Group 1A(1), the alkali metals, includes lithium, sodium, and potassium. Group 7A(17) the halogens, includes chlorine, bromine, and iodine. hope this helps:)
Answer:
8.44 atm
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 2.25 L
Initial temperature (T₁) = 350 K
Initial pressure (P₁) = 1.75 atm
Final volume (V₂) = 1 L
Final temperature (T₂) = 750 K
Final pressure (P₂) =?
The final pressure of the gas can be obtained as illustrated below:
P₁V₁/T₁ = P₂V₂/T₂
1.75 × 2.25 / 350 = P₂ × 1 / 750
3.9375 / 350 = P₂ / 750
Cross multiply
350 × P₂ = 3.9375 × 750
350 × P₂ = 2953.125
Divide both side by 350
P₂ = 2953.125 / 350
P₂ = 8.44 atm
Thus, the final pressure of the gas is 8.44 atm.
Answer:
the answer of the questionis D Photons
<em>ANSWER - 6 MOLES OF </em><em>IRON</em>
Fe2O3(s) + 3H2(g) → 2Fe(s) + 3H2O(1)
One moles of Fe2O3 forming 2 moles of Fe
3 moles of Fe2O3 will form 2×3 = 6 moles of iron
Answer:
M(Fe₂O₃) = 159.70 g/mol
M(CO) = 28.01 g/mol
M(Fe) = 55.85 g/mol
M(CO₂) = 44.01 g/mol
Explanation:
We can calculate the molar mass of a compound by summing the molar masses of the elements that form it.
Fe₂O₃
M(Fe₂O₃) = 2 × M(Fe) + 3 × M(O) = 2 × 55.85 g/mol + 3 × 16.00 g/mol = 159.70 g/mol
CO
M(CO) = 1 × M(C) + 1 × M(O) = 1 × 12.01 g/mol + 1 × 16.00 g/mol = 28.01 g/mol
Fe
M(Fe) = 1 × M(Fe) = 1 × 55.85 g/mol = 55.85 g/mol
CO₂
M(CO₂) = 1 × M(C) + 2 × M(O) = 1 × 12.01 g/mol + 2 × 16.00 g/mol = 44.01 g/mol
