Answer:
30.62 L
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 55 L
Initial pressure (P₁) = 3.2 atm
Initial temperature (T₁) = 520 K
Final temperature (T₂) = 760 K
Final pressure (P₂) = 8.4 atm
Final volume (V₂) =?
The final volume of the gas can be obtained as follow:
P₁V₁ / T₁ = P₂V₂ / T₂
3.2 × 55 / 520 = 8.4 × V₂ / 760
176 / 520 = 8.4 × V₂ / 760
Cross multiply
520 × 8.4 × V₂ = 176 × 760
4368 × V₂ = 133760
Divide both side by 4368
V₂ = 133760 / 4368
V₂ = 30.62 L
Therefore, the new volume of the gas is 30.62 L
Answer: There is one way to write it but i’ll also provide an unbalanced equation and a balanced one.
Explanation:
Unbalanced : Ba (aq) + Cl2 (aq)—-> BaCl (aq)
Balanced : 2Ba (aq) + Cl2 (aq)—> 2BaCl(aq)
Molar mass NO₂ = 46.0 g/mol
1 mole -------- 46.0 g
2.0 moles ----- ?
Mass (NO₂) = 2.0 x 46.0 / 1
=> 92.0 g
hope this helps!
