Answer:
V2= 1.03L
Explanation:
Start off with what you are given.
V^1: 1.00L
T^1: 23°C
V^2?
T^2: 33°C
If you know your gas laws, you have to utilise a certain gas law called Charles' Law:
V^1/T^1 = V^2/T^2
Remember to convert Celsius values to Kelvin whenever you are dealing with gas problems. This can be done by adding 273 to whatever value in Celsius you have.
(23+273 = 296) (33+273 = 306)
Multiply crisscross
1.00/296= V^2/306
296V^2 = 306
Dividing both sides by 296 to isolate V2, we get
306/296 = 1.0337837837837837837837837837838
V2= 1.03L
Here you go! Feel free to ask questions!
Answer:
Thorium-234 option a .......
Answer:
[Zn²⁺] = 4.78x10⁻¹⁰M
Explanation:
Based on the reaction:
ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)
The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:
1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]
We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:
<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>
6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺
<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>
0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻
Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =
0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]
Replacing in Ksp expression:
1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]
<h3>[Zn²⁺] = 4.78x10⁻¹⁰M</h3>
Answer:
8.3 × 10³ mL
Explanation:
Step 1: Calculate the mass of water that contains 100 mg of Pb
The concentration of Pb in the sample is 0.0012% by mass, that is, there are 0.0012 g of Pb every 100 g of water. The mass of water that contains 100 mg (0.100 g) of Pb is:
0.100 g Pb × 100 g Water/0.0012 g Pb = 8.3 × 10³ g Water
Step 2: Calculate the volume corresponding to 8.3 × 10³ g of water
Since the solution is diluted, we will assume the density of the sample is equal than the density of water (1 g/mL).
8.3 × 10³ g × 1 mL/1 g = 8.3 × 10³ mL
