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>
Answer:
phenotype,phenotype,genotype,genotype,
genotype
Explanation:
phenotype is physical appearance and genotype is just like
yy Tt
Answer:
See Explanation
Explanation:
moles of NH₃ = 11.9g/17.03 g/mol = 0.699 mole
moles of CN₂OH₄ = 1/2(0.699) mole =0.349 mole
Theoretical yield of CN₂OH₄ = (0.349 mole)(60 g/mole) = 20.963 grams
%Yield = Actual Yield/Theoretical Yield x 100%
= 18.5g/20.963g x 100% = 88.25%
Answer:
the difference is tyat eruptions of less gassy and more gassy is that the less gassy doesnt retain as much gas as the more gassy one and thus the eruption of the less gassy is less damage to the more gassy
Answer:
V₂ = 0.6 V.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n is constant, and have different values of P, V and T:
<em>(P₁V₁T₂) = (P₂V₂T₁).</em>
<em></em>
V₁ = V, P₁ = P, T₁ = T.
V₂ = ??? V, P₂ = 1.25 P, T₂ = 0.75 T.
<em>∴ V₂ = (P₁V₁T₂)/(P₂T₁) =</em> (P)(V)(0.75 T)/(1.25 P)(T)<em> = 0.6 V.</em>
