and oxygen at the denominator since water is liquid so it is not included in the shown below equation:
![Kc=\frac{[Cl]^2}{[HCl]^4[O_2]}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BCl%5D%5E2%7D%7B%5BHCl%5D%5E4%5BO_2%5D%7D)
Therefore the answer is: Kc=[Cl2]2/[HCl]4[O2].
Best regards.
Itz defs B and E it might be D also bit iffy on that
YEET GOOD LUCK DADDY
