Answer:
3.35 atm
Since P₂ > 3.00 atm, the lighter would explode.
Explanation:
Step 1: Given data
- Initial pressure of butane gas (P₁): 2.50 atm
- Initial temperature of butane gas (T₁): 293 K
- Final pressure of butane gas (P₂): ?
- Final temperature of butane gas (T₂): 393 K
Step 2: Calculate the final pressure of butane gas
If we assume ideal behavior, we can calculate the final pressure of butane gas using Gay Lussac's law.
P₁/T₁ = P₂/T₂
P₂ = P₁ × T₂/T₁
P₂ = 2.50 atm × 393 K/293 K = 3.35 atm
Since P₂ > 3.00 atm, the lighter would explode.
Answer:
Explanation:
The temperature and amount of gas are constant, so we can use Boyle’s Law.
Data:
Calculations:
They have 6 electron in their outer orbital. That means they want 2 electrons to be stable. Since metals have the tendency to give electron oxygen receives 2 electrons. The answer is -2
Answer:
53.7 grams of HNO3 will be produced
Explanation:
Step 1: Data given
Mass of NO2 = 59.0 grams
Molar mass NO2 = 46.0 g/mol
Step 2: The balanced equation
3NO2 + H2O→ 2HNO3 + NO
Step 3: Calculate moles NO2
Moles NO2 = 59.0 grams / 46.0 g/mol
Moles NO2 = 1.28 moles
Step 4: Calculate moles HNO3
For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO
For 1.28 moles NO2 we'll have 2/3 * 1.28 =0.853 moles HNO3
Step 7: Calculate mass HNO3
Mass HNO3 = 0.853 moles * 63.01 g/mol
Mass HNO3 = 53.7 grams
53.7 grams of HNO3 will be produced
