I think it’s hydrothermal vents, rifts and subduction trench
Here's the equation:
<span>Fe2 O3 + 2Al → 2Fe + Al2 O3
</span>
Here's the question.
What mass of Al will react with 150g of Fe2 O3?
<span>In every 2 moles Al you need 1 mole Fe2O3 </span>
<span>moles = mass / molar mass </span>
<span>moles Fe2O3 = 150 g / 159.69 g/mol </span>
<span>= 0.9393 moles </span>
<span>moles Al needed = 2 x moles Fe2O3 </span>
<span>= 2 x 0.9393 mol </span>
<span>= 1.879 moles Al needed </span>
<span>mass = molar mass x moles </span>
<span>mass Al = 26.98 g/mol x 1.879 mol </span>
<span>= 50.69 g </span>
<span>= 51 g (2 sig figs)
</span>
So the <span>mass of Al that will react with 150g of Fe2 O3 is 51 grams.</span>
37 grams of NaCl (when I mean equivalent I mean the ratio of the equation is 1:2 for moles or Cl2 and NaCl
Answer:
Without corals, the algae are not protected and cannot perform photosynthesis.
Answer:
Explanation:
MnO₂(s) + 4 HCl(aq) = MnCl₂(aq) + 2 H₂O(l) + Cl₂
87 g 22.4 x 10³ mL
volume of given chlorine gas at NTP or at 760 Torr and 273 K
= 175 x ( 273 + 25 ) x 715 / (273 x 760 )
= 179.71 mL
22.4 x 10³ mL of chlorine requires 87 g of MnO₂
179.4 mL of chlorine will require 87 x 179.4 / 22.4 x 10³ g
= 696.77 x 10⁻³ g
= 696.77 mg .
