We are using the General gas equation P x V/K = P x V/K
1. P = 1atm V=500ml so PxV= 500 at 6.5km P = 0.5atm V = ? so P xV = 0.5 x V
(We don't have to worry about temperature!) 500 = 0.5 x V so V = 1000ml
2. NO CHANGE in pressure here so we have V/K V=2.75 K = 20 + 273=293 so V/K= 2.75/293
Next set V = 2.46 K = ? so V/K = 2.46/K then 2.75/293= 2.46/K so K=(293/2.75)x2.46
=262 K
Convert back to Celsius 262 - 273 = -11 C
It's raining so I have to rescue the laundry!
Laundry rescued!
3.Now we have to use all three variables. I am using 273K and 100kPa for STP.
P = 100 V = 700 K = 273 These are altered P - unknown, V = 200 K = 273+30=303
!00 x 700/273 = 256.4 this is equal to P x 200/303 = P x 0.66
so P = 256.4/0.66 = 388.48kPa
Answer:
1200 mL
Explanation:
Given data
- Initial pressure (P₁): 600.0 mmHg
- Initial volume (V₁): 400.0 mL
- Final pressure (P₂): 200.0 mmHg
For a gaseous sample, there is an inverse relationship between the pressure and the volume. If we consider the gas as an ideal gas, we can find the final volume using Boyle's law.
Answer:
- What is the limiting reactant?: HCl is the limitng reactant.
- How many moles of H₂ are formed?: 6.5 moles of H₂ are formed.
Explanation:
Part A: <em>what is the limiting reactant?</em>
1) <u>Balanced chemical equation</u>: given
- 2Al + 6HCl → 2AlCl₃ + 3H₂
2)<em> </em><u>Stoichiometric mole ratio:</u>
Use the coefficients of the balanced equation:
- 2 mol Al : 6 mol HCl : 2 mol AlCl₃ : 3H₂
3) <u>Compare the stoichiometric mole ratio of the reactants with their actual ratio</u>:
- Theoretical ratio: 2 mol Al / 6 mol HCl ≈ 0.33 mol Al / mol HCl
- Actual ratio: 6.0 mol Al / 13 mol Cl ≈ 0.46 mol Al / mol Cl
Since the actual ratio indicates that there is a greater number of moles of Al (0.46) per mol of Cl than what is required by the stoichiometric ratio(0.33), Al is in excess and HCl is the limiting reactant.
Answer: the limiting reactant is HCl.
Part B. <em>How many moles of H₂ are formed?</em>
3. <u>Determine how many moles of H₂ can be formed</u>
- Theoretical ratio using limiting reactant:
6 mol HCl / 3 mol H₂ = 13 mol HCl / x
⇒ x = 13 mol HCl × 3 mol H₂ / 6 mol HCl = 6.5 mol H₂.
The answer must be reported with two significant digits, such as the data are given.
Answer: 6.5 moles of H₂ are formed
Answer:
1.20 M
Explanation:
Convert grams of Na₂CO₃ to moles. (50.84 g)/(105.99 g/mol) = 0.4797 mol
Molarity is (moles of solute)/(liters of solvent) = (0.4797 mol)/(0.400 L) = 1.20 M